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# JEE Important Chapter - Electrostatics

Get interactive courses taught by top teachers ## Electrostatics

Last updated date: 20th Sep 2023
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Electrostatics is the branch of physics that deals with the study of charges at rest and their interaction with other charges. This section consists of concepts and advanced problems related to electrostatics. It is a very important chapter for JEE in terms of weightage.

The chapter Electrostatics begins by introducing what is electrostatics, what is electrostatic charge, Coulomb’s law of electrostatics and then problems related to Coulomb's law of electrostatics. Then, we will see the electric potential and electric field due to point charges and continuous charge distribution of different shapes such as rings, discs, cylinders etc. We can see many examples of electrostatic effects in our life.

The torque experienced by an electric dipole and its potential energy is discussed in this chapter along with applications of Gauss law to calculate the net electric flux linked with a closed surface. Frequent questions are asked for JEE about types of capacitors and problems related to capacitors in series and parallel combinations. The effect of the dielectric medium in a capacitor is also discussed in this section.

Let us see what electrostatics is and define electrostatic force. We will also cover the important formulas needed for JEE Main and JEE Advanced related to Electrostatics with solved examples.

## JEE Main Physics Chapter-wise Solutions 2022-23

### Important Topics in Electrostatics

• Coulomb’s Law.

• Electric field and electric lines of force.

• Electric field due to continuous charge distribution.

• Electric dipole and dipole moment.

• Gauss Law.

• Electric potential and Equipotential surface.

• Electric potential due to the various charge distributions.

• Capacitance of a capacitor.

• Grouping of Capacitance.

### Solved Examples

1. The electric potential in a region is given by V=(6x-8xy2-8y+6yz-4x2) volt. Then the electric field acting on a point charge of 2C placed at the origin will be …

Ans: Given,

The electric potential in a region , V=(6x-8xy2-8y+6yz-4x2) volt

The electric field in the region can be obtained from the electric potential using the formula

$\vec E=\dfrac{\partial V}{\partial x} \hat i+\dfrac{\partial V}{\partial y} \hat j+\dfrac{\partial V}{\partial z} \hat k$

$\vec E=(6-8y^2-8x) \hat i+(-16xy-8+6z) \hat j+6y \hat k$...(1)

To find the electric field at origin, put x=0, y=0 and z=0 in equation (1)

$\vec E=6 \hat i+-8 \hat j+0 \hat k$

$|\vec E|=\sqrt{6^2+(-8)^2}$

$|\vec E|=10~N/C$

The force acting on the charge q due to electric E is given by

$F=qE$

$F=2C\times 10~N/C=20~N$

Key point: Electric field at a point can be obtained by the differentiation of electric potential with respect to distance.

2. A parallel plate capacitor with oil between plates (dielectric constant of oil k=2) has a capacitance C. If the oil is removed, then the capacitance of the capacitor becomes…

Ans: Let A be the area of the capacitor and d be the distance between the plates of the capacitor.

The formula for the  capacitance of the parallel plate capacitor with oil as a dielectric is given  by,

$C=\dfrac{kA\epsilon_0}{d}$...(1)

The capacitance of the parallel plate capacitor when oil is removed is

$C'=\dfrac{A\epsilon_0}{d}$...(2)

Divide equation (2) by (1) to obtain the new capacitance C’ when oil is removed in terms of initial capacitance C.

$\dfrac{C'}{C}=\dfrac{\left(\dfrac{A\epsilon_0}{d}\right)}{\left(\dfrac{kA\epsilon_0}{d}\right)}$

$\dfrac{C'}{C}=\dfrac{1}{k}$

$C'=\dfrac{C}{k}$

When the oil is removed from the parallel plate capacitor, its capacitance is reduced by a factor of k.

Key point:  The capacitance of a parallel plate capacitor depends on the dielectric medium and the dimensions of the capacitor. It does not depend on the charge and potential difference across the capacitor.

### Previous Year Questions from JEE Paper

1. Two equal capacitors are first connected in series and then in parallel. The ratio of equivalent capacitances in the two cases will be (JEE 2021)

Sol:

Let the capacitances of each capacitor be C.

In the series connection of two capacitors, the equivalent capacitances is given by $\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$

$\dfrac{1}{C_{12}}=\dfrac{1}{C}+\dfrac{1}{C}$

$C_{12}=\dfrac{C}{2}$...(1)

For the capacitors in parallel connection, the equivalent capacitance is given by $C_{eq}=C_{3}+C_{4}$

$C_{34}=C+C$

$C_{34}=2C$...(2)

The ratio of  the equivalent capacitances in series and the parallel connection is obtained by dividing equation (1) and equation (2).

$\dfrac{C_{12}}{C_{34}}=\dfrac{\left(\dfrac{C}{2}\right)}{2C}$

$\dfrac{C_{12}}{C_{34}}=\dfrac{1}{4}$

The ratio of the equivalent capacitances of series and parallel connection is 1:4

Key point:  In series connection, equivalent capacitance is equal to the reciprocal of the sum of the reciprocal of each capacitance. For parallel connection, equivalent capacitance is equal to the sum of each individual capacitance.

2. A point charge of +12 μC  is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in the figure. The magnitude of electric flux through the square will be _____ ✕ 103 Nm2/C. (JEE 2021 Feb) Sol. To find the electric flux, we use Gauss law and assume a Gaussian surface in the form of a cube of side length 12 cm in such a way that charge is located at the centre of the cube. Applying Gauss law, total flux linked with the cube is given by

$\phi_{total}=\dfrac{q}{\epsilon_0}$

The electric flux(ɸ) through the square we want to find is 1/6 times the total flux.

$\phi=\dfrac{q}{6\epsilon_0}$

$\phi=\dfrac{12\times10^{-6}}{6\times8.85\times10^{-12}}$

$\phi=226\times10^{3}~Nm^2/C$

Therefore, the flux through the square will be 226 × 103 Nm2/C.

Key point: Gauss Law is applicable to any Gaussian surface irrespective of size and shape.

### Practice Questions

1. There is 10 mC of charge at the centre of a circle of radius 10 cm. The work done in moving a charge of 1mC around the circle once is

2. A parallel plate capacitor having a plate separation of 2 mm is charged by connecting it to a 300 V supply. The energy density is

Ans: 0.1 J/m3

### Conclusion

In this article, we define electrostatic force, related definitions and formulae. We also solved some JEE numeral problems to understand the types of questions asked in the exam.  You should memorise all the formulae to solve numerical problems. Numerical problems will help you to score good marks in JEE exams.

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See More ## FAQs on JEE Important Chapter - Electrostatics

FAQ

1. What is the weightage of the electrostatics in JEE?

Electrostatics is a vast chapter compared to others. Around 2-3 questions covering about 8-12 marks are asked in JEE from various parts of the chapter.

2. Which section has to be given more importance in the chapter on electrostatics?

Since electrostatic is a vast topic, there are many important concepts here in this chapter. However, more focus should be given to the electric fields of various charge distribution, Gauss law and capacitance to score better marks in JEE

3. How to score good marks in JEE from the chapter electrostatics?

If we go through the last 20 years’ question papers, we can understand the important concepts of the chapter from where the questions are asked frequently. Understanding these concepts in-depth and then solving practice problems is the best way to score good marks in JEE for the chapter electrostatics. ## Notice board

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