Question

The electric field in a region is radially outward with magnitude $E=A{\gamma }_0$ . The charge contained in a sphere of radius ${\gamma }_0$ centered at the origin is

• A. $\frac{1}{4\pi {\epsilon }_0}A{\gamma }_0^3$
• B. $4\pi {\epsilon }_{0}A{\gamma }_0^3$
• C. $\frac{4\pi {\epsilon }_{0}A}{{\gamma }_0}$
• D. $\frac{1}{4\pi {\epsilon }_0}\frac{A}{{\gamma }_0^3}$

Answer 1

The correct option is B $4\pi {\epsilon }_{0}A{\gamma }_0^3$

From Gauss law we know that,
${\varphi }_{net}=\frac{Q_{in}}{{ϵ}_0}$
$⟹Q_{in}={\varphi }_{net}\times {ϵ}_0$

As the electric field through the surface is radially outward so, the angle between electric field and area vector must be $0^{\circ }$.
$⟹$ $\overrightarrow{d\varphi }=\overrightarrow{E}.\overrightarrow{dS}$
or, $\varphi =EScos\left(0^{\circ }\right)$
or, ${\varphi }_{net}=A{\gamma }_0\times 4\pi {{\gamma }_0}^2$
$\therefore {\varphi }_{net}=A4\pi {\gamma }_0^3$
So, $Q_{net}=4\pi {ϵ}_{0}A{\gamma }_0^3$