10—Partial Differential Equations 261Multiply bysinmφ and integrate overφ. The trigonometric functions are orthogonal, so all that
survives is
0 = (Cm′ Rm+Dm′ R−m)π allm≥ 1
That gets rid of all the rest of the sine terms as promised:D′m= 0for allmbecauseCm′ = 0for all
m. Now repeat forcosmφ.
0 =C 0 +D 0 lnR (m= 0) and 0 = (CmRm+DmR−m)π (m >0)
All of theCm = 0form > 1 , so this says that the same applies toDm. Them= 1equation
determinesD 1 in terms ofC 1 and thenE 0.
D 1 =−C 1 R^2 = +E 0 R^2
Only theC 0 andD 0 terms are left, and that requires another boundary condition. When specifying
the problem initially, I didn’t say whether or not there is any charge on the wire. In such a case you
could naturally assume that it is zero, but you have to say so explicitly because that affects the final
result. Make it zero. That kills thelnrterm. The reason for that goes back to the interpretation of
this term. Its negative gradient is the electric field, and that would be− 1 /r, the field of a uniform line
charge. If I assume there isn’t one, thenD 0 = 0and so the same forC 0. Put this all together and
V(r,φ) =−E 0 rcosφ+E 0
R^2
r
cosφ (10.53)
The electric field that this presents is, from Eq. (9.24)E~=−∇V =E 0 (ˆrcosφ−φˆsinφ)−E 0 R^2
(
−ˆr
1
r^2
cosφ−φˆ
1
r^2
sinφ
)
=E 0 xˆ+E 0
R^2
r^2
(
rˆcosφ+φˆsinφ
)
As a check to see what this looks like, what is the electric field at the surface of the cylinder?
E~(R,φ) =E 0 (rˆcosφ−φˆsinφ)−E 0 R^2
(
−ˆr
1
R^2
cosφ−φˆ
1
R^2
sinφ
)
= 2E 0 ˆrcosφ
It’s perpendicular to the surface, as it should be. At the left and right,φ= 0,π, it is twice as large as
the uniform field alone would be at those points.