JEE Main 2021 — Electrostatics Question with Solution

Given, the side of the cube, s = 0.5 m

Electric field, E = 150 y²$$\hat{j}$$

The direction of electric field is as shown in the below figure,


At bottom surface, y = 0

As we know that, the expression of electric flux,

$$\varphi$$ = E . A cos$$\theta$$

Here, E is the electric field passing through the cube and A is the surface area of the cube.

Substituting the values in the above equations, we get

$$\varphi$$ = 150y² . (0.5 $$\times$$ 0.5) $$\times$$ cos180$${}^{\circ }$$

= 150(0)² . (0.25) $$\times$$ ($$-$$1) = 0

Hence, the electric flux is zero at the bottom surface.

At the top surface, y = 0.5 m

Electric field, E = 150 y²$$\hat{j}$$ = 150(0.5)² = 37.5 N/C

Electric flux at the top surface,

$$\varphi$$ = E . A cos$$\theta$$

= (37.5) . (0.5 $$\times$$ 0.5) cos0$${}^{\circ }$$

= 9.375 N / C - m²

By using the Gauss's law, $$\varphi =\frac{Q_{in}}{{\epsilon }_0}$$

Here, Q_in = net charge enclosed in the cube and $${\epsilon }_0$$ = permittivity of the free space.

Substituting the values in the above equation, we get

$$9.375=\frac{Q_{in}}{8.85\times {10}^{-12}}$$

Q_in = 8.3 $$\times$$ 10^-11 C

The charge inside the cube is 8.3 $$\times$$ 10^-11 C.