In Fig. 24-71, a metal sphere with charge$\mathbf{\text{q=5.00μC}}$and radius $\mathbf{\text{r=3.00cm}}$ is concentric with a larger metal sphere with charge $\mathbf{\text{Q=15.0μC}}$ and radius $\mathbf{\text{R=6.00cm}}$ .

(a) What is the potential difference between the spheres? If we connect the spheres with a wire, what then is the charge on

(b) the smaller sphere and

(c) the larger sphere?

The Coulomb’s constant, $k=9.0{\text{×10}}^{\text{9}}\text{F}/\text{m}$

The charge,$q=5.0\text{μC}=5.0\times {\text{10}}^{\text{-6}}\text{C}$

The radius,$r=3.0\text{cm}=3.0\times 1{\text{0}}^{\text{-2}}\text{m}$

The radius,$R=6.0\text{cm}=6.0\times {10}^{-2}\text{m}$

The charge,$Q=15.0\text{μC}=15.{\text{0×10}}^{\text{-6}}\text{C}$

The innerelectricpotential,$V_{inner}=375\text{0kV}$

The outer electric potential, $V_{outer}=3\text{000kV}$

Use the electric potential formula to calculate the potential difference.

Formulae are as follow:

The electric potential is defined by,

$\mathit{V}\mathbf{=}\mathit{k}\frac{\mathbf{q}}{\mathbf{r}}$

Where, V is the electric potential,K is the electric constant, q is the electric charge, and r is the radius of the sphere.

The net electric potential of the inner sphere, if the inner sphere is of radius r of charge q and outer sphere is of radius R and of charge Q, can be written as,

$\begin{array}{c}{V}_{inner}=V_{duetosmallersphere}+V_{duetolargersphere}\\ =k\frac{q}{r}+k\frac{Q}{R}\\ =k\left(\frac{q}{r}+\frac{Q}{R}\right)\end{array}$

$\begin{array}{c}{V}_{inner}=(9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2)\left(\frac{(5.0\times {10}^{-6}\text{C})}{(3.0\times {10}^{-2}\text{m})}\text{+}\frac{(15.0\times {10}^{-6}\text{C})}{(6.0\times {10}^{-2}\text{m})}\right)\\ =\text{3750kV}\end{array}$

The net electric potential of the outer sphere, if the inner sphere is of radius r of charge q and outer sphere is of radius R and of charge Q , can be written as,

$\begin{array}{c}{V}_{outer}=k\left(\frac{q+Q}{R}\right)\\ =(9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2)\left(\frac{(5.0\times {10}^{-6}\text{C})+(15.0\times {10}^{-6}\text{C})}{(6.0\times {10}^{-2}\text{m})}\right)\\ =3000\text{kV}\end{array}$

The potential difference can be written as,

$\begin{array}{c}\Delta V=V_{inner}-V_{outer}\\ =37\text{50kV}-\text{3000kV}\\ =7\text{50kV}\end{array}$

Hence, the potential difference is$75\text{0kV}$ .

When the two spheres are connected to each other through a wire, then the potential of both the spheres reaches a common value. The spheres are essentially capacitors connected in a parallel combination.

This common potential can be found using the formula,

$V_{common}=\frac{q+Q}{C_1+C_2}$

Where, q and Q are the charges on the inner and outer spheres respectively and $C_1$ and$C_2$are their respective capacitances.

The capacitances of the spheres are given by,

$C_1=4\pi {\epsilon }_{0}r$

And

$C_2=4\pi {\epsilon }_{0}R$

The common potential when the two spheres are connected by a wire is thus given by,

$\begin{array}{c}{V}_{common}=\frac{q+Q}{4\pi {\epsilon }_0(r+R)}\\ =(9\times {10}^9)\frac{(5.0\times {10}^{-6})+(15.0\times {10}^{-6})}{((3.0\times {10}^{-2})+(6.0\times {10}^{-2}))}\\ =2000\text{kV}\end{array}$

The charge on the inner sphere would be the product of its capacitance and this common-potential.
Thus,

$\begin{array}{c}{Q}^{\text{'}}{}_{inner}=C_1{V}_{common}\\ =\left(4\pi {\epsilon }_{0}r\right)V_{common}\\ =(4\times 3.14\times 8.85\times {10}^{-12}\times 3.0\times {10}^{-2})\times 2000000\\ =6.67\text{μC}\end{array}$

Hence, the charge on the inner sphere is $\text{6.67μC}$.

The charge on the outer sphere is the product of its capacitance and the common potential.
Thus,

$\begin{array}{c}{Q}_{outer}^{\text{'}}=C_2{V}_{common}\\ =\left(4\pi {\epsilon }_{0}R\right)V_{common}\\ =(4\times 3.14\times 8.85\times {10}^{-12}\times 6.0\times {10}^{-2})\times 2000000\\ =\text{13.33μC}\end{array}$

Hence, the charge on the outer sphere is $\text{13.33μC}$.