Spherical Integrals MJM PH 316 September 14, 2006
Griffiths pp. 85,6 does an integral for electric
potential for a spherical shell. There d
is a shell of radius R, having surface
charge density . We must integrate
over the shell to find the potential at
a distance s from the origin.
s p
Since
dVp = k dq/r,
we will add up all the dq's on the spherical shell.
(I'll use the symbol instead of Griffiths' curly r.)
The radius of the shell is R, and we want
to find the area of a circular strip whose
radius is R sin and whose width is R d.
The area of this strip is the width Rd times its length (circumference) 2R sin , or 2 R2 sin d.
The tiny element of charge on the strip is
dq = dA = 2 R2 sin d.
The distance curly r, or is given by
= ( R2 + s2 - 2Rs cos ) .
Then the integral for V at p is
Vp = dVp = kdq/ = k 2 R2 sin d / ( R2 + s2 - 2Rs cos )
The limits on are 0 to :
Vp = 2k R2 0 sin d/ ( R2 + s2 - 2Rs cos ).
Now we change variables to x = cos , and we note that dx = -sin d. Then the integral becomes
Vp = 2k R2 1-1 -dx/ ( R2 + s2 - 2 R s x).
Interchanging the limits gives us
Vp = 2k R2 -1+1 dx/ ( R2 + s2 - 2 R s x).
The integral is of the form dx/(a-bx), where a = R2 + s2 and b = 2Rs. If we let
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