Answer to Question #187373 in Electric Circuits for Chitengi Kapalu

Question

A sphere of radius a has a charge + q uniformly distributed throughout its volume.The sphere

is concentric with a spherical shell of inner radius b and outer radius c ,as shown in the diagram.

The shell has a net charge of−q (a)What are the net charged on the inner and outer surfaces of

the shell?(b)Find the expression for the electric field,as a function of radius r, (i) within the

sphere,(ii)between the sphere and the shell(a<r<b),(iii)inside the shell(b<r<c),(iv)outside the

shell(r>c).(c)Find an expression for the capacitance of the system.

Answer 1

Explanations & Calculations

• Refer to the figure attached.

1)

• Charges induced on the shell due to the charges of the sphere are shown in parentheses.
• The -q charge which was initially given to the shell naturally rests on the outer surface and with the induced charge +q, the net charge on the outer shell turns out to be zero.
• There is no influence on the inner surface of the shell from the charge initially given to the shell. therefore, it contains only the induced -q charge.

b)1)

• The +q is held by the entire volume of the sphere. If a volume of radius r is selected that would be containing,

$\qquad\qquad \begin{aligned} \small Q&=\small \frac{+q}{(\frac{4}{3}\pi a^3)}\times\big(\frac{4}{3} \pi r^3\big)\\ &=\small +q.\frac{r^3}{a^3} \end{aligned}$

• Since we know the charge contained by that volume/ included in that area, the flux across that area can be found by Gaussian principle & then the electric field across that area.
• By all of these, it is already derived,

$\qquad\qquad \begin{aligned} \small \vec{E} &=\small \frac{q}{A\epsilon} \end{aligned}$

• Then the electric field within the share is

$\qquad\qquad \begin{aligned} \small E&=\small \frac{Q}{A\epsilon}=\frac{+q.\frac{r^3}{a^3}}{4\pi r^2.\epsilon}\\ &=\small \frac{+q}{4\pi \epsilon a^3}\cdot r \end{aligned}$

• Within the sphere it is a linear variation.

2)

• If we selected a Gaussian surface/area of radius r (a<r<b) again it contains the same amount of charge +q, therefore, using the above equation yeilds,

$\qquad\qquad \begin{aligned} \small E_1&=\small \frac{+q}{A\epsilon}=\frac{+q}{4\pi r^2.\epsilon}\\ &=\small \frac{+q}{4\pi \epsilon}\cdot\frac{1}{r^2} \end{aligned}$

• Within that region, the field is inversely proportional to the square of the distance measured from the center.

3)

• If we select a Gaussian surface of radius r within that region (b<r<c), the charge contained is [+q+(-q)] = 0
• Therefore, there is no effective electric field within that region.

4)

• A selected Gaussian surface within that region contains (+q+(-q)+(+q)+(-q)) = 0, again resulting in a zero field across that region.

5)

• Here the inner surface of the shell and the inner sphere behave as the storage plates of a parallel plate capacitor. (refer to the figure)
• Capacitance can be found by the definition to it: $\small C= \large \frac{q}{\Delta V}$
• To calculate the difference between the potentials of the two surfaces (inner sphere and shell's inner surface) the potential equation can be used

$\qquad\qquad \begin{aligned} \small V_{sphere.inner}&=\small \frac{q}{4\pi \epsilon a}+\frac{(-q)}{4\pi \epsilon b}\\ &=\small \frac{q}{4\pi \epsilon}\bigg[\frac{1}{a}-\frac{1}{b}\bigg]\\\\ \small V_{shell.inner}&=\small \frac{q}{4\pi\epsilon b}+\frac{(-q)}{4\pi \epsilon b }=0 \end{aligned}$

• Therefore, the potential difference is just the potential of the inner sphere.
• Then the capacitance is

$\qquad\qquad \begin{aligned} \small C&=\small \frac{q}{\frac{q}{4\pi \epsilon}\big[\frac{1}{a}-\frac{1}{b}\big]}\\ &=\small 4\pi \epsilon \cdot\frac{ab}{(b-a)} \end{aligned}$