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[YMP/EM-04006] Point Charge and an Infinite Plane

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Problem

A point charge is placed  at a distance \(d\) from an infinite grounded
conducting plane sheet. Assuming the standard image charge, find

  1. induced surface charge density. Show that the total induced charge is equal to−q
  2. find the potential due to the induced charge;
  3. using Coulomb's law, find the force between the charge and conducting sheet making use of the induced surface charge density on the plane.

(a) Surface charge density and induced charge:
Choose the axis the origin lies on the plane and charge is on the \((X\)- axis at position \((d,0,0)\).

The potential at a point \((x,y,z)\) due to the charge \(q\) and its image is given by \begin{equation} V(x,y,z) =\frac{1}{4\pi\epsilon_0}\left[ \frac{q}{\sqrt{(x-d)^2+y^2+z^2}} - \frac{q}{\sqrt{(x+d)^2+y^2+z^2}}\right], \qquad x\ge 0. \end{equation} The induced surface charge density is given by \begin{equation} \sigma = -\epsilon_0 \pp[V]{n} = -\epsilon_0 \pp[V]{x}. \end{equation} Differentiating the potential w.r.t. \(x\) gives

\begin{eqnarray}
 \pp[V]{x}
 &=& \frac{q}{4\pi\epsilon_0}\left[\frac{-(x-d)}{((x-d)^2+y^2+z^2))^{3/2}}
             - \frac{-(x+d)}{((x+d)^2+y^2+z^2)^{3/2}}  \right]_{x=0}\\
 &=& \frac{q}{4\pi\epsilon_0}\frac{2d}{(d^2+y^2+z^2)^{3/2}}            
\end{eqnarray}
Hence the surface charge density is given by
\begin{equation}\label{EQ05}
 \sigma = - \epsilon_0 \pp[V]{x} = \frac{-qd}{2\pi (d^2+y^2+z^2)^{3/2}}
\end{equation}
(b) Total induced charge
The total induced charge is given as double integral of the surface charge density over the \(YZ\) plane.
\begin{eqnarray}
 \text{Induced charge} &&b \\
 &=&  \iint_{-\infty}^{\infty} \frac{-qd}{2\pi (d^2+y^2+z^2)^{3/2}}            
 =  \int _0^\infty\int_0^\infty \rho d\rho d\phi  
\frac{-qd}{(\rho^2+d^2)^3 }              \\[3mm]
 &=& \int_0^\infty \frac{-q d\rho}{(\rho^2+d^2)^{3/2}}
 = qd \left.\frac{1}{(\rho^2+d^2)^{2}}\right|_0^\infty\\[3mm]
 &=& \frac{-q^2}{(4\pi\epsilon_0)}\Big( \frac{1}{4d^2}\Big).
\end{eqnarray}
(c) Total force of attraction due the induced charge induced on the sheet
By symmetry the force will have only \(x\) component. Consider an infinitesimal small area element \(dxdy\) on the plane located at \((0,y,z)\). z component of the  force due to this charge area element is \begin{equation} \Delta F_z = \frac{q}{4\pi\epsilon_0}  \frac{-q}{4\pi\epsilon_0} \frac{\sigma dydz}{(d^2+y^2+z^2)}\times \frac{d}{\sqrt{d^2+^2+z^2}} \end{equation} Therefore the total force is given by \begin{equation} F_x= \iint \frac{q}{4\pi\epsilon_0} \frac{\sigma dydz}{(d^2+y^2+z^2)}\times \frac{d}{\sqrt{d^2+^2+z^2}} \end{equation} Substituting for the surface charged  density \(\sigma\), from \EqRef{EQ05} we get
\begin{eqnarray}
 F_x&=&\frac{-q^2d^2}{8\pi\epsilon_0} \iint  \frac{dy\,  dz}{(d^2+^2+z^2)^3}
 =\frac{-q^2d^2}{8\pi\epsilon_0} \int_0^\infty \frac{ 2\pi\rho d\rho}  
{(d^2+\rho^2)2}\\
&=&-\frac{q^2d^2}{4\pi \epsilon_0}\int_0^\infty\frac{\rho
d\rho}{(d^2+\rho^2)^2 }
 =-\frac{q^2d^2}{4\pi\epsilon_0} \int_0^\infty
\frac{1}{2}\frac{dt}{(d^2+t)^2}\\
&=& \left.
-\frac{q^2d^2}{4\pi\epsilon_0}\frac{1}{4}\frac{1}{(d^2+t)^2}\right|_0^\infty
= -\frac{q^2}{4\pi\epsilon_0}\frac{1}{4d^2}
\end{eqnarray}
Comment
1) Why the force on the charge is not calculated using the expression for the electric field? 2) It appears to be  difficult to compute the electrostatic energy using the energy density \(\frac{\epsilon_0}{2}|\vec{E}|^2\)?  Try it now!

 

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