Find the potential outside a charged metal sphere (charge Q, radius R) placed in an otherwise uniform electric field E₀ . Explain clearly where you are setting the zero of potential.

Consider the potential outside a charged metal sphere (charge Q, radius R) placed in an otherwise uniform electric field E₀.

Write the formula of the total potential at a distance r outside the metal sphere.

$\mathit{V}\mathbf{(}\mathit{r}\mathbf{,}\mathit{\theta }\mathbf{)}\mathbf{=}\mathbf{\sum }_{\mathbf{1}\mathbf{-}\mathbf{0}}^{\mathbf{\infty }}\left(A_1{r}^I+\frac{B_{I+1}}{r^{I+1}}\right){\mathit{P}}_{\mathbf{1}}\left(cos\theta )\right)$ …… (1)

Here, r is radius

The superposition of the potential of a point charge with charge Q centred at origin and the potential owing to induced charges is used to determine the electric potential outside of a charged metal sphere.

An external electric field causes a metal sphere to transfer its positive charge toward the northern surface and its negative charge toward the southern surface when placed in the field. As a result, the metal sphere is seen as a sphere with a radius R and a charge Q.

Determine the electric potential due to sphere having radius r is given as follows:

$V_1\left(r,\theta \right)=\frac{kQ}{r}$ …… (2)

When you are distant from the induced charges, treat potential as zero. When the generated charges are far from the electric potential,

$V=-E_{0}z+C$

Here, E₀ is the external field.

Since, the potential is zero in the equatorial plane (z = 0). Therefore, substitute 0 for V and 0 for z into equation (3).

0 = 0 + C

C = 0

The boundary conditions are as follows:

$$\begin{gathered} V=0(r=R) \\ V=-E_{0}z(r\square R) \\ V=-E_0\cos \theta \end{gathered}$$

Determine the general solution of electric potential is,

Substitute 0 for I into equation (1)

$\begin{array}{rcl}{A}_I{r}^I+\frac{B_I}{r^{I+1}}& =& 0\\ B_I& =& -A_I{r}^{2I+1}\end{array}$

Substitute -A_Ir^2I+1 for B_I into equation (1).

$V(r,\theta )=\sum \left(A_I\left(r^I-\frac{r^{2I+1}}{r^{I+1}}\right)\right)P_1\left(\cos \theta \right)$

The second term in the above expression tends to zero for $r\square R$. Substitute $-E_{0}r\cos \theta$for V and zero for second term in equation (1).

$-E_{0}r\cos =\sum _{I=0}^{\infty }\left(A_I{r}^I\right)p_1\left(\cos \theta \right)$

On comparing the both the sides of above equation,

$\begin{array}{rcl}{A}_1& =& -E_0\\ I& =& 0\\ P_1\left(\cos \theta \right)& =& \cos \theta \end{array}$

All the other terms are zero.

The potential due to induced charges is,

$V_2(r,\theta )=-E_0\left(r-\frac{R^3}{r^2}\right)\cos \theta$ …… (4)

Determine the total potential at a distance r outside the metal sphere is,

$V(r,\theta )=V_1(r,\theta )+V_2(r,\theta )$ …… (5)

Substitute (2) and (4) equation in (5).

role="math" localid="1658727025685" $\begin{array}{rcl}V(r,\theta )& =& \frac{kQ}{r}-E_0\left(r-\frac{R^3}{r^2}\right)\cos \theta \\ & =& -E_0\left(r-\frac{R^3}{r^2}\right)\cos \theta +\frac{1}{4\mathrm{\pi }{\in }_0}\frac{Q}{r}\end{array}$

Therefore, the value of total potential at a distance r outside the metal sphere is $\begin{array}{rcl}& & -E_0\left(r-\frac{R^3}{r^2}\right)\cos \theta +\frac{1}{4\mathrm{\pi }{\in }_0}\frac{Q}{r}\end{array}$.