Q 1. A parallel plate capacitor has plates of area A separated by distance d between them. It is filled with a dielectric which has a dielectric constant that varies as k(x) = K(1+αx ) where x is the distance measured from one of the plates. If (αd ) << 1, the total capacitance of the system is best given by the expression 
Given that, parallel plate capacitor filled with dielectric constant k(x) which varies as, k(x) = K(1+αx)
where x is distance measured from one of the plates.
Now, capacitance of a small element dx is given by
So, total capacitance is given by
For αd << 1 , we have
Now, using binomial expansion, we get
Q 2. A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time and charge is redistributed between them is (in nJ) _________. 
Given that, C = 60 pF = 60 × 10−12 F and V = 20 V.
We know that, energy when capacitor is fully charged is given by
Energy when capacitor is disconnected and connected to uncharged capacitor is
Where C' = C+C = 2C and V' = V/2
So, energy lost is
Q 3. Effective capacitance of parallel combination two capacitors C1 and C2 is 10 μF. When these capacitors are individually connected to a voltage source of 1 V, the energy stored in the capacitor C2 is 4 times that of C1. If these capacitors are connected in series, their effective capacitance will be 
(1) 4.2 μF
(2) 3.2 μF
(3) 1.6 μF
(4) 8.4 μF
CP = C1 + C2 = 10μF ...(1)
Energy stored in capacitor is given by
We know that
⇒C2 = 4C1 (since V = 1V) ...(2)
Solving Eqs. (1) and (2), we get
C1 = 2μF, C2 = 8μF
Effective capacitance of series combination is
Q 4. Consider two charged metallic spheres S1 and S2 of radii R1 and R2 respectively. The electric fields E1 (on S1) and E2 (on S2) on their surface are such that E1/E2 = R1/R2. Then the ratio V1 ( on S1) /V2 ( on S2) of the electrostatic potentials on each sphere is 
(1) R1/ R2
(2) (R1/ R2)2
(3) R2/ R1
(4) (R1/ R2)3
We know that, electric filed is given by
We know that, electrostatic potential is given by
Q 5. A capacitor is made of two square plates each of side α making a very small angle α between them, as shown in figure. The capacitance will be close to 
Let small element dx at distance x has capacitance dC.
We know that, capacitance of parallel plat capacitor is given by
Since α is very small so tan α ≈ α .
Q 6. Two identical capacitors A and B, charged to the same potential 5 V are connected in two different circuits as shown below at time t = 0. If the charge on capacitors A and B at time t = CR is QA and QB respectively, then (Here e is the base of natural logarithm) 
(1) QA = VC/e, QB = CV/2
(2) QA = VC, QB = CV
(3) QA = VC, QB =VC/e
(4) QA = CV/2, QB = VC/e
In first circuit diode is reversed biased so it is open circuit.
so, QA = CV
In second circuit diode is forward biased so it it short circuit.
So, current flow in the circuit at time t is given by
At t = RC, we get
Q 7. Two infinite planes each with uniform surface charge density +σ are kept in such a way that the angle between them is 30º. The electric field in the region shown between them is given by 
Electric filed due to infinite plane is given by
For first infinite plane, electric filed is given by
For second infinite plane, electric filed is given by
So, resultant electric field in region is
Q 8. Three charged particles A, B and C with charges −4q, 2q and −2q are present on the circumference of a circle of radius d. The charged particles A, C and Centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x-direction is 
QA = -4q, QB = 2q, QC = -2q
x component of electric field due to point charge A at centre O is given by
x component of electric field due to point charge B at centre O is given by
x component of electric field due to point charge C at centre O is given by
Total x component electric field due to point charges A, B, and C at centre O is given by
ET = EA + EB + EC
Q 9. In finding the electric field using Gauss law the formula is applicable. In the formula ε0 is permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?
(1) Only when the Gaussian surface is an equipotential surface.
(2) Only when the Gaussian surface is an equipotential surface and is constant on the surface.
(3) Only when = constant on the surface.
(4) For any choice of Gaussian surface. 
Gauss law is given by
Gauss law only holds when gaussian surface is an equipotential surface and electric field is constant throughout the surface.
Q 10. A particle of mass m and charge q is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed v on the distance x travelled by it is correctly given by (graphs are schematic and not drawn to scale).
Given that vi = 0
Coulomb force is given by F = qE
⇒ ma = qE
⇒ a = qE/m
Final velocity of charge particle in electric filed is
Only option (2) graph satisfy the condition.
Q 11. Consider a sphere of radius R which carries a uniform charge density ρ. If a sphere of radius R/2 is carved out of it, as shown, the ratio of magnitude of electric field respectively, at points A and B due to the remaining portion is 
So, cavity in solid sphere has density
Electric filed of a solid sphere is given by
Electric filed at point A is only due to cavity given by
Electric filed at point B is due to solid sphere and cavity given by
From Eqs. (1) and (2), we get
Q 12. An electric dipole of moment cm is at the origin (0, 0, 0). The electric field due to this dipole at (note that is parallel to 
Since are perpendicular to each other so, point lies on the equatorial plane.
Therefore, electric field at the point will be antiparallel to the dipole moment
Q 13. An electric field passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as ϕI and ϕII respectively. The difference between (ϕI - ϕII) is (in Nm2 /C) ________. 
Electric flux through closed area is given by
Electric flux through surface ABCD is
Electric flux through surface BCGF is
Here x = 3 so ϕII = 16 x 3 = 48 Nm2 /C
ϕI - ϕII = 0 - 48 = -48Nm2/C
Q 14. For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its center. Then value of h is 
Let the total charge on the ring is Q then, electric field at the axis of a charged ring is given by
For maximum electric field,
Q 15. Three charges +Q, q, +Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by +Q, placed at x = 0, is zero, then value of q is 
Charge q is placed at center so, it is equilibrium condition for +Q placed at x = 0 For equilibrium,
Q 16. Two point-charges and q2 (−25μC) are placed on the x-axis at x = 1 m and x = 4m, respectively. The electric field (in V/m) at a point y = 3 m on y-axis is
We know that electric field vector is
Let E1 and E2 are the electric field of q1 and q2.
For charge q1 = √10 μC
Therefore, total electric field vector is
Q 17. Charge is distributed within a sphere of radius R with a volume charge density where A and a are constants. If Q is the total charge of this charge distribution, the radius R is 
Let Q be charge is distribution within a sphere of radius R
Q 18. Two electric dipoles- A, B with respective dipole moments are placed on the x-axis with a separation R, as shown in the figure.
The distance from A at which both of them produce the same potential is 
On the x axis in left on A potential due to A and B. But potential due to A is higher than that of B, between sign of potential due to A and B is opposite. So, potential can be same only in right of B on x-axis.
Q 19. Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, −2) and (0, −2). The work required to put a fifth charge Q at the origin of the coordinate system will be 
Potential at origin is
Now, work done, W = qΔV
Q 20. Charges –q and +q located at A and B, respectively, constitute an electric dipole. Distance AB = 2a, O is the mid-point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >> 2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P′ such that OP' = (y/3), the force on Q will be close to 
Electric field on equatorial plane of dipole is
Dividing Eq. (2) by Eq. (1), we get
⇒ F' = 27F
Q 21. Three charges Q, +q and +q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is 
(4) -√2q/(√2+ 1)
Let a be the two equal sides of isosceles triangle.
Thus, net electrostatic energy is given as
Since, the net electrostatic energy is zero.
Q 22. Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle, as shown in the figure: 
Dipole moment is P1 = ql cos θ
= ql cos 30° (1)
P2 = ql cos θ
= ql cos 30° (2)
Now, net dipole moment is
Since, direction of dipole moment is negative y axis.
Q 22. A parallel plate capacitor is made of two square plates of side ‘a’, separated by a distance d (d ≪ a). The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. The value of capacitance of this arrangement is 
Consider a small element dx at a distance x.
Integrating both the sides, we get
Q 23. A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K1, K2, K3, K4 arranged as shown in the figure. The effective dielectric constant K will be 
Answer does not match with any given options
The arrangement of capacitor shown in the figure
This Answer does not match with any given options.
Q 24. A parallel plate capacitor is of area 6 cm2 and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants K1 = 10, K2 = 12 and K3 = 14. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be 
Let K be the dielectric constant of material
Q 25. A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be 
Let potential at distance r from then common center is
But surface change density for all are same.
Put these values in Eq. (1), we get
Q 26. A parallel-plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is 
(1) 692 pJ
(2) 508 pJ
(3) 560 pJ
(4) 600 pJ
Initial energy of capacitor
Since battery is disconnected so charge remains same.
Final energy of capacitor is
Therefore, work done is
W = Ui – Uf
= 600 – 92 = 508 pJ
Q 27. The given graph shows variation (with distance r from centre) of 
(1) electric field of a uniformly charged sphere.
(2) potential of a uniformly charged spherical shell.
(3) potential of a uniformly charged sphere.
(4) electric field of a uniformly charged spherical shell.
Growth and decay of current in LR circuit is
Therefore, the given charge shows the potential of a uniformly charged spherical shell.
Q 28. In the figure shown below, the charge on the left plate of the 10 μF capacitor is −30 μC. The charge on the right plate of the 6 μF capacitor is 
(1) + 18 μC
(2) + 12 μC
(3) −18 μC
(4) −12 μC
6 μF and 4 μF are parallel to each other. Total charge is 30 μF.
Therefore, charge on right plate is +18 μC.
Q 29. An electric field of 1000 V/m is applied to an electric dipole at angle of 45°. The value of electric dipole moment is 10−29 C m. What is the potential energy of the electric dipole? 
(1) −20 × 10−18 J
(2) −7 × 10−27 J
(3) −10 × 10−29 J
(4) −9 × 10−20 J
Potential energy is
Q 30. Seven capacitors, each of capacitance 2 μF, are to be connected in a configuration to obtain an effective capacitance of . Which of the combinations, shown in figures below, will achieve the desired value? 
Solution: Effective capacitance is
Three capacitors are in parallel.
Q 30. There is a uniform, spherically symmetric surface charge density at a distance R0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that best represents the speed V(R(t)) of the distribution as a function of its instantaneous radius R(t) is 
At any instant time t total energy of charge distribution is constant.
Also the slope of V-R curve will go on decreasing.
Q 31. In the circuit shown, find C if the effective capacitance of the whole circuit is to be 0.5 μF. All values in the circuit are in μF. 
From given circuit
Q 32. A parallel-plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is 
(1) 7.85 × 10−10 C
(2) 6.85 × 10−10 C
(3) 8.85 × 10−10 C
(4) 9.85 × 10−10 C
By Gauss law electric field is