Question

A spherical volume contains a uniformly distributed charge of density $2·0\times {10}^{-4}\mathrm{C}{\mathrm{m}}^{-3}$. Find the electric field at a point inside the volume at a distance 4⋅0 cm from the centre.

Answer 1

Given :
Volume charge density, ρ = $2\times {10}^{-4}\mathrm{C}/{\mathrm{m}}^3$

Let us assume a concentric spherical surface inside the given sphere with radius =$4\mathrm{cm}=4\times {10}^{-2}\mathrm{m}$

The charge enclosed in the spherical surface assumed can be found by multiplying the volume charge density with the volume of the sphere. Thus,

$$\begin{gathered} q=\rho \times \frac{4}{3}\mathrm{\pi }{r}^3 \\ \Rightarrow q=\left(2\times {10}^{-4}\right)\times \frac{4}{3}\mathrm{\pi }{r}^3 \end{gathered}$$
The net flux through the spherical surface,
$\varphi =\frac{q}{{\in }_0}$
The surface area of the spherical surface of radius r cm:
A = $4\pi r^2$
Electric field,
$$\begin{gathered} E=\frac{q}{{\in }_0\times A} \\ E=\frac{2\times {10}^{-4}\times 4\mathrm{\pi }{r}^3}{{\in }_0\times 3\times 4\mathrm{\pi }{r}^2} \\ E=\frac{2\times {10}^{-4}\times r}{3\times {\in }_0} \end{gathered}$$

The electric field at the point inside the volume at a distance 4⋅0 cm from the centre,

$$\begin{gathered} E=\frac{(2\times {10}^{-4})\times (4\times {10}^{-2})}{3\times (8.85\times {10}^{-12})}\mathrm{N}/\mathrm{C} \\ E=3.0\times {10}^5\mathrm{N}/\mathrm{C} \end{gathered}$$