Question

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

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

Answer 1

Answer: (C)

$2.22$ x ${10}^{-10}C$

According to the question, $\overrightarrow{E}=250rV/m$

where $r$ is the distance from centre

Let's take a sphere of radius $20cm$

or $r=20\times {10}^{-2}m$

So, Electric field from any point on sphere is given by

$=250\times 20\times {10}^{-2}V/m$

$=50V/m$

$\therefore$ we can deduce electric flux from the whole sphere, i.e. of radius $r$.

$\int \varphi =\int E.da=EA$

$=E4\pi r^2$

$=50\times 4\times \frac{22}{7}\times {(20\times {10}^{-2})}^2$

$=25.14Vm$

from gauss law, flux through any symmetrical closed surface is given by $q_{inc}/{\epsilon }_0$

$\Rightarrow \varphi =\frac{q_{inc}}{{\epsilon }_0}$ {$q_{inc}$ : the charge enclosed by the sphere}

$\therefore 25.14\times 8.85\times {10}^{-12}=q_{inc}$

$\Rightarrow q_{inc}=222.51\times {10}^{-12}C$

$=2.2251\times {10}^{-10}C$

$\therefore$ Option (C) is the correct answer.