Question

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. $3.8 \times 10^{-11} C$
• B. $8.3 \times 10^{-11} C$
• C. $3.8 \times 10^{-12} C$
• D. $8.3 \times 10^{-12} C$

Answer 1

Answer: (B)

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 $y =0.5\, m$
$\Rightarrow E =150(.5)^{2}=\frac{150}{4}$
Now flux $\phi= EA =\frac{150}{4}(.5)^{2}=\frac{150}{16}$
By Gauss's law $\phi=\frac{Q_{\text {in }}}{\epsilon_{0}}$
$\frac{150}{16}=\frac{Q_{\text {in }}}{\epsilon_{0}} $
$Q_{\text {in }}=\frac{150}{16} \times 8.85 \times 10^{-12}$
$=8.3 \times 10^{-11} C$