Question

The half-speed filled with uniform isotronic dielectric with permittivity $$\epsilon$$ has the conducting boundary plane. Inside the dielectric, at a distance $$l$$ from this plane, there is a small metal ball possessing a charge $$q$$. Find the surface density of the bound charges at the boundary plane as a function of distance $$r$$ from the ball.

Answer 1

To calculate the electric field, first we note that an image charge will be needed to ensure that the electric field on the metal boundary is normal to the surface.

The image charge must have magnitude $$-\dfrac {q}{\epsilon}$$

so that the tangential component of the electric field may vanish. Now,

$$E_{n} = \dfrac {1}{4\pi \epsilon_{0}}\left (\dfrac {q}{\epsilon r^{2}}\right ) 2\cos \theta = \dfrac {ql}{2\pi \epsilon_{0} \epsilon r^{3}}$$

Then $$P_{n} = D_{n} - \epsilon_{0} E_{n} = \dfrac {(\epsilon - 1)ql}{2\pi \epsilon r^{3}} = \sigma'$$

This is the density of bound charge on the surface.