91Ó°ÊÓ

Let’s consider that the direction of electric filed is in x-direction. The metal pipe placed in yz plane, because electric filed and metal pipe should be in right angles.

Assume V = 0 in yz plane.

Find out the potential due to pipe at distance S.

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{V}\mathbf{=}\mathbf{0} \\ WhenS=R \\ \mathit{V}\mathbf{=}\mathbf{-}{\mathit{E}}_{\mathbf{0}}\mathit{x} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{T}\mathbf{-}{\mathit{E}}_{\mathbf{s}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{φ}}fors>>>R \end{gathered}$$ …… (1)

The figure shows the electric filed lines.

As, $x=s\cos \mathrm{φ}$

Then, write the expression for the potential.

$$\begin{gathered} V=-E_{0}s\cos \mathrm{φ} \\ \frac{V}{\cos \mathrm{φ}}=-E_{0}s \end{gathered}$$ …… (2)

Write the expression for the potential in cylindrical co-ordinates.

$V\left(s,\mathrm{φ}\right)=a_0+b_{0}Ins+\underset{k=1}{\overset{∞}{∑}}\left[\begin{array}{c}{s}^k\left(a_k\cos K\mathrm{φ}+b_k\sin K\mathrm{φ}\right)\\ +s^{-k}\left(c_k\cos K\mathrm{φ}+d_k\sin K\mathrm{φ}\right)\end{array}\right]$ …… (3)

From the above two equations,

$$\begin{gathered} K=1 \\ {a}_0=b_0 \\ =c_0 \\ =d_{} \\ =0 \end{gathered}$$

$$\begin{gathered} a_k=c_k \\ =0forKotherthan1 \end{gathered}$$

Now, substitute 1 for K in equation (3)

$$\begin{gathered} V\left(s,\mathrm{φ}\right)=\left(a·s+\frac{C_1}{S}\right)\cos \mathrm{φ} \\ \frac{V\left(s,\mathrm{φ}\right)}{\cos \mathrm{φ}}=a_{1}s+\frac{C_1}{S} \\ -E_{0}s=a_{1}s+\frac{C_1}{S} \end{gathered}$$

Now, substitute 0 for V in $V\left(s,\mathrm{φ}\right)=\left(a_{1}s+\frac{C_1}{S}\right)\cos \mathrm{φ}$and substitute R for S.

$0=\left(a_{1}R+\frac{C_1}{R}\right)\cos \mathrm{φ}$

Solve as further,

$$\begin{gathered} a_{1}R+\frac{C_1}{R}=0 \\ \frac{C_1}{R}=-a_{1}R \\ {C}_1=-E_0 \end{gathered}$$

Then, the potential is,

$$\begin{gathered} V\left(s,\mathrm{φ}\right)=\left(-E_{0}s+\frac{E_0{R}^2}{s}\right)\cos \mathrm{φ} \\ =-E_{0}s\left[1-\frac{R^2}{s^2}\right]\cos \mathrm{φ} \end{gathered}$$ …… (4)

Hence, the potential outside an infinitely long metal pipe is $-E_{0}s\left[1-\frac{R^2}{s^2}\right]\cos \mathrm{φ}$.

Write the expression for induced surface charge density.

$\mathrm{σ}=-{\mathrm{Î\mu }}_0{\left(\frac{∂V}{∂s}\right)}_{s=R}$ …… (5)

Substitute $-E_{0}s\left[1-\frac{R^2}{S^2}\right]\cos \mathrm{φ}$for V in equation (5).

role="math" localid="1655736461654" $$\begin{gathered} {\mathrm{σ}}_0=-{\mathrm{Î\mu }}_{}\frac{∂}{∂s}\left[-E_{0}s\left[1-\frac{R^2}{s^2}\right]\cos \mathrm{φ}\right] \\ =E_0{\mathrm{Î\mu }}_0\cos \mathrm{φ}\frac{∂}{∂s}\left[1-\frac{R^2}{s^2}\right] \\ =E_0{\mathrm{Î\mu }}_0\cos \mathrm{φ}{\left[1+\frac{R^2}{s^2}\right]}_{s=R} \\ =2E_0{\mathrm{Î\mu }}_0\cos \mathrm{φ} \end{gathered}$$

Therefore, the surface charge induced on pipe is $2E_0{\mathrm{Î\mu }}_0\cos \mathrm{φ}$.