Question

A uniform line charge $\mathit{\lambda }$is placed on an infinite straight wire, a distanced above a grounded conducting plane. (Let's say the wire runs parallel to the x-axis and directly above it, and the conducting plane is the $\mathit{x}\mathit{y}$plane.)

1. Find the potential in the region above the plane. [Hint: Refer to Prob. 2.52.]
2. Find the charge density $\mathit{\sigma }$ induced on the conducting plane.

Short answer

Answer

1. The total potential $\mathit{p}$is $\mathit{V}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\left(Z+d\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}{{\left(Z-d\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathbf{\right)}$.

2. The charge density $\mathit{\sigma }$induced on conducting plate is $\frac{\mathbf{-}\mathbf{\lambda d}}{\mathbf{\pi }\mathbf{(}{\mathbf{d}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{)}}$.

Step-by-step solution

Define functions

Write the expression for the potential for the infinite straight wire.

$\mathit{V}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\left(\frac{S}{d}\right)$ …… (1)

Here, $\mathit{\lambda }$is the linear charge density, $\mathit{d}$is the distance and ${\mathit{\epsilon }}_{\mathbf{0}}$is the permittivity for the free space.

Determine the potential in the region above the plane

a)

The potential at distance $\mathit{s}$from an infinitely straight long wire that carries uniform the line charge density is given as,

$\mathit{V}\mathbf{=}\frac{\mathbf{-}\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{\mathbf{s}}{\mathbf{d}}\mathbf{\right)}$ …… (2)

Now let’s consider that, ${\mathit{S}}_{\mathbf{1}}$is the distance from $\mathbf{+}\mathit{\lambda }$to point $\mathit{P}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}$where the potential is calculated and ${\mathit{S}}_{\mathbf{-}}$be the distance from $\mathbf{-}\mathit{\lambda }$to point $\mathit{P}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}$where potential is calculated.

Write the electric potential at point $\mathit{P}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}$as shown in above figure,

$$\begin{gathered} {\mathit{s}}_{\mathbf{+}}\mathbf{=}\sqrt{{\mathbf{(}\mathbf{Z}\mathbf{-}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}} \\ {\mathit{s}}_{\mathbf{-}}\mathbf{=}\sqrt{{\mathbf{(}\mathbf{Z}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}} \end{gathered}$$

Write the potential at $\mathit{P}$due to $\mathbf{+}\mathit{\lambda }$.

$\mathit{V}\mathbf{=}\frac{\mathbf{+}\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{s}}_{\mathbf{-}}}{\mathbf{d}}\mathbf{\right)}$ …… (3)

Write the potential at $\mathit{P}$due to $\mathbf{-}\mathit{\lambda }$.

${\mathit{V}}_{\mathbf{-}}\mathbf{=}\frac{\mathbf{+}\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{s}}_{\mathbf{-}}}{\mathbf{d}}\mathbf{\right)}$ …… (4)

Write the expression for the total potential at point $\mathit{p}$.

$\mathit{V}\mathbf{=}\mathit{V}\mathbf{+}\mathit{V}$ …… (5)

Substitute $\frac{\mathbf{-}\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{s}}_{\mathbf{+}}}{\mathbf{d}}\mathbf{\right)}$for ${\mathit{V}}_{\mathbf{1}}$and $\frac{\mathbf{+}\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{S}}_{\mathbf{-}}}{\mathbf{d}}\mathbf{\right)}$for ${\mathit{V}}_{\mathbf{-}}$in equation (5).

$$\begin{gathered} \mathit{V}\mathbf{=}\frac{\mathbf{-}\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{S}}_{\mathbf{-}}}{\mathbf{d}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{S}}_{\mathbf{-}}}{\mathbf{d}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{[}\mathbf{In}\mathbf{\left(}\frac{{\mathbf{S}}_{\mathbf{-}}}{\mathbf{d}}\mathbf{\right)}\mathbf{-}\mathbf{In}\mathbf{\left(}\frac{{\mathbf{S}}_{\mathbf{+}}}{\mathbf{d}}\mathbf{\right)}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{\left({\displaystyle \frac{S_{-}}{d}}\right)}{{\displaystyle \frac{S_{+}}{d}}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{S}}_{\mathbf{-}}}{{\mathbf{S}}_{\mathbf{+}}}\mathbf{\right)} \end{gathered}$$

Multiply and divide the above equation by 2.

$$\begin{gathered} \mathit{V}\mathbf{=}\frac{\mathbf{-}\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{S}}_{\mathbf{-}}}{{\mathbf{d}}_{\mathbf{+}}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}{\mathbf{\left(}\frac{{\mathbf{S}}_{\mathbf{-}}}{{\mathbf{S}}_{\mathbf{+}}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{{\mathbf{S}}_{\mathbf{-}}}^{\mathbf{2}}}{{{\mathbf{S}}_{\mathbf{+}}}^{\mathbf{2}}}\mathbf{\right)} \end{gathered}$$

Substitute $\sqrt{{\left(z+d\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}$for ${\mathit{S}}_{\mathbf{+}}$and $\sqrt{{\mathbf{(}\mathbf{Z}\mathbf{-}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}\mathbf{y}}$for ${\mathit{S}}_{\mathbf{-}}$in above equation.

$$\begin{gathered} \mathit{V}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{\left(}\sqrt{{\mathbf{(}\mathbf{Z}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{\left(}\sqrt{{\mathbf{(}\mathbf{Z}\mathbf{-}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathbf{\right)}}^{\mathbf{2}}}\mathbf{\right)} \\ \mathit{V}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{(}\mathbf{Z}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{(}\mathbf{Z}\mathbf{-}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathbf{\right)} \end{gathered}$$

Therefore, the total potential $\mathit{P}$is $\mathit{V}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{(}\mathbf{Z}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{(}\mathbf{Z}\mathbf{-}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathbf{\right)}$.

Determine the charge density induced on the conducting plane

b)

Write the expression for the induced charge density on the conducting plane.

$\mathit{\sigma }\mathbf{=}\mathbf{-}{\mathit{\epsilon }}_{\mathbf{0}}\frac{\mathbf{\partial }\mathbf{V}}{\mathbf{\partial }\mathbf{n}}$

But, $\frac{\mathbf{\partial }\mathbf{V}}{\mathbf{\partial }\mathbf{n}}\mathbf{=}\frac{\mathbf{\partial }\mathbf{V}}{\mathbf{\partial }\mathbf{z}}$evaluated as $\mathit{z}\mathbf{=}\mathbf{0}$.

Thus the charge density $\mathit{\sigma }$is given as,

$\mathit{\sigma }\mathbf{=}\mathbf{-}{\mathit{\epsilon }}_{\mathbf{0}}\frac{\mathbf{\partial }\mathbf{V}}{\mathbf{\partial }\mathbf{z}}$ …… (6)

Substitute $\frac{\mathbf{\lambda }}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{(}\mathbf{Z}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{(}\mathbf{Z}\mathbf{-}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathbf{\right)}$for $\mathit{V}$in above equation.

$$\begin{gathered} \mathit{\sigma }\mathbf{=}\mathbf{-}{\mathit{\epsilon }}_{\mathbf{0}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{z}}\mathbf{\left[}\frac{\mathbf{\lambda }}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{In}\mathbf{\left(}\frac{{\mathbf{(}\mathbf{Z}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{(}\mathbf{Z}\mathbf{-}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathbf{\right)}\mathbf{\right]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{\lambda }}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{z}}\mathbf{[}\mathbf{In}\mathbf{(}{\mathbf{(}\mathbf{z}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{)}\mathbf{-}\mathbf{In}\mathbf{(}{\mathbf{(}\mathbf{Z}\mathbf{-}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{)}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{\lambda }}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\{}\frac{\mathbf{1}}{{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{Z}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}}\mathbf{2}\mathbf{(}\mathbf{z}\mathbf{+}\mathbf{d}\mathbf{)}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{Z}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}}\mathbf{2}\mathbf{(}\mathbf{z}\mathbf{-}\mathbf{d}\mathbf{)}\mathbf{\}}\mathit{x}\mathit{y} \end{gathered}$$

Given that, the conducting plane is the role="math" localid="1657082025144" $\mathit{x}\mathit{y}$planes means that plane lies at $\mathit{Z}\mathbf{=}\mathbf{0}$. Thus, the term $\mathit{Z}\mathbf{+}\mathit{d}\mathbf{=}\mathit{d}$.

$$\begin{gathered} \mathit{\sigma }\mathbf{=}\frac{\mathbf{-}\mathbf{2}\mathbf{\lambda }}{\mathbf{4}\mathbf{\pi }}\mathbf{(}\frac{\mathbf{d}}{{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{d}}^{\mathbf{2}}}\mathbf{+}\frac{(-d)}{{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{d}}^{\mathbf{2}}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}\mathbf{2}\mathbf{\lambda }}{\mathbf{4}\mathbf{\pi }}\mathbf{(}\frac{\mathbf{d}}{{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{d}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{d}}{{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{d}}^{\mathbf{2}}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}\mathbf{\lambda }\mathbf{4}\mathbf{d}}{\mathbf{4}\mathbf{\pi }\mathbf{(}{\mathbf{d}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{)}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}\mathbf{\lambda d}}{\mathbf{\pi }\mathbf{(}{\mathbf{d}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{)}} \end{gathered}$$

Therefore, the charge density role="math" localid="1657082132658" $\mathit{\sigma }$induced on conducting plate is $\frac{\mathbf{-}\mathbf{\lambda d}}{\mathbf{\pi }\mathbf{(}{\mathbf{d}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{)}}$.