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Conducting sphere in uniform electric field

Consider a conducting sphere placed in otherwise uniform electric field. The field near the sphere gets modified by the charge induced on the sphere. We want to solve the problem by method of images. For this, consider two charges $-Q$ and $Q$ placed at $Z \hat k$ and $-Z \hat k$ respectively. The electric field of these charges at $\vec r$ is

\begin{displaymath}\vec E(\vec r) = - \frac{Q \vec r - Z \hat k}{\vert\vec r - Z...
...}
+ \frac{Q \vec r + Z \hat k}{\vert\vec r + Z \hat k\vert^3}.\end{displaymath}

For $Z » \vert\vec r\vert$, $\vec
E(\vec r) = \frac{2Q}{Z^2} \hat k$. Thus, in the limit of $Z$ going to infinity with $\frac{2Q}{R^2}=E_0$ we have uniform electric field $E_0$ along z-axis. We compute the effect of the presence of a conducting sphere of radius $a$ in uniform electric field by computing the electric field of a system consisting of a conducting sphere of radius $a$ with it's center at the origin and two charges $-Q$ and $Q$ placed at $Z \hat k$ and $-Z \hat k$ respectively and then taking the limit of $Z \rightarrow \infty$ with $\frac{2Q}{Z^2}=E_0$.

Using method of images, the two image charges are at $a^2/Z$ and $-a^2/Z$ and their strengths are $aQ/Z$ and $-aQ/Z$ respectively. Note that we don't need a charge at the origin as the total image charge is zero. Note that this configuration of the image charges forms a dipole and in the limit of $Z \rightarrow \infty$ the dipole moment is $\vec d = \mathrm {Lim}_{Z \rightarrow \infty} \frac{aQ}{Z} \frac{2 a^2}
{Z} \hat k = a^3 E_0 \hat k$. The electric field of the system is then the field due to this dipole, which is placed at the origin, added to the uniform electric field. We therefore have

$\displaystyle \vec E(\vec r) = E_0 \hat k + a^3 E_0 \left ( \frac{3 z \vec r}{r^5} - \frac{\hat k}
{r^3} \right )$     (50)

The corresponding electric potential is
$\displaystyle \phi(\vec r) = - E_0 z + a^3 E_0 \frac{\hat k \cdot \vec r}{r^3}$     (51)

As expected the potential at the surface of the sphere, $\vert\vec r\vert = a$ is constant ( zero in this case ). The surface charge density at the surface can be computed from the normal component of the electric field at the surface of the sphere.


next up previous
Next: Planar interface of two Up: Solutions of Poisson's equation Previous: A point charge in
S.C.Phatak 2007-02-20