Electrostatics is one of the most useful topics of JEE Main Physics. Every year , in JEE main there are two or three questions asked from this topic . Basically , Electrostatics deals with the study related to electric charges . It Includes some important physics laws like Coulomb's Law , law of electric field etc.The weightage of this topic in JEE Main is up to 3 or 4 questions that mean 12 to 16 marks out of 360 marks .

JEE Mains Electrostatics Year-wise Weightage

Here , year wise Electrostatics weightage trend is shown below for the last five years of JEE Main Question Paper

• The difficulty level of this type of question is moderate. It means students get the solution after applying some logic and correct formulation . 
• Now , the Syllabus of Electrostatics chapter is shown below for JEE Mains as well as JEE Advanced examination . 

1. Conservation of Charge
2. Coulomb's law 
3. Forces between multiple charges
4. Electric Field
5. Electric Field lines
6. Electric Dipole
7. Electric charges due to dipole
8. Torque
9. Electric Flux
10. Gauss’s Law
11. Electrical Potential energy of system
12. Electrostatic field
13. Conductors and insulators
14. Dielectrics and electric polarization
15. Capacitors and combination of capacitors
16. Capacitance
17. Energy

Electrostatics Test Series : Previous Year Asked Questions and Solutions

Here , Some Previous year questions are shown below with their respective solutions .

Que.1) For the circuit shown in figure, which of the following statements is true ? [ JEE 1999 - 2 Marks ]

A)with S1 closed, V1 = 15 V, V2 = 20 V

B)with S3 closed, V1 = V2 = 25 V

C)with S1 & S2 closed, V1 = V2 = 0

D)with S1 & S3 closed, V1 = 30 V, V2 = 20 V

Solution : (D) No change of any potential across any capacitor will occur unless the two capacitors Cj and C2 are either connected in series or in parallel.Charges would be maintained as they are unless Sj, S2 and S3, all are closed.This condition is given in none of the options. Hence the potentials of C1 and C2 will be maintained as V1 = 30 volt and V2 = 20 volt, with S, and S3 closed.

Que.2 ) A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral. [JEE 2007- 3 marks]

A) A potential difference appears between the two cylinders when a charge density is given to the inner cylinder.

B)A potential difference appears between the two cylinders when a charge density is given to the outer cylinder.

C)No potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders.

D)No potential difference appears between the two cylinders when the same charge density is given to both the cylinders.

Solution :(A) When a charge density is given to the inner cylinder, the potential developed at its surface is

different from that on the outer cylinder. This is because the potential decreases with distance for a

charged conducting cylinder when the point of consideration is outside the cylinder.

But when a charge density is given to the outer cylinder, it will change its potential by the same amount as that of the inner cylinder. Therefore no potential difference will be produced between the cylinders in this case.

Que.3) A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path(s) [JEE 2003 - 3 Marks]

A)1

B)2

C)3

D)4

Solution : (D) The lines of force never enter a metallic sphere because intensity is zero inside a conductor/metal. Hence options (B) and (C) are incorrect.The lines of force are at normal to the surface. Hence option (A) is incorrect.Option (D) represents the correct answer.

Que.4) 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

(A) +Q/4

(B) –Q/2

(C) +Q/2

(D) –Q/4

Solution : (D)

QQ/d2 + Qq/(d/2)2 =0

Q + 4q = 0

or q = -Q/4

Que.5) 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 [JEE 2013 - 4 Marks]

(A)508 pJ

(B)692 pJ

(C)560 pJ

(D)600 pJ

Solution : (A)

Work done = Change in energy stored in the system

W = (Q2/2C1) – (Q2/2C2)

Q = C1V = (12 pF)(10 V) = 120 x 10-12 C

C2 = 6.5

C1 = 6.5 x 120 x 10-12 F

W = (Q2/2C1)(1 – (⅙.5))

W = (C1V2/2)(1- [2/13]) = (100 x 12 x 10-12/2)(1-[2/13]) = 508 pJ

Que.6) Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 106 V m–1. The plate area is 10–4 m2. What is the dielectric constant if the capacitance is 15 pF? (given ε0 = 8.86 × 10–12 C2 N–1 m–2) [JEE 2011-4 Marks]

(A) 3.8

(B) 8.5

(C) 6.2

(D) 4.5

Solution : (B)

C = Kε0A/d

Or K = CV/ε0AEmax

K = (15 x 10-12 x 500)/(8.86 x 10-12 x 10-4 x 106) = 8.5

Que.7) There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at P, in the region, is found to vary between the limits 589.0 V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of 60° with the direction of the field? [JEE 2004 - 3 Marks]

(A) 589.2 V

(B) 589.6 V

(C) 589.5 V

(D) 589.4 V

Solution : (D)

ΔV = E.d

ΔV = Edcosθ = 0.8 x cos 600

ΔV = 0.4

Hence the new potential at the point on the sphere is

589.0 + 0.4 = 589.4 V

Que.8) Two capacitors C1 and C2 are charged to 120 V and 200 V, respectively. It is found that by connecting them together the potential on each one can be made zero. Then… [JEE 2014 - 4 Marks]

(A) 9C1 = 4C2

(B) 5C1 = 3C2

(C) 3C1 = 5C2

(D) 3C1 + 5C2 = 0

Solution : (C)

For potential to be made zero, after connection

120C1 = 200C2

6C1 = 10C2

3C1 = 5C2

Que.9) An electric dipole is placed at an angle of 30º to a non-uniform electric field. The dipole will experience [JEE 2016 - 4 Marks]

(a) a torque only

(b) a translational force only in the direction of the field

(c) a translational force only in a direction normal to the direction of the field

(d) a torque as well as a translational force

Solution : (D)

In a non-uniform electric field, the dipole will experience a torque as well as a translational force.

Quick Formulas For Revision

In the Physics Subject Formulas are the most important thing to solve any question because without applying any formula we can’t formulate the question . Here some basics and very useful formulas and laws are shown below . 

• Coulomb’s Law:- 

F = Kq1q2/r2

Here, K = 1/4πε0 = 9×109 Nm2C-2 (in free space)

• Relative Permittivity (εr):- εr = ε/ε0
• Unit of Dipole Moment:- coulomb meter (S.I), stat coulomb cm (non S.I)
• Potential at point due to several charges:-

V = (1/4π ε0) [q1/r1 + q2/r2 + q3/r3]

= V1+V2+ V2+….

• Potential due to charged spherical shell:-

(a) Outside, Vout = (1/4π ε0) (q/r)

(b) Inside, Vin = - (1/4π ε0) (q/R)

(c) On the surface, V_Surface = (1/4π ε0) (q/R)

• Capacity of a parallel plate capacitor:- Cair = ε0A/d

 Cmed = Kε0A/d

Here, A is the common area of the two plates and d is the distance between the plates.