### A cube is placed inside an electric field, $\vec E = 150 y^2 \hat j$ ....

A cube is placed inside an electric field, $\vec E = 150 y^2 \hat j$. The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is:

(A) $8.3 \times 10^{-11} C$
(B) $8.3 \times 10^{-12} C$
(C) $3.8 \times 10^{-12} C$
(D) $3.8 \times 10^{-11} C$

Solution

We have, $\phi = \frac {q_{in}}{\epsilon_0}$

Since electric field is upwardly directed, it will only cut the top and bottom surfaces.

Since y = 0 at the bottom surface makes electric field 0 there, no flux is present there. The only flux through the cube is contributed by the flux through the top surface.

Flux through the top surface $=150 \times 0.5^2 \times 0.5^2$

So, $\phi =150 \times 0.5^2 \times 0.5^2 = \frac {q_{in}}{8.85 \times 10^{-12}}$

$\therefore q_{in} \approx 8.3 \times 10^{-11} C$