Question

A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has a radius 15.0 cm and the capacitance is 116 pF.

(a) What is the radius of the outer sphere?

(b) If the potential difference between the two spheres is 220 V, what is the magnitude of the charge on each sphere?

Short answer

The radius of the outer sphere is 17.5cm.

The magnitude of the charge Q on each sphere is 25.52nC.

Step-by-step solution

Capacitance of spherical body.

We were given a spherical capacitor where the radius of the first sphere is

$C=4\pi {ϵ}_{\circ }\frac{r_a{r}_b}{r_b-r_a}$

a) The capacitance C is the magnitude Q of the charge on either sphere divided by the potential difference $V_{ab}$between the spheres where we can get the capacitance as the terms of the radius of each sphere

$$\begin{gathered} C=4\pi {ϵ}_{\circ }\frac{r_a{r}_b}{r_b-r_a} \\ {r}_{b}C=r_{a}C=4\pi {ϵ}_{\circ }{r}_a{r}_b \\ {r}_{a}C=r_b(C-4\pi {ϵ}_{\circ }{r}_a) \\ {r}_a=\frac{r_{b}C}{\left(C-4\pi {ϵ}_{\circ }{r}_a\right)} \end{gathered}$$

Outer radius

Now Substitute our values for C, $r_a$into the above equation to get the radius of the outer sphere where

$$\begin{gathered} r_b=\frac{r_{a}C}{\left(C-4\pi {ϵ}_{\circ }{r}_a\right)} \\ =\frac{(0.15m)(116\times {10}^{-12}F)}{116\times {10}^{-12}F-4\pi \left[8.854\times {10}^{-12}F/m\right]\left[0.15m\right]} \\ =0.175m \\ =17.5cm \end{gathered}$$

The radius of the outer sphere is 17.5cm

Charge on the sphere.

b) The two spheres have the same magnitude of the charge Q but in the opposite sign. So, we can use the value of the capacitance and the potential difference $V_{ab}$to get the charge by applying the equation

$Q=C{V}_{ab}$

where the charge is directly proportional to the capacitance C. Now substitute our values to get the charge Q

$$\begin{gathered} Q=C{V}_{ab} \\ =(116\times {10}^{-12}F)(220.0V) \\ =25.52\times {10}^{-9}C \\ =25.52nC \end{gathered}$$

Hence the magnitude of the charge Q on each sphere is 25.52nC