Question

Find the potential outside an infinitely long metal pipe, of radius R, placed at right angles to an otherwise uniform electric field ${\mathit{E}}_{\mathbf{0}}$. Find the surface charge induced on the pipe. [Use your result from Prob. 3.24.]

Short answer

Answer

The potential outside an infinitely long metal pipe is ${\mathit{E}}_{\mathbf{0}}\mathit{s}\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{R}}^{\mathbf{2}}}{{\mathbf{S}}^{\mathbf{2}}}\mathbf{]}\mathit{c}\mathit{o}\mathit{s}\mathit{\phi }$.

The surface charge induced on pipe is $\mathbf{2}\mathit{E}{\mathit{\epsilon }}_{\mathbf{0}}\mathit{c}\mathit{o}\mathit{s}\mathit{\phi }$.

Step-by-step solution

Define functions

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.

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

The figure shows the electric filed lines.

Determine potential

As, $x=s\cos \phi$

Then, write the expression for the potential.

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

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

$V\left(s,\phi \right)=a_0+b_{0}Ins+\sum _{k=1}^{\infty }\left[\begin{array}{c}{s}^k\left(a_k\cos K\phi +b_k\sin K\phi \right)\\ +s^{-k}\left(c_k\cos K\phi +d_k\sin K\phi \right)\end{array}\right]$ …… (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)

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

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

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

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,

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

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

Determine induced surface charge density

Write the expression for induced surface charge density.

$\sigma =-{\epsilon }_0{\left(\frac{\partial V}{\partial s}\right)}_{s=R}$ …… (5)

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

role="math" localid="1655736461654" $$\begin{gathered} {\sigma }_0=-{\epsilon }_{}\frac{\partial }{\partial s}\left[-E_{0}s\left[1-\frac{R^2}{s^2}\right]\cos \phi \right] \\ =E_0{\epsilon }_0\cos \phi \frac{\partial }{\partial s}\left[1-\frac{R^2}{s^2}\right] \\ =E_0{\epsilon }_0\cos \phi {\left[1+\frac{R^2}{s^2}\right]}_{s=R} \\ =2E_0{\epsilon }_0\cos \phi \end{gathered}$$

Therefore, the surface charge induced on pipe is $2E_0{\epsilon }_0\cos \phi$.