JEE Physics 2023 Question with Solution

Given: $\vec{E} = E_0 x \hat{i}$, $E_0 = 4 \times 10^4 \; \text{N C}^{-1} \text{m}^{-1}$, $a = 2 \times 10^{-2} \, \text{m}$, and $\epsilon_0 = 9 \times 10^{-12} \, \text{C}^{2}/\text{N m}^{2}$.

Find: The value of $Q$ if enclosed charge is written as $Q \times 10^{-14} \, \text{C}$.

The electric field is along the $x$-direction, so only the two faces perpendicular to the $x$-axis contribute to net flux. At $x = 0$, the field is zero. At $x = a$, the field is $\vec{E} = E_0 a \hat{i}$.

Using electric flux,

so the net flux through the cube is

Now apply Gauss's law,

Therefore,

Substituting the given values,

Hence, in the form $Q \times 10^{-14} \, \text{C}$, the value of $Q$ is $288$.

Given: $\vec{E} = E_0 x \hat{i}$.

Find: The enclosed charge inside the cube.

For the faces at $y = 0, y = a, z = 0, z = a$, the area vectors are along $\pm \hat{j}$ or $\pm \hat{k}$, while the field is along $\hat{i}$. Hence $\vec{E} \cdot \vec{A} = 0$ on these four faces.

For the face at $x = 0$,

so flux through this face is zero.

For the face at $x = a$,

and the outward area vector has magnitude $a^2$ in the $+\hat{i}$ direction. Therefore,

Using Gauss's law,

This gives

Therefore, the required value is $288$.