Question

A charge Q is distributed uniformly within the material of a hollow sphere of inner and outer radii r1 and r2 (figure 30-E4). Find the electric field at a point P at a distance x away from the centre for r1 < x < r. Draw a rough graph showing the electric field as a function of x for 0 < x < 2r2 (figure 30-E4).
Figure

Answer 1

Explanation:

• Magnitude Q’s charge is inside the hemisphere, which is not enclosed by a single hemisphere. So, another part of the sphere can be imagined and completed. Now, inside the sphere, the charge is Q.
• So, in two hemispheres, the total flux is distributed equally. So, the flux via one hemisphere will be half of the total sphere’s flux.

Sphere’s volume with radius x = 4/3π$\times$3.

Having radius, the charge enclosed by the sphere is

$\chi=\left(\frac{4}{3} \pi \chi^{3}-\frac{4}{3} \pi r_{1}^{3}\right) \times \frac{Q}{\frac{4}{3} \pi r_{2}^{3}-\frac{4}{3} \pi r_{1}^{3}}=Q\left(\frac{\chi^{3}-r_{1}^{3}}{r_{2}^{3}-r_{1}^{3}}\right)$

$\begin{aligned}&\text { Applying Gauss's law }-E \times 4 \pi \chi^{2}=\frac{q \text { enclosed }}{\varepsilon_{0}}\\&\Rightarrow E=\frac{Q}{\varepsilon_{0}}\left(\frac{\chi^{3}-r_{1}^{3}}{r_{2}^{3}-r_{1}^{3}}\right) \times \frac{1}{4 \pi \chi^{2}}=\frac{Q}{4 \pi \varepsilon_{0} \chi^{2}}\left(\frac{\chi^{3}-r_{1}^{3}}{r_{2}^{3}-r_{1}^{3}}\right)\end{aligned}$

According to Gauss's Law,

$\oint \mathrm{E} \cdot \mathrm{d} \mathrm{s}=\mathrm{q} \in 0$

• On the sphere, the surface integral is performed, which has radius x. Enclosed by the sphere, the charge is q. The graph can be shown as below: