Question

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

• A. $8{\epsilon }_0$
• B. $2{\epsilon }_0$
• C. $3{\epsilon }_0$
• D. $5{\epsilon }_0$

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 ${\epsilon }_0.$." Then
$\oint \vec{E}\cdot \overrightarrow{dS}=\frac{q_{\text{in }}}{{\epsilon }_0}$
$\therefore 100\times (0.2{)}^2-(-100)(0.2{)}^2+0+0=\frac{q_{in}}{{\epsilon }_0}$
$\Rightarrow 4-(-4)=\frac{q_{\text{in }}}{{\epsilon }_0}$
$\therefore q_{\text{in }}=8{\epsilon }_0$