Electrostatics – JEE First

Electric flux through a hemisphere

March 24th, 2022jeefirst0 Comments

JEE Advanced 2017 Paper 2, Question 10

A point charge $+Q$ is placed just outside an imaginary hemispherical surface of radius ${\rm R}$ as shown in the figure. Which of the following statements is/are correct?

Rendered by QuickLaTeX.com

The electric flux passing through the curved surface of the hemisphere is $-\frac{Q}{2 \varepsilon_{0}}\left(1-\frac{1}{\sqrt{2}}\right)$
Total flux through the curved and the flat surfaces is $\frac{Q}{\varepsilon_{0}}$
The component of the electric field normal to the flat surface is constant over the surface
The circumference of the flat surface is an equipotential

Solution

In problems like these a beginner may …

A spherical capacitor

January 20th, 2022jeefirst0 Comments

A spherical conducting shell with radius $b$ is concentric with a conducting ball with radius $a$ , with $a<b$ .

Compute the capacitance $C = Q / \Delta \phi$ when the shell is grounded and the ball has charge $Q$ .
Compute the capacitance when the ball is grounded and the shell has charge $Q$ .
Compute the full matrix of coefficients of capacitance for the two conductors.
Considering these conductors as a capacitor, determine its capacitance. That is, assign equal and opposite charges $\pm Q$ to the shell and the ball, and compute $C = Q / \Delta \phi$ .

Related Problem: Insulating spherical shell with a hole

Solution

(a) First, we ground the shell and give the …

Dipole in a uniform electric field

January 18th, 2022jeefirst0 Comments

JEE Advanced 2019 Paper 2, Question 4

An electric dipole with dipole moment $\frac{p_{0}}{\sqrt{2}}(\hat{i}+\hat{j})$ is held fixed at the origin $O$ in the presence of an uniform electric field of magnitude $E_{0}$ . If the potential is constant on a circle of radius $R$ centered at the origin as shown in figure, then the correct statement(s) is/are:

( $\varepsilon_{0}$ is permittivity of free space. $R \gg$ dipole size)

Rendered by QuickLaTeX.com

$R=\left(\frac{p_{0}}{4 \pi \epsilon_{0} E_{0}}\right)^{1 / 3}$
Total electric field at point $A$ is ${\bf E}^A=\sqrt{2} E_{0}(\hat{i}+\hat{j})$
Total electric field at point $B$ is ${\bf E}^B=0$
The magnitude of total electric field on any two points of the circle will be same.

Solution

The potential due to the dipole kept at the origin …

Electric field in a hollow region

January 9th, 2022jeefirst0 Comments

IIT-JEE 2007 Paper 2, Question 6

(This problem is was repeated in JEE Advanced 2015 Paper 2, Question 11)

A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

Rendered by QuickLaTeX.com

zero everywhere
non-zero and uniform
non-uniform
zero only at its center

Solution

As explained in a related problem we can think of the given charge distribution as a …

Charge at one corner of a cube

January 8th, 2022jeefirst0 Comments

A charge $q$ sits at one corner of a cube of side length $a$ as shown in the figure below. What is the electric flux through the shaded side?

Rendered by QuickLaTeX.com

Solution

Since the problem is asking us to find the electric flux it is natural to guess that we need to apply Gauss’s Law. However, we need a closed surface with appropriate symmetry to be able to use Gauss’s Law (see article). So we consider the situation shown below.

Rendered by QuickLaTeX.com

We have placed …