A cube is placed inside an electric field, vector E = 150 y^2 j. he side of the cube is 0.5 m and is placed in the field as shown in the given figure.

Question

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

(1) 3.8 x 10^-11 C

(2) 8.3 x 10^–11 C

(3) 3.8 x 10^–12 C

(4) 8.3 x 10^–12 C

Answer 1

Correct option is (2) 8.3 x 10^–11 C

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 × 0.5² × 0.5²

So, ϕ = 150 × 0.5² × 0.5² 

= \(\frac{q_{in}}{8.85 \times 10^{-12}}\)

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

Answer 2

Correct Option is (2) 8.3 × 10^–11 C

As electric field is in y-direction so electric flux is only due to top and bottom surface Bottom surface y = 0

\(\Rightarrow\) E = 0 \(\Rightarrow\) \(\phi\) = 0

Top surface y = 0.5 m

\(\Rightarrow\) E = 150 (.5)² = 150/4

Now flux \(\phi\) = EA = \(\frac{150}4(.5)^2=\frac{150}{16}\)