Question

Electric field at $x=10\, cm$ is $100\, V / m$ and at $x=-10\, cm$ is $-100 \,V / m .$ The magnitude of charge enclosed by the cube of side $20 \,m$ is

• A. $ 8\,{{\varepsilon }_{0}} $
• B. $ 2\,{{\varepsilon }_{0}} $
• C. $ 3\,{{\varepsilon }_{0}} $
• D. $ 5\,{{\varepsilon }_{0}} $

Answer 1

Answer: (A)

From Gauss's law, "the net electric flux through any closed surface is equal to the net charge inside the surface divided by $\varepsilon_{0} .$." Then
$\oint \vec{ E } \cdot \vec{ d S }=\frac{q_{\text {in }}}{\varepsilon_{0}} $
$\therefore 100 \times(0.2)^{2}-(-100)(0.2)^{2}+0+0=\frac{q_{i n}}{\varepsilon_{0}} $
$\Rightarrow 4-(-4)=\frac{q_{\text {in }}}{\varepsilon_{0}}$
$\therefore q_{\text {in }}=8 \varepsilon_{0}$