Question

A spherical volume contains a uniformly distributed charge of volume density $2.0\times {10}^{-4}C{m}^{-3}$. Find the strength of electric field at a point inside the volume at a distance $4.0cm$ from the centre.

• A. $3\times {10}^4\frac{N}{C}$
• B. $3\times {10}^5\frac{N}{C}$
• C. $3\times {10}^6\frac{N}{C}$
• D. $3\times {10}^7\frac{N}{C}$

Answer 1

The correct option is B

$3\times {10}^5\frac{N}{C}$

Consider a spherical surface of radius $4cm$ as a Gaussian surface

The electric field intensity at all points on the Gaussian surface should have the same magnitude and its direction radially outwards at each point as there cannot be any preferred direction other than radial. {Symmetry and uniform charge distribution}

Let the magnitude of electric field strength be E.

According to Gauss’ law,

$\oint \overrightarrow{E}.\overrightarrow{ds}=\phi =\frac{q_{in}}{{ϵ}_0}$

Again considering radially outward direction of elementary normal at each point on the Gaussian surface, we can claim,

Angle between $\overrightarrow{E}$ at $\overrightarrow{ds}$ each small area element = $0^0$

$\Rightarrow E\left(4\pi r^2\right)=\frac{q_{in}}{{ϵ}_0}=\frac{\rho \left(\frac{4}{3}\right)\pi r^3}{{ϵ}_0}$

Where is $\rho$ the given volume charge density.

$E=\frac{\rho r}{3{ϵ}_0}=\frac{2\times {10}^{-4}\times 4}{3\times 100\times 8.85\times {10}^{-12}}\frac{N}{C}$

$E=3\times {10}^5\frac{N}{C}$