Question

Electric field in a certain region is acting radially outward and the variation of electric field with distance from the origin is represented below. The slope of the graph is $A^{\prime }$.


The charge contained in a sphere of radius $a$ centred at the origin will be given by

• A. $4\pi {ϵ}_0{A}^{\prime }{a}^2$
• B. $\pi {ϵ}_0{A}^{\prime }{a}^2$
• C. $8\pi {ϵ}_0{A}^{\prime }{a}^3$
• D. $4\pi {ϵ}_0{A}^{\prime }{a}^3$

Answer 1

The correct option is D $4\pi {ϵ}_0{A}^{\prime }{a}^3$

Given that,
The slope of the graph, $m=A^{\prime }$
$\Rightarrow E=mr=A^{\prime }r$
$\Rightarrow E=A^{\prime }r\dots \left(i\right)$
(Analogous to $y=mx$)

Let us consider a Gaussian sphere of radius $a$ centred at the origin.

Applying Gauss's law,
$\oint \overrightarrow{E}.\overrightarrow{dA}=\frac{q_{en}}{{ϵ}_0}$
where electric field $\left(\overrightarrow{E}\right)$ is radially outwards, $\overrightarrow{dA}$ is an elemental Gaussian surface area and charge enclosed is $q_{en}$.

$\Rightarrow EA=\frac{q_{en}}{{ϵ}_0}$
$\Rightarrow \left(A^{\prime }r\right)\left(4\pi a^2\right)=\frac{q_{en}}{{ϵ}_0}$ (by using equation ($i$))
$\Rightarrow 4\pi a^3{A}^{\prime }=\frac{q_{en}}{{ϵ}_0}$ (at $r=a$)
$\therefore q_{en}=4\pi {ϵ}_0{A}^{\prime }{a}^3$
Hence, option (b) is correct.