Q145 / 160 · TS EAMCET 2019 (06 May Shift 1) · Physics
A spherical volume contains a uniformly distributed charge of density $1.0 \times 10^{-6} \mathrm{C} / \mathrm{m}^3$.
Find the electrical field (in N/C) at a point inside the volume at a distance $1 \mathrm{~mm}$ from the centre. $\left(\right.$ Let $\left.\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)$
- A$\frac{8}{\pi}$
- B$6 \pi$
- C$\frac{\pi}{6}$
- Correct answerD$12 \pi$
Solution
Given, the volume charge density of sphere,
$\begin{aligned} \rho_v & =1 \times 10^{-6} \mathrm{C}^2 \mathrm{~m}^3, \\ \frac{1}{4 \pi \varepsilon_0} & =9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2} \\ r & =1 \mathrm{~mm}=10^{-3} \mathrm{~m}\end{aligned}$
Charge on sphere,
$q=$ charge density $\times$ volume
$=\rho_v \times \frac{4}{3} \pi r^3 \quad\left(\because V=\frac{4}{3} \pi r^3\right)$
$\therefore$ Electric field intensity at a distance $r$,
$\begin{aligned} & E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} \\ & E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\rho_v \times \frac{4}{3} \pi r^3}{r^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\rho_v \times 4 \pi r}{3}\end{aligned}$
Putting the given values, we get
$\begin{aligned} & =9 \times 10^9 \times \frac{1 \times 10^{-6} \times 4 \times 10^{-3} \times \pi}{3} \\ E & =12 \pi\end{aligned}$
Hence, the electric field inside the sphere at a distance $1 \mathrm{~mm}$ from the centre is $12 \pi \mathrm{N} / \mathrm{C}$.
Hence, the correct answer is option: 'D'.