Potential Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 389

Solution. Put a point electric charge of capacity ε at M 0 ( x 0 , y 0 , z 0 ) inside Ω . Locate

the symmetric point M 1 of M 0 with respect to the spherical surface ∂Ω ( r = R ) along

the ray OM 0 and outside the spherical surface such that

R 2

r 0 r 1 = r OM 0 r OM 1 = . (7.111)

Here r 0 = r OM 1 are the distances of M 0 and M 1 from the center O of

the sphere, respectively. (Fig. 7.2)
For any point P on

r OM 0 and r 1 =

∂Ω ,the Δ OPM 0 and the Δ OPM 1 have a common angle

∠ POM 1 . Its sides are also, by Eq. (7.111), proportional to each other. Therefore,

Δ OPM 0 U Δ OM 1 P . Thus

R
r 0 r M 0 P

r M 1 P =

or

1
r M 0 P −

R
r 0

1
r M 1 P =

0 .

Therefore, we have

1 R 1

r M 1 P =

r M 0 P −

0 . (7.112)

4 π 4 π r 0

This holds for all P on ∂Ω , and thus the induced electric field by a point electric

charge of capacity r 0 ε

at M 1 can completely neutralize that due to the charge at M 0

∈ Ω

on ∂Ω . The electric potential generated by such a charge is, for M ,

R 1

r M 1 M .

g ( M , M 0 )=

4 π r 0

Fig. 7.2 Symmetric point M 1 of M 0