91Ó°ÊÓ

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

The charge,$q=5.0\text{μ°ä}=5.0×{\text{10}}^{\text{-6}}\text{C}$

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

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

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

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,

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

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

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,

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

The potential difference can be written as,

$\begin{array}{c}\mathrm{Δ}V=V_{inner}−V_{outer}\\ =37\text{50kV}−\text{3000kV}\\ =7\text{50kV}\end{array}$

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,

$C_1=4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r$

And

$C_2=4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R$

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

$\begin{array}{c}{V}_{common}=\frac{q+Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0(r+R)}\\ =(9×{10}^9)\frac{(5.0×{10}^{-6})+(15.0×{10}^{-6})}{((3.0×{10}^{-2})+(6.0×{10}^{-2}))}\\ =2000\text{kV}\end{array}$

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

$\begin{array}{c}{Q}^{\text{'}}{}_{inner}=C_1{V}_{common}\\ =\left(4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r\right)V_{common}\\ =(4×3.14×8.85×{10}^{-12}×3.0×{10}^{-2})×2000000\\ =6.67\text{μ°ä}\end{array}$

Hence, the charge on the inner sphere is $\text{6.67μ°ä}$.

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

$\begin{array}{c}{Q}_{outer}^{\text{'}}=C_2{V}_{common}\\ =\left(4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R\right)V_{common}\\ =(4×3.14×8.85×{10}^{−12}×6.0×{10}^{−2})×2000000\\ =\text{13.33μ°ä}\end{array}$

Hence, the charge on the outer sphere is $\text{13.33μ°ä}$.