Question

A spherical volume contains a uniformly distributed charge of density $1.0 \times 10^{-6} \,C / m ^{3}$ Find the electrical field (in $N / C )$ at a point inside the volume at a distance $1\, mm$ from the centre. $\left(\right.$ Let $\left.\frac{1}{4 \pi \in_{0}}=9 \times 10^{9} \,Nm ^{2} C ^{-2}\right)$

• A. $\frac{8}{\pi}$
• B. $6 \pi$
• C. $\frac{\pi}{6}$
• D. $12 \pi$ $\pi$

Answer 1

Answer: (D)

Given, the volume charge density of sphere,
$\rho_{v} =1 \times 10^{-6} \,C / m ^{3} $
$\frac{1}{4 \pi \varepsilon_{0}} =9 \times 10^{9} \,Nm ^{2}\, C ^{-2}$
$r =1 mm =10^{-3} \,m$
Charge on sphere, $q=$ charge density $\times$ volume
$=\rho_{v} \times \frac{4}{3} \pi r^{3} \,\,\,\left(\because V=\frac{4}{3} \pi r^{3}\right)$
$\therefore $ Electric field intensity at a distance $r$,
$E=\frac{1}{4 \\,pi \,\varepsilon_{0}} \cdot \frac{q}{r^{2}} $
$E=\frac{1}{4 \,\pi \,\varepsilon_{0}} \cdot \frac{\rho_{v} \times \frac{4}{3} \pi r^{3}}{r^{2}}=\frac{1}{4 \,\pi \,\varepsilon_{0}} \cdot \frac{\rho_{v} \times 4 \,\pi\, r}{3}$
Putting the given values, we get
$=9 \times 10^{9} \times \frac{1 \times 10^{-6} \times 4 \times 10^{-3} \times \pi}{3}$
$E=12\, \pi$
Hence, the electric field inside the sphere at a distance $1 mm$ from the centre is $12 \,\pi \,N / C$.