- #1

- 100

- 0

This question is from Griffiths', Intro. to Electrodynamics, Example 3.2, to wit:

"A point charge q is situated a distance 'a' from the center of a grounded conducting sphere of radius 'R' (Fig. 3.12). Find the potential outside the sphere."

Note: a > R. Griffiths' solution does not provide the explanation on how the value of 'b' (the distance of the image charge, q', which is inside the sphere b < R) was obtained. The potential I started with is:

V(r) = k(q'/r_1' + q/r_2),

where r'_1 and r_2 are found using the law of cosines from their respective charges to the observation point outside the sphere and k := 1/4pi*epsilon_o

The image charge, q', can be found using the fact that the sphere is grounded:

q' = - (R - a)/( b - R)

I tried using V = 0 at r = infinity, but I get a wrong answer:

b = 2*R - a, (2 times R)

whereas, the correct result is:

b = R^2/a (R-squared).

Thanks in advance for any help you may give me,

-LD