Question

The electric field in a region is radially outward and is given by $E=250r\text{V/m}$ at a point (where $r$ is the distance of the point from the origin). Calculate the charge contained in a sphere of radius $20\text{cm}$ centred at the origin.

• A. $2.22\times {10}^{-6}\text{C}$
• B. $2.22\times {10}^{-8}\text{C}$
• C. $2.22\times {10}^{-10}\text{C}$
• D. Zero

Answer 1

The correct option is C $2.22\times {10}^{-10}\text{C}$


Here, $E=250r$

At the surface, $r=R=0.02\text{m}$

$\Rightarrow E=250\times R=250\times 20\times {10}^{-2}\text{V/m}$

Since electric field is radially outwards, the value of $\oint \overrightarrow{E}.d\overrightarrow{A}$ at the Gaussian surface is :

$\oint \overrightarrow{E}.d\overrightarrow{A}=EA$

$\Rightarrow \oint \overrightarrow{E}.d\overrightarrow{A}=E\left(4\pi R^2\right).....\left(i\right)$

On applying Gauss`s law,

${\varphi }_{net}=\frac{q_{inside}}{{\epsilon }_0}$

$\Rightarrow \oint \overrightarrow{E}\cdot d\overrightarrow{A}=\frac{q_{inside}}{{\epsilon }_0}$

$(\because {\varphi }_{net}=\oint \overrightarrow{E}\cdot d\overrightarrow{A})$

Substituing from Eq.$\left(i\right)$,

$\Rightarrow E\left(4\pi R^2\right)=\frac{q_{inside}}{{\epsilon }_0}$

$\Rightarrow (250\times 20\times {10}^{-2})\times (20\times {10}^{-2}{)}^2=\frac{q_{inside}}{4\pi {\epsilon }_0}$

$\Rightarrow 50\times 400\times {10}^{-4}=(9\times {10}^9)q_{inside}$

$\Rightarrow q_{inside}=\frac{2}{9\times {10}^9}=\frac{20}{9}\times {10}^{-10}$

$\therefore q_{inside}=2.22\times {10}^{-10}\text{C}$

This is the enclosed charge in the given sphere.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: Choosing a spherical Gaussian surface at a distance}\\ \text{of}R=20\text{cm}\text{from origin will enclose all the contained}\\ \text{charge and radial direction of}\overrightarrow{E}\text{at spherical surface}\\ \text{will yield}\oint \overrightarrow{E}\cdot d\overrightarrow{A}=EA=E\left(4\pi R^2\right).\end{array}$