Uploaded
Lectures
Electrostatics Introduction
Charge (EntryTest Concepts)
Coulomb Law (magnitude)
(Force In A Vacuum)
(Force In A Medium)
[IMP MCQ’s Key Points]
Coulomb Law (Vector Form)
[IMP MCQ’s Key Points]
Electric field Intro and
Electric field lines
Electric field strength and
Zero Field Point (Important)
Important MCQ’s Key Points
Electric flux (Detailed)
Favorite EntryTest Topic
Gauss law
Applications of Gauss Law
Field Flux Passing Through A Closed Charged Body
Electric Field Due To Infinite Sheet Of Charges
Electric Field Due To Oppositely Charged Plates
Potential difference and Electrical Energy
Potential at single point,
Net Potential,
Conversion of Electrical Energy
Potential Gradient and Electron Volt
Capacitor and Capacitance
Energy stored in a Capacitor and QV Curve
Combination of Capacitors
_________________________________
Practice Tests
Electrostatics MDCAT 2025 (Test-1)
Quiz-summary
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Question 1 of 30
1. Question
1 pointsThe slope of the graph shown in figure is
Correct
Explanation A
Slope = Q/V = C
Incorrect
Explanation A
Slope = Q/V = C
-
Question 2 of 30
2. Question
1 pointsCoulomb’s force is
Correct
Explanation D
Coulomb’s force is a mutual force. It means that if a charge ‘q1’ exerts a force on charge ‘q2’. Then ‘q2’ also exerts an equal and opposite force on ‘q1’.
If charge ‘q1’ exerts an electrostatic force ‘F12’ on charge ‘q2’ and ‘q2’ exerts electrical force ‘F21’ on charge ‘q1’ and then
F12 = –F21
Incorrect
Explanation D
Coulomb’s force is a mutual force. It means that if a charge ‘q1’ exerts a force on charge ‘q2’. Then ‘q2’ also exerts an equal and opposite force on ‘q1’.
If charge ‘q1’ exerts an electrostatic force ‘F12’ on charge ‘q2’ and ‘q2’ exerts electrical force ‘F21’ on charge ‘q1’ and then
F12 = –F21
-
Question 3 of 30
3. Question
1 pointsThe main function of a capacitor is to
Correct
Explanation: B
FUNCTION
i. Capacitors store electrical charge.
ii. They are used in timing circuits because it takes time for a capacitor to fill with charge.
iii. They are used to smooth varying DC supplies by acting as a reservoir of charge.
iv. They are used in filter circuits because capacitors easily pass AC (changing) signals but they block DC (constant) signals.Incorrect
Explanation: B
FUNCTION
i. Capacitors store electrical charge.
ii. They are used in timing circuits because it takes time for a capacitor to fill with charge.
iii. They are used to smooth varying DC supplies by acting as a reservoir of charge.
iv. They are used in filter circuits because capacitors easily pass AC (changing) signals but they block DC (constant) signals. -
Question 4 of 30
4. Question
1 pointsWhich of the following figures represent the electric field lines due to a single negative charge?
Correct
Explanation: B
The field lines of a single negative charge are radially inwardIncorrect
Explanation: B
The field lines of a single negative charge are radially inward -
Question 5 of 30
5. Question
1 pointsThe number of electrons in one coulomb charge is equal to
A. 6.25×1018
B. 6.25×1021
C. 1.6×1019
D. 1.6×10-27Correct
Explanation A
Total charge required 1 Coulomb
∴ q = 1C
since q = ne
⟹ n = q/e
⟹ n = 1/1.6×10-19
⟹ 6.25×1018Incorrect
Explanation A
Total charge required 1 Coulomb
∴ q = 1C
since q = ne
⟹ n = q/e
⟹ n = 1/1.6×10-19
⟹ 6.25×1018 -
Question 6 of 30
6. Question
1 pointsThe value of the capacitance depends upon the
Correct
Explanation C
The capacitance depends on the geometry of the capacitor and capacitance of a capacitor is affected by the area of the plates, the distance between the plates, and the ability of the dielectric to support electrostatic forces.Incorrect
Explanation C
The capacitance depends on the geometry of the capacitor and capacitance of a capacitor is affected by the area of the plates, the distance between the plates, and the ability of the dielectric to support electrostatic forces. -
Question 7 of 30
7. Question
1 pointsAn electric charge at rest produces
Correct
Explanation B
Charge at rest produces only electric fieldIncorrect
Explanation B
Charge at rest produces only electric field -
Question 8 of 30
8. Question
1 pointsThe capacity of a parallel plate capacitor increases with the
Correct
Explanation C
C = Aεo/d, increases of its area.
C ∝ AIncorrect
Explanation C
C = Aεo/d, increases of its area.
C ∝ A -
Question 9 of 30
9. Question
1 pointsAn alpha particle is accelerated through a potential difference of 200V. The increase in its kinetic energy is
Correct
Explanation C
K.E = q∆V = 2e × 200V = 400eVIncorrect
Explanation C
K.E = q∆V = 2e × 200V = 400eV -
Question 10 of 30
10. Question
1 pointsValue of constant k in Coulombs law has value of
A. 9×103 Nm2/C2
B. 9×105 Nm2/C2
C. 9×107 Nm2/C2
D. 9×109 Nm2/C2Correct
Explanation D
k is a constant of proportionally known as the Coulomb constant, having the value 9×109 Nm2/C2 .Incorrect
Explanation D
k is a constant of proportionally known as the Coulomb constant, having the value 9×109 Nm2/C2 . -
Question 11 of 30
11. Question
1 pointsWhat is the unit of ‘k’ in Coulomb’s law?
A. Nm2 C-2
B. NmC-1
C. N/m2
D. CCorrect
Explanation: A
Coulomb’s law is given by equation
F = kq1q2/r2
from this we can obtain the unit of k as Nm2C-2Incorrect
Explanation: A
Coulomb’s law is given by equation
F = kq1q2/r2
from this we can obtain the unit of k as Nm2C-2 -
Question 12 of 30
12. Question
1 pointsThe magnitude of electric field strength E such that an electron placed in it would experience an electric force equal to its weight is given by
A. mge
B. mg/e
C. e/mg
D. e2g/2mCorrect
Explanation B
Incorrect
Explanation B
-
Question 13 of 30
13. Question
1 pointsIf a unit charge is taken from one point to another over an equipotential surface
Correct
Explanation D
On equipotential surface work done always zeroIncorrect
Explanation D
On equipotential surface work done always zero -
Question 14 of 30
14. Question
1 pointsThe electric potential at a point 1.6 cm away from a proton will be
A. 9×10-10 V
B. 9×10-8 V
C. 9×10-6 V
D. 9×10-4 VCorrect
Explanation B
Incorrect
Explanation B
-
Question 15 of 30
15. Question
1 pointsThe capacitance of a capacitor does not depend on
Correct
Explanation D
C depends on Area.
C = Aεo/dIncorrect
Explanation D
C depends on Area.
C = Aεo/d -
Question 16 of 30
16. Question
1 pointsHow fast or how slow a capacitor is charging and discharging, depends upon.
Correct
Explanation: A
RC value is called time constant and determine the time how slow or fast a capacitor will charge or dischargeIncorrect
Explanation: A
RC value is called time constant and determine the time how slow or fast a capacitor will charge or discharge -
Question 17 of 30
17. Question
1 pointsWhen an electron is accelerated through a potential difference of one volt, it will acquire energy equal to
Correct
Explanation C
K.E = q∆V = 1e × 1V = 1eVIncorrect
Explanation C
K.E = q∆V = 1e × 1V = 1eV -
Question 18 of 30
18. Question
1 pointsAmount of work done on a charge from one point to other in an electric field is called
Correct
Explanation: B
Electric potential is equals to the work done in moving a charge from one point to another in an electric field.Incorrect
Explanation: B
Electric potential is equals to the work done in moving a charge from one point to another in an electric field. -
Question 19 of 30
19. Question
1 pointsThe flux through any closed surface is___________ times the total charge enclosed in it.
A. ∈o
B. 1/∈°2
C. 1/∈o
D. ∈°2Correct
Explanation: C
Gauss’s law states that the flux through any closed surface is 1/∈o times the total charge enclosed in it.Incorrect
Explanation: C
Gauss’s law states that the flux through any closed surface is 1/∈o times the total charge enclosed in it. -
Question 20 of 30
20. Question
1 pointsOne of the plates X of a capacitor is connected to a source of +10V. The other plate Y is earthed. What is the potential of the Y?
Correct
Explanation C
Due to earthed V=0Incorrect
Explanation C
Due to earthed V=0 -
Question 21 of 30
21. Question
1 pointsAn electric field can deflect
Correct
Explanation: D
Electric field exerts force only on charge particles.Incorrect
Explanation: D
Electric field exerts force only on charge particles. -
Question 22 of 30
22. Question
1 pointsElectric potential is an
Correct
Explanation: A
Vectors and Scalars
Some quantities in physics, such as time, distance, mass, speed, temperature, etc., just need one number to specify them. These are called scalar quantities.
On the other hand, many quantities are fully specified only if, in addition to a number, a direction is given. Examples are velocity, acceleration, force, etc. These are called vector quantities.
Incorrect
Explanation: A
Vectors and Scalars
Some quantities in physics, such as time, distance, mass, speed, temperature, etc., just need one number to specify them. These are called scalar quantities.
On the other hand, many quantities are fully specified only if, in addition to a number, a direction is given. Examples are velocity, acceleration, force, etc. These are called vector quantities.
-
Question 23 of 30
23. Question
1 pointsThe electric intensity at infinite distance from the point charge is
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 24 of 30
24. Question
1 pointsCharges of +2μc and -2μc are placed at points (P) and (Q) respectively. Tell the location at which electric potential is zero
Correct
Explanation: D
Let is separation between +2μC and -2μC. Net potential due to both charges at mid-way between “P” and “Q” is
Incorrect
Explanation: D
Let is separation between +2μC and -2μC. Net potential due to both charges at mid-way between “P” and “Q” is
-
Question 25 of 30
25. Question
1 pointsThe S.I unit of relative permittivity is
A. Nm-1 A-1
B. Nm2 C2
C. N-1 m1C2
D. No unitCorrect
Explanation: D
Incorrect
Explanation: D
-
Question 26 of 30
26. Question
1 pointsThe volt is equal to:
A. C/q
B. N/m2
C. Joule/coulomb
D. Nm2/CCorrect
Explanation: C
Voltage(V)
◈ Measured in volts (V) using a voltmeter
◈ 1 volt = 1 joule/coulombIncorrect
Explanation: C
Voltage(V)
◈ Measured in volts (V) using a voltmeter
◈ 1 volt = 1 joule/coulomb -
Question 27 of 30
27. Question
1 pointsS.I unit of permittivity of free space is
A. farad
B. C2/Nm2
C. Weber
D. C2/NmCorrect
Explanation: B
In units, the value of permittivity of free space is
ε0 = 8.85 × 10-12 C2 N-1 m-2Incorrect
Explanation: B
In units, the value of permittivity of free space is
ε0 = 8.85 × 10-12 C2 N-1 m-2 -
Question 28 of 30
28. Question
1 pointsThe constant “k” in Coulomb’s Law depends upon
Correct
Explanation: D
“k” is the constant of proportionality Its value depends upon the nature of the medium between the two charges and the system of unitsIncorrect
Explanation: D
“k” is the constant of proportionality Its value depends upon the nature of the medium between the two charges and the system of units -
Question 29 of 30
29. Question
1 pointsThe ratio of Cvac and Cmed is equal to
A. εr
B. 1/εr
C. εo
D. 1/εoCorrect
Explanation: B
Incorrect
Explanation: B
-
Question 30 of 30
30. Question
1 pointsThe ratio of the force between two small spheres with constant charges A in air, B in a medium of dielectric constant K is:
A. K2:1
B. K:1
C. 1:K
D. 1:K2Correct
Explanation: B
Incorrect
Explanation: B
________
Electrostatics MDCAT 2025 (Test-2)
Quiz-summary
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- No Negative Marking
So, Attempt all the Questions. - The Explanations of each MCQ’s will be provided to you with Your Right & Wrong Answers after each MCQ.
- Bismillah & Darood Sharif Phr k, Start your Test/Quiz, and Attempt it with Full Concentration.
You have already completed the quiz before. Hence you can not start it again.
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Question 1 of 30
1. Question
1 pointsIf area of plates is halved and their separation is reduced to one third. The capacitance will be
Correct
Explanation: B
C = A∈0/d
above relation describes that if area of plates is halved and their separation is reduced to one third, the capacitance will become 1.5 times.Incorrect
Explanation: B
C = A∈0/d
above relation describes that if area of plates is halved and their separation is reduced to one third, the capacitance will become 1.5 times. -
Question 2 of 30
2. Question
1 pointsIf charges are doubled and distance between them is halved then Coulomb force will become?
Correct
Explanation: C
Coulomb’s law is given by
F = kq1q2/r2
Applying the conditions
F = k × 2q1 × 2q2/ (1/2r2)
F = 16 (kq1q2/r2)Incorrect
Explanation: C
Coulomb’s law is given by
F = kq1q2/r2
Applying the conditions
F = k × 2q1 × 2q2/ (1/2r2)
F = 16 (kq1q2/r2) -
Question 3 of 30
3. Question
1 pointsThe expression of energy stored in a capacitor is given by
A. E = CV2
B. E = ½ C2V
C. E = ½ CV2
D. E = ½ CV2Correct
Explanation: C
Incorrect
Explanation: C
-
Question 4 of 30
4. Question
1 pointsThe force between two point charges separated by air is 4N. When separated by a medium of relative permittivity 2, the force between them becomes
Correct
Explanation: C
Fmed = Fvac/εr = 4/2 = 2NIncorrect
Explanation: C
Fmed = Fvac/εr = 4/2 = 2N -
Question 5 of 30
5. Question
1 pointsA 5MΩ resistor is connected with a 2μf capacitor. The time constant of the circuit is
Correct
Explanation: D
t = RC = (5×106) (2×10-6) = 10sIncorrect
Explanation: D
t = RC = (5×106) (2×10-6) = 10s -
Question 6 of 30
6. Question
1 pointsUnit of 1eV is
Correct
Explanation: B
Unit of 1eV is joule.Incorrect
Explanation: B
Unit of 1eV is joule. -
Question 7 of 30
7. Question
1 pointsWhich one of the following can be taken as measure of electric field intensity.
Correct
Explanation: C
Φe = EA ⇒ E = Φe/AIncorrect
Explanation: C
Φe = EA ⇒ E = Φe/A -
Question 8 of 30
8. Question
1 pointsThe diagram shows the electric field lines due to two charged parallel metal plates. We conclude that
Correct
Explanation: B
Point X, Y and Z lie in the uniform electric field. So, at these points electric force on any charge particle will be same.Incorrect
Explanation: B
Point X, Y and Z lie in the uniform electric field. So, at these points electric force on any charge particle will be same. -
Question 9 of 30
9. Question
1 pointsOne of materials listed below is to be placed between two identical metal sheets, with no air gap, to form a parallel-plate capacitor. Which produces the greatest capacitance?
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 10 of 30
10. Question
1 pointsCharge “Q” is spread uniformly along the circumference of a circle of radius “R”. A point particle with charge “q” is placed at the center of this circle. The total force exerted on the particle can be calculated
Correct
Explanation: D
The vector sum of all the forces acting on a charge placed at the center of the circle is zero.Incorrect
Explanation: D
The vector sum of all the forces acting on a charge placed at the center of the circle is zero. -
Question 11 of 30
11. Question
1 pointsThe electric field at a distance of 10cm from an isolated point particle with a charge of 2×10-9 C is
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 12 of 30
12. Question
1 pointsIf force between point charges of 10μC and 40μC is 9000N, then distance between them is:
Correct
Explanation: A
Incorrect
Explanation: A
-
Question 13 of 30
13. Question
1 pointsThe diagram shows four pairs of large parallel conducting plates. The value of the electric potential is given for each plate. Rank the pairs according to the magnitude of the electric field between the plates, least to greatest.
Correct
Explanation: D
E = ΔV/d ⇒ E ∝ ΔV
ΔV1 = 90 V
ΔV2 = 50 V
ΔV3 = 100 V
ΔV4 = 60 V
E2 < E4 < E1 < E3Incorrect
Explanation: D
E = ΔV/d ⇒ E ∝ ΔV
ΔV1 = 90 V
ΔV2 = 50 V
ΔV3 = 100 V
ΔV4 = 60 V
E2 < E4 < E1 < E3 -
Question 14 of 30
14. Question
1 pointsThe work required to carry a particle with a charge of 6.0C from a 5.0 V equi-potential surface to a 6.0 V equi-potential surface and back again to the 5.0 V surface is
A. 0
B. 3.0×10-5 J
C. 1.2×10-5 J
D. 6.0×10-5 JCorrect
Explanation: A
W = q∆V
on equi-potential surface ΔV = 0
W = 0
(Electric filed is a conservative field. So, work done along closed path is zero.)Incorrect
Explanation: A
W = q∆V
on equi-potential surface ΔV = 0
W = 0
(Electric filed is a conservative field. So, work done along closed path is zero.) -
Question 15 of 30
15. Question
1 pointsA capacitor is charged by using a battery, which is then disconnected. A dielectric slab is inserted between the plates, which results in
Correct
Explanation: C
when a dielectric inserted between the plates of charged parallel plates capacitor
i. Charge remains same
ii. Electric field decreases
iii. Potential difference decreases
iv. Capacitance increases
. Energy stored in capacitor decreasesIncorrect
Explanation: C
when a dielectric inserted between the plates of charged parallel plates capacitor
i. Charge remains same
ii. Electric field decreases
iii. Potential difference decreases
iv. Capacitance increases
. Energy stored in capacitor decreases -
Question 16 of 30
16. Question
1 pointsTwo point particles, one with charge +8×10-9 C and the other with charge -2×10-9 C, are separated by 4m. The electric field in N/C midway between them is:
A. 9×109
B. 135,000
C. 13,500
D. 22.5Correct
Explanation: D
Incorrect
Explanation: D
-
Question 17 of 30
17. Question
1 pointsA 20μF capacitor is charged to 200V. Its stored energy is
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 18 of 30
18. Question
1 pointsAn electron, a proton and an α-particle are displaced through a p.d. of 1V when placed plates of a capacitor. The gain in K.E. will be highest for
Correct
Explanation: C
K.E = q∆V ⇒ K.E ∝ q
(for same potential difference) As
qα > qP = qeIncorrect
Explanation: C
K.E = q∆V ⇒ K.E ∝ q
(for same potential difference) As
qα > qP = qe -
Question 19 of 30
19. Question
1 points1KJ work is required to traverse a charged particle through a potential difference of 20V. The magnitude of charge on the particle is
Correct
Explanation: B
W = qV
⇒ q = W/V = 1000/20 = 50CIncorrect
Explanation: B
W = qV
⇒ q = W/V = 1000/20 = 50C -
Question 20 of 30
20. Question
1 pointsWhat are the SI units of electric field intensity?
Correct
Explanation A
E = F/q. So its units are N/C.Incorrect
Explanation A
E = F/q. So its units are N/C. -
Question 21 of 30
21. Question
1 pointsIf the potential difference on a surface is equal to zero between any two points, then surface is said to be?
Correct
Explanation B
any surface over which the potential is constant is called an equipotential surface. In other words, the potential difference between any two points on an equipotential surface is zero. Work done in moving a charge over an equipotential surface is zero.Incorrect
Explanation B
any surface over which the potential is constant is called an equipotential surface. In other words, the potential difference between any two points on an equipotential surface is zero. Work done in moving a charge over an equipotential surface is zero. -
Question 22 of 30
22. Question
1 pointsGravitational force does not depends on?
Correct
Explanation C
Electric force depends on medium while gravitational force does not.
Gravitational force between two point objects is directly proportional to the product of masses, inversely proportional to square of distance between them and acts along the line joining them.
G is known as the universal gravitational constant.
◈ Does not depend on the nature of bodies
◈ Does not depend on temperature
◈ Does not depend on the medium between the particlesIncorrect
Explanation C
Electric force depends on medium while gravitational force does not.
Gravitational force between two point objects is directly proportional to the product of masses, inversely proportional to square of distance between them and acts along the line joining them.
G is known as the universal gravitational constant.
◈ Does not depend on the nature of bodies
◈ Does not depend on temperature
◈ Does not depend on the medium between the particles -
Question 23 of 30
23. Question
1 pointsIf σ is the uniform surface charged density then electric field intensity due to an infinite sheet of charge is?
A. σ * 2ε0
B. σ * ε0
C. σ / 2ε0
D. σ / ε0Correct
Explanation C
Through the application of Gauss’s law we get the intensity = σ / 2ε0.
Incorrect
Explanation C
Through the application of Gauss’s law we get the intensity = σ / 2ε0.
-
Question 24 of 30
24. Question
1 pointsIf work done on moving a unit positive charge from point A to B in an electric field is 5J the potential difference between A and B is?
Correct
Explanation B
V = J/C , if J = 5, C = 1 then V = 5 VIncorrect
Explanation B
V = J/C , if J = 5, C = 1 then V = 5 V -
Question 25 of 30
25. Question
1 pointsCapacity of a capacitor depends upon?
Correct
Explanation D
The capacitance of a capacitor is affected by the area of the plates, the distance between the plates, and the ability of the dielectric to support electrostatic forces.
Larger plates provide greater capacity to store electric charge.Incorrect
Explanation D
The capacitance of a capacitor is affected by the area of the plates, the distance between the plates, and the ability of the dielectric to support electrostatic forces.
Larger plates provide greater capacity to store electric charge. -
Question 26 of 30
26. Question
1 pointsA charge of 0.01C accelerated through a P.d of 1000 V acquires K-E?
Correct
Explanation B
W= q V = 0.01 ⨯ 1000 = 10 J.Incorrect
Explanation B
W= q V = 0.01 ⨯ 1000 = 10 J. -
Question 27 of 30
27. Question
1 pointsThe force per unit charge is known as?
Correct
Explanation C
E = F/q Electric field intensity is force per unit charge.Incorrect
Explanation C
E = F/q Electric field intensity is force per unit charge. -
Question 28 of 30
28. Question
1 pointsAn n number of identical capacitors are first connected in series and then in parallel such that the ratio of the net capacitance in series to net capacitance in parallel comes out to be 1⁄4. What is the total number of capacitors?
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 29 of 30
29. Question
1 pointsA capacitor may be considered as a device for?
Correct
Explanation A
Capacitors store electric energy.Incorrect
Explanation A
Capacitors store electric energy. -
Question 30 of 30
30. Question
1 pointsTwo charged spheres are separated by 2 mm. Which of the following would yield the greatest attractive force?
Correct
Explanation D
The only pair of charged bodies that would attract one another is the pair of unlike charged spheres. The rest of the pairs are like charges, which repel one another.Incorrect
Explanation D
The only pair of charged bodies that would attract one another is the pair of unlike charged spheres. The rest of the pairs are like charges, which repel one another.
_________
Electrostatics MDCAT 2025 (Test-3)
Quiz-summary
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Question 1 of 30
1. Question
1 pointsPresence of di-electric?
Correct
Explanation A
Presence of di-electric always reduces the electrostatic force between charged particles as compared to free space by a factor known as relative permittivity. Represented by εr.Incorrect
Explanation A
Presence of di-electric always reduces the electrostatic force between charged particles as compared to free space by a factor known as relative permittivity. Represented by εr. -
Question 2 of 30
2. Question
1 pointsIf a mica sheet is place between the plates of a capacitor, its capacitance will?
Correct
Explanation A
Capacitance increases when a di-electric is place between the capacitor plates.Incorrect
Explanation A
Capacitance increases when a di-electric is place between the capacitor plates. -
Question 3 of 30
3. Question
1 pointsWhich of the following is a correct relationship between work done, charge and voltage?
A. J = VC
B. J = V/C
C. JV = C
D. J = VC2Correct
Explanation A
Potential of 1 V exists between two points if work done in moving a unit positive charge from one point to other, keeping equilibrium, is 1 J. or 1 V = 1J/1C or J = VC.Incorrect
Explanation A
Potential of 1 V exists between two points if work done in moving a unit positive charge from one point to other, keeping equilibrium, is 1 J. or 1 V = 1J/1C or J = VC. -
Question 4 of 30
4. Question
1 pointsWhich of the following is true about electric field lines?
(i) Electric field lines start from positive charge and end on negative charge.
(ii) Tangent to the field line gives the direction of field at that point.
(iii) Electric field lines never cross each other.Correct
Explanation D
All these statements are correct. E has only one direction at a given point so the lines never cross each other.Incorrect
Explanation D
All these statements are correct. E has only one direction at a given point so the lines never cross each other. -
Question 5 of 30
5. Question
1 pointsA Capacitor of 2μF is connected with a battery of 12 Volts, the charge stored in the capacitor is?
Correct
Explanation D
Q = CV = 2⨯10-6 ⨯12 = 24 μCIncorrect
Explanation D
Q = CV = 2⨯10-6 ⨯12 = 24 μC -
Question 6 of 30
6. Question
1 pointsThe value of minimum possible charge is?
A. 1 C
B. 1.6 x 10-19 C
C. 3.2 x 10-19 C
D. There is no such value as minimum possible charge.Correct
Explanation B
The minimum possible value of charge is the charge of electron i.e. 1.6×10-19 C.Incorrect
Explanation B
The minimum possible value of charge is the charge of electron i.e. 1.6×10-19 C. -
Question 7 of 30
7. Question
1 pointsThe electric intensity is expressed in unit of N/C or?
Correct
Explanation A
The SI units of the field are N⋅C−1 or, equivalently, volts per meter (V⋅m−1)Incorrect
Explanation A
The SI units of the field are N⋅C−1 or, equivalently, volts per meter (V⋅m−1) -
Question 8 of 30
8. Question
1 pointsOne coulomb is the unit of charge which when placed at 1m from an equal and similar charge repels it with a force?
A. 3 × 109 N
B. 6 × 109 N
C. 9 × 109 N
D. 18 × 109 NCorrect
Explanation C
F = 9 × 109 * q1 * q2 / r2;
q1, q2 and r are all equal to one. So F = 9 × 109 N.Incorrect
Explanation C
F = 9 × 109 * q1 * q2 / r2;
q1, q2 and r are all equal to one. So F = 9 × 109 N. -
Question 9 of 30
9. Question
1 pointsFor the figure below, an insulator plate is slowly placed between the plates of a capacitor and then removed. The current?
Correct
Explanation B
As insulator plate is passed between the plates of the capacitor, its capacity increases first and then decreases as the plate slips out. As a result, positive charge on plate A increases first and then decreases, hence, current in outer circuit flows from B to A and then from A to B.Incorrect
Explanation B
As insulator plate is passed between the plates of the capacitor, its capacity increases first and then decreases as the plate slips out. As a result, positive charge on plate A increases first and then decreases, hence, current in outer circuit flows from B to A and then from A to B. -
Question 10 of 30
10. Question
1 pointsRelative permittivity of three materials is given below. Which one would have highest electrostatic force for charged particles in that medium, keeping all other factors constant?
Mica = 4
Germanium = 16
Glass = 10Correct
Explanation C
Presence of dielectric reduces the electrostatic force by the factor of relative permittivity. So the medium with lowest value of relative permittivity shall have highest force.Incorrect
Explanation C
Presence of dielectric reduces the electrostatic force by the factor of relative permittivity. So the medium with lowest value of relative permittivity shall have highest force. -
Question 11 of 30
11. Question
1 pointsWhat will be the effect on the capacitance of the capacitor if area of each plate is double while separation between the plates is halved?
Correct
Explanation D
by putting area double and separation half, the capacitance becomes 4 as shown below.
C = ε 2A / (1/2)d
= ε 4A/dIncorrect
Explanation D
by putting area double and separation half, the capacitance becomes 4 as shown below.
C = ε 2A / (1/2)d
= ε 4A/d -
Question 12 of 30
12. Question
1 pointsIf two capacitors of value C1 and C2 are attaching in series their total capacitance is?
A. C1 + C2
B. C1C2 / (C1 + C2)
C. C1 * C2
D. (C1 + C2) / C1C2Correct
Explanation B
Total capacitance = C1C2 / (C1 + C2)
for calculating net capacitance, capacitance in series behave just like resistances in parallel.Incorrect
Explanation B
Total capacitance = C1C2 / (C1 + C2)
for calculating net capacitance, capacitance in series behave just like resistances in parallel. -
Question 13 of 30
13. Question
1 pointsTime constant for capacitor is defined as?
Correct
Explanation C
The time constant is defined as the time it takes for the capacitor to deplete 36.8% (for a discharging circuit) of its charge or the time it takes to reach 63.2% (for a charging circuit) of its maximum charge capacity
Incorrect
Explanation C
The time constant is defined as the time it takes for the capacitor to deplete 36.8% (for a discharging circuit) of its charge or the time it takes to reach 63.2% (for a charging circuit) of its maximum charge capacity
-
Question 14 of 30
14. Question
1 pointsIf two capacitors of value 10 mili F and 100 mili F are connected in parallel. Total capacitance is?
Correct
Explanation D
Capacitance is added when connected in parallel. \Incorrect
Explanation D
Capacitance is added when connected in parallel. \ -
Question 15 of 30
15. Question
1 pointsIn the region where the electric field is zero, the electric potential is always?
Correct
Explanation D
if the electric field is zero throughout a region of space, the electric potential must be constant throughout that regionIncorrect
Explanation D
if the electric field is zero throughout a region of space, the electric potential must be constant throughout that region -
Question 16 of 30
16. Question
1 pointsIn RC circuits, smaller values of time constant RC leads to?
Correct
Explanation B
In RC circuits, smaller values of time constant RC leads to more rapid discharging.Incorrect
Explanation B
In RC circuits, smaller values of time constant RC leads to more rapid discharging. -
Question 17 of 30
17. Question
1 pointsWhich of the following statement is true?
Correct
Explanation B
Only i and ii are correct. Electrical force is medium dependent while gravitational force is not. Another difference is that electrical forces can be attractive or repulsive while gravitational force is always attractive.Incorrect
Explanation B
Only i and ii are correct. Electrical force is medium dependent while gravitational force is not. Another difference is that electrical forces can be attractive or repulsive while gravitational force is always attractive. -
Question 18 of 30
18. Question
1 pointsUnit of electric flux is?
A. N/C
B. N2m/C2
C. N2/Cm
D. Nm2/CCorrect
Explanation D
Incorrect
Explanation D
-
Question 19 of 30
19. Question
1 pointsTo measure an electric field, the test charge q0 has to be very small because?
Correct
Explanation A
We do not want the test charge to distort the electric field that it is measuring so it has be very small.Incorrect
Explanation A
We do not want the test charge to distort the electric field that it is measuring so it has be very small. -
Question 20 of 30
20. Question
1 pointsThe medium used between the plates of capacitor is called?
Correct
Explanation A
Incorrect
Explanation A
-
Question 21 of 30
21. Question
1 pointsThe electric field created by positive charge is
Correct
Explanation A
Radially Outward
because an electric field has both magnitude and direction, the direction of the force on a positive charge is chosen arbitrarily as the direction of the electric field. Because positive charges repel each other, the electric field around an isolated positive charge is oriented radially outward
Incorrect
Explanation A
Radially Outward
because an electric field has both magnitude and direction, the direction of the force on a positive charge is chosen arbitrarily as the direction of the electric field. Because positive charges repel each other, the electric field around an isolated positive charge is oriented radially outward
-
Question 22 of 30
22. Question
1 pointsThe amount of work done in carrying a charge q along the path MARYM between the opposite charged plates is:
A. Zero
B. q
C. qE
D. q/E0Correct
Explanation A
E → uniform inside so no work done.
Incorrect
Explanation A
E → uniform inside so no work done.
-
Question 23 of 30
23. Question
1 pointsAs 1 ohm × 1 farad = 1 second. For MΩ × pF =
Correct
Explanation D
MΩ × pF =?
106Ω × 10-12F = µs
µ = 10-6Incorrect
Explanation D
MΩ × pF =?
106Ω × 10-12F = µs
µ = 10-6 -
Question 24 of 30
24. Question
1 pointsThe value of relative permittivity for all the dielectrics other than air or vacuum is always
Correct
Explanation: C
The value of relative permittivity for all the dielectrics other than air or vacuum is always greater than unity.Incorrect
Explanation: C
The value of relative permittivity for all the dielectrics other than air or vacuum is always greater than unity. -
Question 25 of 30
25. Question
1 pointsEnergy density in case of a capacitor is always proportional to
A. E2
B. V2
C. εo
D. CCorrect
Explanation: A
Incorrect
Explanation: A
-
Question 26 of 30
26. Question
1 pointsHow many electrons are there in 1 coulomb of charge?
A. Zero.
B. 3.1 × 1019
C. 1.9 × 1019
D. 6.2 × 1018Correct
Explanation D
If the charge on the electron is e = 1.6 × 10-19 C. Then in 1 coulomb of charge there will be
number of electrons = 1/1.6 × 10-19
which is 6.2 × 1018 electrons.Incorrect
Explanation D
If the charge on the electron is e = 1.6 × 10-19 C. Then in 1 coulomb of charge there will be
number of electrons = 1/1.6 × 10-19
which is 6.2 × 1018 electrons. -
Question 27 of 30
27. Question
1 pointssec/ohm is equal to
Correct
Explanation: A
RC = t ⇒ C = t/RIncorrect
Explanation: A
RC = t ⇒ C = t/R -
Question 28 of 30
28. Question
1 pointsA capacitor “C” has a charge “Q”. The actual charges on its plates are:
Correct
Explanation: B
The amount of charge on both plates of the capacitor will be same. If one plate has +Q charge then other will be -Q.Incorrect
Explanation: B
The amount of charge on both plates of the capacitor will be same. If one plate has +Q charge then other will be -Q. -
Question 29 of 30
29. Question
1 pointsIf the magnitudes of charges and distance between them is doubled then force will be
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 30 of 30
30. Question
1 pointsThe automobiles wind shield wipers works on
Correct
Explanation: C
The automobiles wind shield wipers works on charging and discharging of capacitor.Incorrect
Explanation: C
The automobiles wind shield wipers works on charging and discharging of capacitor.
_______
Electrostatics MDCAT 2025 (Test-4)
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Question 1 of 30
1. Question
1 pointsWhen two capacitors of same capacitance C are connected in parallel and then in series, the capacitance in these two cases is in ratio of?
Correct
Explanation B
Capacitance in parallel is C + C = 2C.
Capacitance in series is = C/2
Ratio = 2C / (C/2) = 4 : 1Incorrect
Explanation B
Capacitance in parallel is C + C = 2C.
Capacitance in series is = C/2
Ratio = 2C / (C/2) = 4 : 1 -
Question 2 of 30
2. Question
1 pointsThe magnitude of the force of a 400 N/C electric field on a 0.02C point charge is
A. 8.0 N
B. 8×10-3 N
C. 8×10-5 N
D. 0.08 NCorrect
Explanation: A
F = qE = (2×10-2 )(400) = 8NIncorrect
Explanation: A
F = qE = (2×10-2 )(400) = 8N -
Question 3 of 30
3. Question
1 pointsA charged capacitor stores 10C at 40V. Its stored energy is
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 4 of 30
4. Question
1 pointsIf 60 capacitors each of capacitance C are connected in series and then in parallel what is the ratio of Cs/Cp = ?
Correct
Explanation: C
Cs = C/60
Cp = 60C
Cs/Cp = 1/3600Incorrect
Explanation: C
Cs = C/60
Cp = 60C
Cs/Cp = 1/3600 -
Question 5 of 30
5. Question
1 pointsTwo-point charges are +2C and +6C repel each other with a force of 12N. If a charge of -4C is given to each of these charges, the force now is
Correct
Explanation: C
New charges
q1 = -2C & q2 = 2C
F ∝ q1q2
F = 4N
as q1q2 are opposite charges their force is attractiveIncorrect
Explanation: C
New charges
q1 = -2C & q2 = 2C
F ∝ q1q2
F = 4N
as q1q2 are opposite charges their force is attractive -
Question 6 of 30
6. Question
1 pointsThe units of are
A. N2 C2
B. N2m2/C2
C. Nm/C
D. Nm2/C2Correct
Explanation: D
Incorrect
Explanation: D
-
Question 7 of 30
7. Question
1 pointsWhich of the following is not the energy stored between the plates of a condenser?
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 8 of 30
8. Question
1 pointsThree capacitor of 4 each are to be connected in such a way that the net capacitance is 6 . Then
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 9 of 30
9. Question
1 pointsIf a metal plate is placed between two-point charges then electrostatic force between them
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 10 of 30
10. Question
1 pointsWhich of the following is equivalent capacitance between P and Q
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 11 of 30
11. Question
1 pointsTwo charges q1 and q2 are kept at a certain distance in air. If a dielectric (glass slab) is introduced between them, the force between the charges will_________
Correct
Explanation: B
Two charges are placed a certain distance apart in air. If a glass slab is introduced between them, the force between them will decreases.Incorrect
Explanation: B
Two charges are placed a certain distance apart in air. If a glass slab is introduced between them, the force between them will decreases. -
Question 12 of 30
12. Question
1 pointsWhen the relative permittivity of the medium is increased, the force between two charges placed at a given distance apart
Correct
Explanation: B
In Coulomb’s law, permittivity constant of material comes in the denominator.
Since a metallic sheet is put in between and metals are good conductors, the permittivity is greatly increased and hence the force between two charges will decrease.Incorrect
Explanation: B
In Coulomb’s law, permittivity constant of material comes in the denominator.
Since a metallic sheet is put in between and metals are good conductors, the permittivity is greatly increased and hence the force between two charges will decrease. -
Question 13 of 30
13. Question
1 pointsTwo charged spheres of radii 10 cm and 15 cm are connected by a thin wire. No current will flow if they have
Correct
Explanation: D
Both the spheres are connected and when both the potentials are equal then there is no current flow between themIncorrect
Explanation: D
Both the spheres are connected and when both the potentials are equal then there is no current flow between them -
Question 14 of 30
14. Question
1 pointsWhich of the following remains unchanged if a dielectric is placed between a charged capacitor?
A. Q
B. E
C. Fe
D. VCorrect
Explanation: A
Variation of different variables (Q, C, V, E and U) of parallel plate capacitor when dielectric (K) is introduced when battery is removed is
C′=KC E′=E/K
Q′=Q U′=U/K
V′=V/K
Only the amount of charge stored in the capacitor will remain conserved when the dielectric is inserted between the plates.Incorrect
Explanation: A
Variation of different variables (Q, C, V, E and U) of parallel plate capacitor when dielectric (K) is introduced when battery is removed is
C′=KC E′=E/K
Q′=Q U′=U/K
V′=V/K
Only the amount of charge stored in the capacitor will remain conserved when the dielectric is inserted between the plates. -
Question 15 of 30
15. Question
1 pointsIf a charge on a capacitor is doubled, then its capacitance will be
Correct
Explanation: C
Explanation: C
When the charge given to a capacitor is doubled its capacitance does not changes. The capacitance depends only on the geometrical configuration of your conductors (capacitors).Incorrect
Explanation: C
Explanation: C
When the charge given to a capacitor is doubled its capacitance does not changes. The capacitance depends only on the geometrical configuration of your conductors (capacitors). -
Question 16 of 30
16. Question
1 pointsThe potential gradient between the two charged plates having separation of 0.5 cm and potential difference of 12 volts is:
A. 240 NC–1
B. 24 NC–1
C. 2.4 NC–1
D. 2400 NC–1Correct
Explanation: D
We know that electric field strength is the negative gradient of potential.
⇒ E = – (dV/dr)
Given that the potential difference is 12 volts and distance between the two charged plates is 0.005 metre.
Substituting we get
⇒ E = – (12/0.005) = -2400 Vm⁻¹
⇒ As Potential Gradient is negative of the Electric field strength.
⇒ The Potential gradient between the two plates = 2400 Vm⁻¹
Incorrect
Explanation: D
We know that electric field strength is the negative gradient of potential.
⇒ E = – (dV/dr)
Given that the potential difference is 12 volts and distance between the two charged plates is 0.005 metre.
Substituting we get
⇒ E = – (12/0.005) = -2400 Vm⁻¹
⇒ As Potential Gradient is negative of the Electric field strength.
⇒ The Potential gradient between the two plates = 2400 Vm⁻¹
-
Question 17 of 30
17. Question
1 pointsIf the distance between two-point charges become half then force between them becomes
Correct
Explanation C
Incorrect
Explanation C
-
Question 18 of 30
18. Question
1 pointsTwo charges are placed at a certain distance. If the magnitude of each charge is doubled the force will become
A. 1/4th of original value
B. 4 times of original value
C. 1/8th of original value
D. 8 times of its original valueCorrect
Explanation B
Incorrect
Explanation B
-
Question 19 of 30
19. Question
1 pointsConcept of field lines of force was introduced by
Correct
Explanation: B
The concept was introduced into physics in the 1830s by the English scientist Michael Faraday, who considered magnetic and electric effects in the region around a magnet or electric charge as a property of the region rather than an effect taking place at a distance from a cause.Incorrect
Explanation: B
The concept was introduced into physics in the 1830s by the English scientist Michael Faraday, who considered magnetic and electric effects in the region around a magnet or electric charge as a property of the region rather than an effect taking place at a distance from a cause. -
Question 20 of 30
20. Question
1 pointsThe potential difference between two points in an electric field is
A. ∆w/∆r
B. ∆V/q₀
C. ∆q/w
D. ∆w/q₀Correct
Explanation: D
Potential difference: The difference of potentials at two points in an electric field is called potential difference. Or, The amount of work done in transferring a unit positive charge from one point to another point in an electric field is called potential difference between the two points.
Let the potentials at two points A and B in an electric field is VA and VB respectively. Then
VB – VA = ∆V = WAB/qo = W/qoIncorrect
Explanation: D
Potential difference: The difference of potentials at two points in an electric field is called potential difference. Or, The amount of work done in transferring a unit positive charge from one point to another point in an electric field is called potential difference between the two points.
Let the potentials at two points A and B in an electric field is VA and VB respectively. Then
VB – VA = ∆V = WAB/qo = W/qo -
Question 21 of 30
21. Question
1 pointsElectric lines of force are
Correct
Explanation: B
Electric Fields
◈ Electric fields are represented by electric field lines (imaginary) and show the direction of the force at that point.
◈ Electric field direction is from a positive charge to a negative charge.
◈ Electric field lines (also known as electric flux) do not touch each other.
◈ Electric field line represents the path a small positive charge would move in the electric field.
Incorrect
Explanation: B
Electric Fields
◈ Electric fields are represented by electric field lines (imaginary) and show the direction of the force at that point.
◈ Electric field direction is from a positive charge to a negative charge.
◈ Electric field lines (also known as electric flux) do not touch each other.
◈ Electric field line represents the path a small positive charge would move in the electric field.
-
Question 22 of 30
22. Question
1 pointsSelenium is
Correct
Explanation: C
Selenium is a good conductor of electricity in presence of light. This phenomenon is called photoconductivity.
Used as a photoconductor in photocopiers and also in an X-ray imaging technique known as xeroradiography for decades.Incorrect
Explanation: C
Selenium is a good conductor of electricity in presence of light. This phenomenon is called photoconductivity.
Used as a photoconductor in photocopiers and also in an X-ray imaging technique known as xeroradiography for decades. -
Question 23 of 30
23. Question
1 pointsPhotocopier and inkjet printer are the application of
Correct
Explanation: D
Electrostatics is the study of electric fields in static equilibrium. In addition to research using equipment such as a Van de Graaff generator, many practical applications of electrostatics exist, including photocopiers, laser printers, ink-jet printers and electrostatic air filters.Incorrect
Explanation: D
Electrostatics is the study of electric fields in static equilibrium. In addition to research using equipment such as a Van de Graaff generator, many practical applications of electrostatics exist, including photocopiers, laser printers, ink-jet printers and electrostatic air filters. -
Question 24 of 30
24. Question
1 pointsϵ₀ is measured in
A. Hm ⁻¹
B. Ωm⁻¹
C. Nm⁻¹
D. Fm⁻¹Correct
Explanation: D
Vacuum permittivity, commonly denoted ε0 (pronounced as “epsilon nought” or “epsilon zero”) is the value of the absolute dielectric permittivity of classical vacuum. … ε0 = 8.8541878128(13)×10−12 F⋅m−1 (farads per meter).
Incorrect
Explanation: D
Vacuum permittivity, commonly denoted ε0 (pronounced as “epsilon nought” or “epsilon zero”) is the value of the absolute dielectric permittivity of classical vacuum. … ε0 = 8.8541878128(13)×10−12 F⋅m−1 (farads per meter).
-
Question 25 of 30
25. Question
1 pointsThe Coulomb’s law is
The units of “εo” are:
A. Nm-2C-2
B. Nm‑2C2
C. N-1m-2C2
D. None of theseCorrect
Explanation: C
The units of “εo” are reciprocal of the units of “k”.Incorrect
Explanation: C
The units of “εo” are reciprocal of the units of “k”. -
Question 26 of 30
26. Question
1 pointsThe study of charges at rest under the ratio of electric forces is called
Correct
Explanation: B
The study of charges at rest under the action of the electric force is named as electrostatics.Incorrect
Explanation: B
The study of charges at rest under the action of the electric force is named as electrostatics. -
Question 27 of 30
27. Question
1 pointsCoulomb’s force
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 28 of 30
28. Question
1 pointsThe ratio of electric force to electric field strength gives the unit of:
Correct
Explanation: B
Electric field strength is defined as:
Incorrect
Explanation: B
Electric field strength is defined as:
-
Question 29 of 30
29. Question
1 pointsFor a capacitor the charge per unit volt is called
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 30 of 30
30. Question
1 pointsAn electron of mass m is accelerated from rest through a potential difference of V volt. The speed of electron will be
A. (eV/2m)1 /2
B. eV/m
C. (2eV/m)1 /2
D. (eV/m)1 /2Correct
Explanation C
Incorrect
Explanation C
________________
Electrostatics Assessment+Revision Test
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- Answered
- Review
-
Question 1 of 71
1. Question
1 pointsWhich of the following among the given is the smallest charge
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 2 of 71
2. Question
1 pointsWhich of the following according to coulombs law for the two charges is incorrect?
Correct
Explanation: A
Incorrect
Explanation: A
-
Question 3 of 71
3. Question
1 pointsThe ratio of electric field produced due to infinite sheet of charges to that produced by parallel plates is equal to
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 4 of 71
4. Question
1 pointswhich of the following statements is not in accordance with gauss law
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 5 of 71
5. Question
1 pointsWhen a test charge is moved in the direction of electric field line its potential
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 6 of 71
6. Question
1 pointsWhen a negative charge is brought near a positive charge, what happens.
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 7 of 71
7. Question
1 pointsWhich of the following is correct when a dielectric is introduced.
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 8 of 71
8. Question
1 pointsWhat is the ratio of energy stored in a capacitor to that of capacitance of capacitor
A. V2/2
B. E2/2
C. V2d/2
D. E2d/2Correct
Explanation: A
Incorrect
Explanation: A
-
Question 9 of 71
9. Question
1 pointsWhich of the following graph describes the correct relation with force and distance between the charges.
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 10 of 71
10. Question
1 pointsTwo charges of 9C and 4C are separated by a distance of 0.5m from each other, At what point the electric field between the two charges is zero.
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 11 of 71
11. Question
1 pointsWhen the surface is placed perpendicularly to a positively charged plate flux passing through it 10units, how much flux will pass through it when it is tilted by 30degrees
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 12 of 71
12. Question
1 points2 charges of magnitude +Q and –Q are separated from each other by a distance r, The electric field at a point midway between them is
A. 2kQ/r2
B. 4kQ/r2
C. kQ/r2
D. 8kQ/r2Correct
Explanation: D
Incorrect
Explanation: D
-
Question 13 of 71
13. Question
1 pointsWhen a negative charge moves in the direction of electric field.
Correct
Explanation: A
Incorrect
Explanation: A
-
Question 14 of 71
14. Question
1 pointsA capacitor has a time constant of 10seconds, the after 20 seconds how much total percent of charge is deposited on a capacitor.
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 15 of 71
15. Question
1 pointsConsider 2 point q1 and q2 charges with q2 being greater than q1 are placed at a distance “r” from each other, if charge q1 exerts a force of F on q2, and charge q2 exerts a force of N q1 then what is true for the forces.
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 16 of 71
16. Question
1 pointsThe coulomb’s force between the charges in air is 2.0N the coulomb’s force between these charges in insulating medium having Er = 3.8 is:
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 17 of 71
17. Question
1 pointsA charge ‘Q’ is divided into two parts ‘q’ and ‘Q-q’ and separated by a distance ‘R’. The force of repulsion between them will be maximum when:
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 18 of 71
18. Question
1 pointsThere are two charges 1μc and 5μc, the ratio of the force acting on them will be:
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 19 of 71
19. Question
1 pointsIn the M.K.S system of units, Ꜫ0 equals:
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 20 of 71
20. Question
1 pointsThere are two charges +3μC and +8μC the ratio of the force acting on them will be:
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 21 of 71
21. Question
1 pointsA point charge at a distance “x” from another point charge experiences a force of repulsion, which one of the following graphs shows. How the force is related to “x”:
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 22 of 71
22. Question
1 pointsThe force between two charged bodies is “F”. If one of the charge is doubled and the distance between them is halved, the force acting on each charged body is:
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 23 of 71
23. Question
1 pointsThe unit of the electric field is:
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 24 of 71
24. Question
1 pointsIf ∆v/∆r is potential gradient, then the intensity of electric field at a point is:
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 25 of 71
25. Question
1 pointsA charge of 10e-5C exerts a force of 10 dynes on a charge of 10e-7C. How much is the electric field.
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 26 of 71
26. Question
1 pointsConsider the following figure of the two charges, if a force of 1dyne is exerted on the test charge, How much is the electric field strength.
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 27 of 71
27. Question
1 pointsThe electric field between the plates of a parallel plate capacitor can be written as.
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 28 of 71
28. Question
1 pointsOil droplets of mass m and charge q are dropped between two horizontal parallel plates. Air resistance is negligible. The droplets are falling at constant velocity when electric field strength between the plates is E. Which of the following is true?
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 29 of 71
29. Question
1 pointsAn electric current of 1A is passing through a cross section of the coil in 1 second. How many electrons are involved in providing a current of 1A? The charge on 1 electron is 1.602 x 10-19C.
A. 3.21 x 1018
B. 2.2 x 1016
C. 1.602 x 1019
D. 6.42 x 1018Correct
Explanation: D
Incorrect
Explanation: D
-
Question 30 of 71
30. Question
1 pointsThe electric field at a certain distance from an isolated alpha particle is 3.0×107 NC-1. What is the force on an electron when at that distance from the alpha particle?
A. 4.8×10-12N
B. 2.6×1012N
C. 3.0×107N
D. 6.0×107NCorrect
Explanation: A
Incorrect
Explanation: A
-
Question 31 of 71
31. Question
1 pointsThe rate of change of electric potential with respect to displacement is equal to:
Correct
Explanation: A
Incorrect
Explanation: A
-
Question 32 of 71
32. Question
1 pointsThe negative gradient of electric potential is also called:
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 33 of 71
33. Question
1 pointsIn the direction indicated by an electric field line:
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 34 of 71
34. Question
1 pointsIf a soap bubble is charged:
Correct
Explanation: B
Because same charge repel and size increases.Incorrect
Explanation: B
Because same charge repel and size increases. -
Question 35 of 71
35. Question
1 pointsA close surface contains equal and opposite charges. The net electric flux through the close surface is:
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 36 of 71
36. Question
1 pointsTwo point particles, one with charge +8×10-9C and the other with charge -2×10-9C, are separated by 4m. The electric field in N/C midway between them is:
A. 9×109
B. 13,500
C. 36×10-9
D. 22.5Correct
Explanation: D
E1 + E2 = KQ1/r2 + KQ2/r2 = (9×109)(8×10-9)/22 + (9×109)(2×10-9/22 = 18 + 4.5 = 22.5Incorrect
Explanation: D
E1 + E2 = KQ1/r2 + KQ2/r2 = (9×109)(8×10-9)/22 + (9×109)(2×10-9/22 = 18 + 4.5 = 22.5 -
Question 37 of 71
37. Question
1 pointsWhen will 1C of charge pass a point in an electrical circuit?
Correct
Explanation: C
Q = It = 5×10-3 x 200 = 5/1000 x 200 = 1CIncorrect
Explanation: C
Q = It = 5×10-3 x 200 = 5/1000 x 200 = 1C -
Question 38 of 71
38. Question
1 pointsBefore a thunder stand on end. A hair with mass 0.50mg and charge 1.0pc is supported by a force other than the weight of the hair and the electric force. What is the electric field strength?
A. 4.9 x 103 NC-1
B. 4.9 x 105 NC-1
C. 4.9 x 106 NC-1
D. 4.9 x 109 NC-1Correct
Explanation: C
Incorrect
Explanation: C
-
Question 39 of 71
39. Question
1 pointsIn a uniform electric field, which statement is correct?
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 40 of 71
40. Question
1 pointsThe number of electrons in one coulomb of charge are:
A. 6.25×1021
B. 1.6×1019
C. 6.25×1018
D. 9.1×1031Correct
Explanation: C
Incorrect
Explanation: C
-
Question 41 of 71
41. Question
1 pointsWhat is the magnitude of a point charge N which produces an electric field of 2NC-1 at a distance of 60cm ?
A. 8×10-11C
B. 2×10-12C
C. 6 x 10-10C
D. 6×10-10CCorrect
Explanation: A
Hints: E = kq/r2⇒ q = Exrr2/E(As E = 2NC-1, d = 0.6 m, q =?) = 2 × (0.6)2/9 ×10-9 = 8 × 10-11 CIncorrect
Explanation: A
Hints: E = kq/r2⇒ q = Exrr2/E(As E = 2NC-1, d = 0.6 m, q =?) = 2 × (0.6)2/9 ×10-9 = 8 × 10-11 C -
Question 42 of 71
42. Question
1 pointsTwo electrically neutral materials are rubbed together. One acquires a net positive charge.
Correct
Explanation: A
Incorrect
Explanation: A
-
Question 43 of 71
43. Question
1 pointsThe electric field strength between a pair of plates is “E”. If the separation of the plates is doubled and potential difference between the plates is increased by factor of four, the new field strength is Med – 20i9
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 44 of 71
44. Question
1 pointsIf two charges experience a force of 10N when medium is air, if medium is changed whose relative permittivity is 2 then force will be:
Correct
Explanation: B
Fair = 10N
Er = 2, F =?
F = Fair/ Er = 10/2 = 5NIncorrect
Explanation: B
Fair = 10N
Er = 2, F =?
F = Fair/ Er = 10/2 = 5N -
Question 45 of 71
45. Question
1 pointsAn isolated charged point particle produced an electric field with magnitude E at point 2 m away at a point 1 m from the particle the magnitude of the field is:
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 46 of 71
46. Question
1 pointsThe particle carrying a charge of (2e) falls through a potential difference of 3V. Energy required by the particle is:
A. 9.6 x 1019J
B. 1.6 x 10-19J
C. 3.2 x 10-19J
D. 6.9 x 10-19JCorrect
Explanation: A
Incorrect
Explanation: A
-
Question 47 of 71
47. Question
1 pointsThe potential gradient between the two charged plates having, separation of 0.50cm and potential difference of 12volts is:
A. 240NC-1
B. 24NC-1
C. 2.4NC-1
D. 2400NC-1Correct
Explanation: D
Incorrect
Explanation: D
-
Question 48 of 71
48. Question
1 pointsThe potential difference between two points is one volt. The work done in moving one coulomb of charge from on point to their point is:
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 49 of 71
49. Question
1 pointsWhich physical quantity would result from a calculation in which a potential difference is multiplied by an electric charge?
Correct
Explanation: D
U (Energy) = q x vIncorrect
Explanation: D
U (Energy) = q x v -
Question 50 of 71
50. Question
1 pointsAn electron when accelerated through a potential difference of one volt will gain an energy equal to:
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 51 of 71
51. Question
1 pointsIf an electron is accelerated from rest through a potential difference of 100 volts. Its final kinetic energy is:
A. 1.6 x 10-18J
B. 1.6 x 1017J
C. 100J
D. 100 electron voltCorrect
Explanation: D
Incorrect
Explanation: D
-
Question 52 of 71
52. Question
1 pointsAn electron has charge e and mass m. A proton has charge e and mass 1840m. A “Proton volt” is equal to:
Correct
Explanation: A
Incorrect
Explanation: A
-
Question 53 of 71
53. Question
1 pointsWhich of the following is correct?
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 54 of 71
54. Question
1 pointsA proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1kV, its kinetic energy will be:
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 55 of 71
55. Question
1 pointsAn electron volt is a unit of:
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 56 of 71
56. Question
1 pointsAn ∝-particle is accelerated through a potential difference of 106 volts. Its K.E is
Correct
Explanation: B
Hints:
K.E = qv
qalpha = 2e
K.E = 2eV
= 2MeVIncorrect
Explanation: B
Hints:
K.E = qv
qalpha = 2e
K.E = 2eV
= 2MeV -
Question 57 of 71
57. Question
1 pointsWhich of the following represents the highest charge as per magnitude.
Correct
Explanation: B
Incorrect
Explanation: B
-
Question 58 of 71
58. Question
1 pointsThree capacitors each of having different capacitance are connected in parallel with a battery of 6V. Then
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 59 of 71
59. Question
1 pointsA particle having the charge of 20 electrons on its falls through a potential difference of 100 volts.Calculate the energy acquired by it in electron volt (eV).
Correct
Explanation: D
Incorrect
Explanation: D
-
Question 60 of 71
60. Question
1 pointsThe quantity 1/2 ∈O∈rE2 has the significant of:
Correct
Explanation: C
Incorrect
Explanation: C
-
Question 61 of 71
61. Question
1 pointsThe correct expression for the energy of the charged capacitor is:
A. 1/2C2 V
B. 1/2Q2/C
C. 1/2/C
D. 1/2C2 V2Correct
Explanation: B
Incorrect
Explanation: B
-
Question 62 of 71
62. Question
1 pointsThe charge on electron is equal to:
A. 1.7588 x 1019 coulomb
B. 1.6022 x 10-19 coulomb
C. 1.2057 x 1019 coulomb
D. 0.6022 x 1019 coulombCorrect
Explanation: B
Incorrect
Explanation: B
-
Question 63 of 71
63. Question
1 pointsThe ratio of the capacitance of the capacitor having dielectric to the capacitance of the capacitor having free space is the dielectric:
Correct
Explanation: A
Er = Cmed/CvacIncorrect
Explanation: A
Er = Cmed/Cvac -
Question 64 of 71
64. Question
1 pointsThe capacitor which charges and discharges quickly will have.
Correct
Explanation: A
Incorrect
Explanation: A
-
Question 65 of 71
65. Question
1 pointsOhm x farad is equivalent to:
Correct
Explanation: A
Incorrect
Explanation: A
-
Question 66 of 71
66. Question
1 pointsWhich of the following is the same unit as the farad?
A. Ωs
B. Ωs-1
C. Ω-1s
D. Ω-1s-1Correct
Explanation: C
C = Q/V => Farad = Coulomb/Voltage = It/Ir = Amp sec/ Amp Ohm = Ohm-1 sec = Ω-1sIncorrect
Explanation: C
C = Q/V => Farad = Coulomb/Voltage = It/Ir = Amp sec/ Amp Ohm = Ohm-1 sec = Ω-1s -
Question 67 of 71
67. Question
1 pointsThe potential difference between a pair of similar. Parallel conducting plates is known. What additional information is needed in order to find the electric field strength between the plates?
Correct
Explanation: A
E = V/d, V is given and d is required i.e. separation between platesIncorrect
Explanation: A
E = V/d, V is given and d is required i.e. separation between plates -
Question 68 of 71
68. Question
1 pointsA charged capacitor stores 10C at 40V. Its stored energy is:
Correct
Explanation: D
U = 1/2 10×40=200JIncorrect
Explanation: D
U = 1/2 10×40=200J -
Question 69 of 71
69. Question
1 pointsTo determine the resistance of a voltmeter by discharging a capacitor through it, the instantaneous voltage is then given by the relation:
A. Voe -t/RC
B. Voet/RC
C. V0 y2
D. V0/√2Correct
Explanation: A
Hints: For discharging =q=qoe-t/RC (Dividing by C) So;q/C =q/C(e-t/RC)⇒V=V_oe-t/RCIncorrect
Explanation: A
Hints: For discharging =q=qoe-t/RC (Dividing by C) So;q/C =q/C(e-t/RC)⇒V=V_oe-t/RC -
Question 70 of 71
70. Question
1 pointsIf there are “n” capacitors each of capacity “C” connected in parallel to” V” volt source then energy stored is equal to
A. CV
B. 1/2nCV2
C. CV2
D. CV2/2nCorrect
Explanation: B
Incorrect
Explanation: B
-
Question 71 of 71
71. Question
1 pointsCapacitance of parallel plate capacitor is independent of:
Correct
Explanation: C
Incorrect
Explanation: C
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