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Previous Year CBSE Questions Page 31 of 484 
 
 SI unit of intensity is N/C. 
 Electrostatic force = charge x electric field. 
20. E at origin, E = 9x109 [4x10-6/ (1)2 + 4x10-6/22] + ---- 
 = 36x103[1+1/22+1/42 +--] 
 (Geometric series) 
 Therefore, Sum = a/1-r 
a= first term = 1 
r = common ratio = ¼ 
E= 36x10 3x1/ (1-1/4) = 48x103 
21. 4th 
 Explanation: When metallic solid sphere is placed in uniform electric field the electrons of 
the sphere move against the direction of electric field. Consequently, the left face acquires 
negative charge while the right face attains +ve charge. The filed lines will terminate at the left 
face of sphere and restart from right face. The electric field inside the sphere is zero. On the 
other surface of the sphere, the filed lines are normal at energy point. i.e., directed towards the 
centre. Therefore, the correct field is represented. 
22. Refer from NCERT textbook 
23. The total charge enclosed by a surface is zero, it doesn’t imply that the electric field everywhere 
on the surface is zero. As ∮ 𝐸.⃗⃗ ⃗
𝑠 
 𝑑𝑆⃗⃗⃗⃗ = q/ϵo, therefore, the field may be normal to the surface. 
Also, the conversely it does imply that net charge inside is zero if electric 
 field everywhere on the surface is zero. 
24. Here, λ= 
𝑑𝑞
 𝑑𝑥
= Kx; ϕ=? 
 dq=Kx dx 
 Total charge on the wire 
 q= ∫ 𝐾𝑥 𝑑𝑥
𝐿
0
= [
𝐾𝑥2
2
]=KL2/2 
 Total electric flux through the Gaussian hollow surface is 
 Φ= q/ϵo= KL2/2ϵo 
 25. (a) d=2.4m r=1.2m 
 Surface charge density, σ =180.0 μ C/m2 =180 x 10-6 C/m2 
 Total charge on surface of sphere, 
Previous Year CBSE Questions Page 32 of 484 
 
 Q =σ×4πr2=180×10-6×4×3.14× (1.2)2 
 =3.25×10-3 C 
 
 (b) Φ Total = Q/ϵo 
 Φ Total =
3.25∗10−3
8.85∗10−12=3.67 X 108 Nm2C-1 
 26. Here, q= 17.7 x 10-4 C, A= 400 cm2, E=?, r =10 cm = 10-1m 
 In case of a large plane sheet, distance of the point (=r) doesn’t matter. 
 E= σ/2ϵo =q/2ϵoA=
17.7∗10−4
2∗(8.85∗10−12)∗400
 =2.5 x 105 N/C 
 27. Here, σ = 5 x 10-16 Cm-2, ϕ=? 
 r= 1 cm =10-2m, θ=60o 
 ϕ= E (ΔS) cos θ= (σ/2ϵo) п r2 cos 60o 
 = 
5∗10−16∗3.14∗(10−4)∗1/2
2∗8.85∗10−12 = 4.44 x 10-9Nm2C-1. 
 
Answer to Three-mark questions 
 28. Refer from NCERT textbook 
 29. For electron 
 Y1= 1.5 cm = 1.5 x10-2m 
 E1= 2x104N/C 
 q0 = (-) 1.6x10-19 C 
 m1 = 9x10-3 kg 
 Y1= u1 t1+ ½ a1𝑡1
2
 
 = 0 + ½ a1𝑡1
2
 
 t1= √
2𝑦1
𝑎1
 = √
2×1⋅5×10−2
3.55×1015 = 2.9 X 10-9sec 
 therefore, 
a1 = F1/m1 = q0 E1/m1 
a1 = 1.6 x 10 -19x 2 x 104/ 9 x 10-37 
a1 =3.55 x 1015 m/s2 
For proton 
Previous Year CBSE Questions Page 33 of 484 
 
When electric field is reversed 
q0 = +1.6 x 10-19C. 
m2= 1.67 x 10-27kg. 
acceleration a2 = F2/m2 = q0E/m2 
a = 1.6 x 10 -19x 2 x 104/ 1.67 x 10-27 = 1.92 x 1012 m/s2 
Similarly, t2 = √
2𝑦2
𝑎2
 = √
2×1⋅5×10−2
1.92×1012 
 t1/t2 = 2.9 x 10-9/1.25 x 10-7 = 2.3 x 10-2 
Observation: 
Acceleration of e- = 1015 m/s2 
Acceleration of p+ = 1012 m/s2 
Acceleration of g = 9.8 m/s2 = 10 m/s2 (negligible), Effect of gravity can be ignored. 
30. a) When we consider the charged particle to be placed at the centre of the cube whose 
 side is 2a, then the charge is equally distributed among 8 cubes. Therefore, the total flux 
 through the faces of the cube=q/8 ϵo. 
 b) When the charge is placed at B, the charge is equally distributed among the 4 cubes. 
 Therefore, the total flux through the four faces is given as = q/4 ϵo. 
 c) When the charge is placed at C, the charge is shared among 2 cubes equally. Therefore, 
 the total flux through these faces is given as = q/2 ϵo. 
 d) When the charge is place at D, the charge is distributed among two cubes and 
 therefore, the total flux is given as = q/2 ϵo. 
31. Here, �⃗� = 30 x103 𝑖 ̂NC-1 
 A= 10-2 m2 
(a)As normal to the area is in the direction of electric field, therefore θ=0o 
Φ= EA cos θ= 30x103x10-2 cos 00 
 = 300Nm2C-1 
(b)In this case, θ=60o 
Φ`= EA cos θ= 30x103x10-2 cos 600 
 = 150 Nm2C-1. 
Previous Year CBSE Questions Page 34 of 484 
 
(c)In this case, θ=90o 
 Φ= EA cos θ= 30x103x10-2 cos 900 
= 0 Nm2C 
32. Through the left face 
 Φ1= Ex. A cos 180o 
 =500 x 0.1 x 10-2(-1) = -0.5 
 Through the right face 
 Φ2= Ex. A cos 0o 
 =500 x 0.2 x 10-2= 1.0 
 ∴ Net flux through the cube 
 Φ=Φ1+Φ2= 0.5 Nm2C-1 
 Charge inside the cube =ϵoΦ 
 = 8.85 x 10-12 x 0.5= 4.425 x 10-12 C. 
Answer to Five-mark questions 
Q33. Refer from NCERT textbook 
Q34. Refer from NCERT textbook 
 1. b 2. a 3. d 4. a 5. c 
Q35. 1. d 2. b 3. d 4. c 5. a 
 
 
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Previous Year CBSE Questions Page 35 of 484 
 
QUESTION BANK 
 Chapter:2 Electric Potential and Capacitance 
 
SECTION A 
(1 MARK QUESTIONS) 
Q1. What is the net charge on a charged capacitor? 
Q2. What is an equipotential surface. Give an example. (CBSE 2003) 
Q3. What is the geometrical shape of equipotential surfaces due to a single isolated charge? 
 (CBSE 2013) 
Q4. Why are electric field lines are perpendicular at a point on an equipotential surface of a 
conductor? (CBSE 2015) 
Q5. Define dielectric constant in terms of the capacitance of a capacitor. 
Q6. What may be a possible reason of water having a much greater dielectric constant (=80) than 
say mica (=6)? 
Q7.In what form is the energy stored in a charged capacitor? 
 
MULTIPLE CHOICE QUESTIONS 
Q8. If voltage applied on a capacitor is increased from V to 2V, choose the correct conclusion. 
(a) Q remains the same, C is doubled 
(b) Q is doubled, C doubled 
(c) C remains same, Q doubled 
(d) Both Q and C remain same 
Q9. A parallel plate capacitor is charged. If the plates are pulled apart 
(a) the capacitance increases 
(b) the potential differences increase 
(c) the total charge increases 
(d) the charge & potential difference remain the same 
Q10.Which of the following is an example of a molecule whose centre of mass of positive and 
negative charges coincide each other? 
 (a)CO2 (b) CO (c) CH3OH (d) NH3 
Previous Year CBSE Questions Page 36 of 484 
 
Q11.What is the angle between electric field and equipotential surface? 
(a) 90o always (b)0o always (c)0o to 90 (d)0o to 180o 
Q12.If we carry a charge once around an equipotential path, then work done by the charge is: 
(a) Infinity (b) Positive (c) Negative (d) Zero 
 
2 MARKS QUESTIONS 
Q13.Sketch equipotential surfaces for 
(a) A negative point charge (CBSE 2001) 
(b) Two equal and positive charges separated by a small distance. (CBSE 2015) 
Q14.Deduce the expression for the potential energy of an electric dipole placed with its axis at an 
angle θ to the external field �⃗� .Hence discuss the conditions of its stable and unstable equilibrium. 
 (CBSE 2008,2019,2021 Compt.) 
 
3 MARKS QUESTIONS 
Q15. Obtain the expression for the resultant capacitance when three capacitors C1 , C2 and C3 are 
connected (i) in series (ii) in parallel. 
Q16. Define the capacitance of a capacitor. Obtain the expression for the capacitance of a 
 parallel plate capacitor in vacuum in terms of plate area A and separation d between the plates. 
5 MARKS QUESTIONS 
Q17. (a) Definethe SI unit of capacitance. 
 (b) Obtain the expression for the capacitance of a parallel plate capacitor. 
Q18. (a) Define potential energy of a system of two charges. 
 (b)Two-point charges q1 and q2, separated by a distance r12 are kept in an external electric 
field. Derive an expression for the potential energy of the system of two charges in the field. 
 
*********** 
 
 
 
 
Previous Year CBSE Questions Page 37 of 484 
 
SECTION-A Answer Key 
1. Zero, because the two plates have equal & opposite charges. 
2. A surface with a constant value of potential at all points on the surface. Example: Surface of a 
charged conductor. 
3.For the point charge, the equipotential surfaces are concentric spherical shells with their centre 
at the point charge. 
4. If it were not so, the presence of a component of the field along the surface would destroy its 
equipotential nature 
5. The ratio of the capacitance of the capacitor completely filled with dielectric material to the 
capacitance of the same capacitor with vacuum between the platesis called dielectric constant. 
6. Water molecules have permanent dipole moment. 
7. Electrostatic potential energy 
MCQ 
8. (c ) 9. (b) V = Ed , V increases as distance increases 
10. (a) 11. (a) 12. (d) 
 2 MARKS ANSWERS 
13. 
 
 
14. Work done in moving the charge q1 at the point A, 
W1 = q1V (r1) Work done in moving the charge q1 at the point B, 
 
W2 = q2V (r2) + 
1
4πє0
q1𝑞2
𝑟12
 
Total work done in assembling this configuration, 
W = W1 + W2 
W = q1V (r1) + q2V (r2) + 
1
4πє0
q1𝑞2
𝑟12
 
Previous Year CBSE Questions Page 38 of 484 
 
 3 MARKS ANSWERS 
15. Derivation from ncert book 
16. Derivation from ncert book 
5 MARKS ANSWERS 
17.When a charge of one coulomb produces a potential difference of one volt between the plates 
of capacitor, the capacitance is one farad. 
Capacity of a parallel plate capacitor. A 
parallel plate capacitor consists of two 
large plane parallel conducting plates 
separated by a small distance. We first 
take the intervening medium between the 
plates to be vaccum. Let A be the area of 
each plate and d the separation between 
them. The two plates have charges Q and 
– Q. Since d is much smaller than the 
linear dimension of the plates (d2 << A), 
we can use the result on electric field by 
an infinite plane sheet of uniform surface 
charge density. Plate 1 has surface charge density σ = Q/A and Plate 2 has a surface charge density 
-σ, the electric field in different region is: 
In the inner region between the plates 1 and 2, the electric fields due to the two charged plates add 
up, giving 
 
The capacitance C of the parallel plate capacitor is then 
 
18. a) The potential energy of a system of two charges is the amount of work done in assembling 
the charges at their locations by bringing them in, from infinity. 
Previous Year CBSE Questions Page 39 of 484 
 
(b) Work done in moving the charge q1 at the point A, 
W1 = q1V (r1) 
Work done in moving the charge q1 at the point B, 
W2 = q2V (r2) + 
1
4πє0
q1𝑞2
𝑟12
 
Electric energy of the system, 
U = Total work done in assembling this configuration, 
U = W1 + W2 
 U = q1V (r1) + q2V (r2) + 
1
4πє0
q1𝑞2
𝑟12
 
 
********** 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Previous Year CBSE Questions Page 40 of 484 
 
SECTION-B 
1 MARK QUESTIONS 
Q1. An air capacitor is given a charge of 2µC raising its potential to 200 V. If on inserting a 
dielectric medium, its potential falls to 50 V. What is the dielectric constant of the medium. 
Q2. A parallel plate capacitor with air between the plates has a capacitance of 8pF. What will be 
the capacitance if the distance between the plates by reduced by half and the space between them 
is filled with a substance of dielectric constant k =6. 
Q3. What is the equivalent capacitance, C, of the five capacitors connected as shown in the figure 
 
 
ASSERTION – REASON QUESTIONS 
Directions: These questions consist of two statements, each printed as Assertion and Reason. 
While answering these questions, you are required to choose any one of the following four 
responses. 
(a) Both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion. 
(b) Both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion. 
(c) Assertion is correct, Reason is incorrect 
(d) Both Assertion and Reason are incorrect. 
 
Q4 Assertion: A spherical equipotential surface is not possible for a point charge. 
Reason: A spherical equipotential surface is not possible inside a spherical capacitor. 
Q5.Assertion: The equatorial plane of a dipole is an equipotential surface. 
Reason: The electric potential at any point on equatorial plane is zero. 
Q6.Assertion: Electric potential and electric potential energy are different quantities.