Question 6.14: A point charge Q is located at point (a, 0, b) between two s...

The image configuration is shown in Figure 6.24. Three image charges are necessary to satisfy the two conditions listed at the beginning of this section. From Figure 6.24(a), the potential at point P(x,y,z) is the superposition of the potentials at P due to the four point charges; that is

V=\frac{Q}{4\pi\varepsilon_{o}}\left[\frac{1}{r_{1}}-\frac{1}{r_{2}}+\frac{1}{r_{3}}-\frac{1}{r_{4}}\right]

where

r_{1}=[(x-a)^{2}+y^{2}+(z-b)^{2}]^{1/2}

r_{2}=[(x+a)^{2}+y^{2}+(z-b)^{2}]^{1/2}

r_{3}=[(x+a)^{2}+y^{2}+(z+b)^{2}]^{1/2}

r_{4}=[(x-a)^{2}+y^{2}+(z+b)^{2}]^{1/2}

From Figure 6.24(b), the net force on Q is

F=F_{1}+F_{2}+F_{3}

=-\frac{Q^{2}}{4\pi\varepsilon_{o}(2b)^{2}}a_{z}-\frac{Q^{2}}{4\pi\varepsilon_{o}(2a)^{2}}a_{x}+\frac{Q^{2}(2aa_{x}+2ba_{z})}{4\pi\varepsilon_{o}[(2a)^{2}+(2b)^{2}]^{3/2}}

=\frac{Q^{2}}{16\pi\varepsilon_{o}}\left\{\left[\frac{a}{(a^{2}+b^{2})^{3/2}}-\frac{1}{a^{2}}\right]a_{x}+\left[\frac{b}{(a^{2}+b^{2})^{3/2}}-\frac{1}{b^{2}}\right]a_{z}\right\}

The electric field due to this system can be determined similarly, and the charge induced on the planes can also be found.

In general, when the method of images is used for a system consisting of a point charge between two semi-infinite conducting planes inclined at an angle \phi (in degrees), the number of images is given by

N=\left(\frac{360^{\circ}}{\phi}-+\right)

because the charge and its images all lie on a circle. For example, when \phi=180^{\circ},  N=1 as in the case of Figure 6.22; for \phi=90^{\circ},  N=3 as in the case of Figure 6.23; and for \phi=60^{\circ}, we expect N=5 as shown in Figure 6.25.