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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

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 $

Answer: (A)