3

b) Let us compute the charge distribution induced on the sphere. We call on the simple symmetry of the charge-image-charge system noting that this is identical to the charge distribution induced by the image charge on the outside of the sphere. Because we have already derived this expression in class—and in Jackson—up to a redeﬁnition of r and r , we simply have that σ(x) = − ∂ϕ 0 ∂x =

|x|=a

q 4πar 1 +

1− a2 r2

a2 r2 3/2

− 2 a cos θ r

,

where θ is the angle between r and x. c) Let us compute the force acting on the charge q. Using our results above, we see that F = q2 ar r. ˆ 4π 0 (a2 − r2 )2

d) We are to discuss how the work above is altered, if at all, if the sphere were kept at ﬁxed potential V or if there were total charge Q on its inner and outer surfaces. Neither of the two situations alters the work above because neither would eﬀect the interior of the sphere—only the outside. By Gauß’ law, we know that the electric ﬁeld inside a charged or ﬁxed-potential sphere is identically zero. Because electrostatics is linear, the ﬁeld inside the sphere will be the linear sum of that described above and that caused by the sphere—which is vanishing. Hence, there is no alteration. 2.7 Let us consider the space ½R3 satisfying Dirichlet boundary conditions on the plane ∂½R3 . a) We are to ﬁnd the appropriate Green’s function describing this system. In many ways, this problem is similar to that describing a point charge and an inﬁnite conducting plane. Speciﬁcally, we see that the Green’s function given by 1 1 G(x, x ) = − , 1/2 1/2 ((x1 − x1 )2 + (x2 − x2 )2 + (x3 − x3 )2 ) ((x1 − x1 )2 + (x2 − x2 )2 + (x3 + x3 )2 ) is of the correct form and satisﬁes the Dirichlet boundary conditions. In particular, we manifestly have that G(x, x ) = 0 ∀x |x3 = 0. Hence, this is our required Green’s function. b) Let us say that the potential on the plane x3 = 0 is speciﬁed to be ϕ = V inside a circle of radius a and vanish outside the circle. We are to ﬁnd an integral expression for the potential in cylindrical coordinates. In general, we know that the potential function for a problem with a Green’s function satisfying Dirichlet boundary conditions is given by ϕ(x) = 1 4π 0 ρ(x )G(x, x )d3 x −

½R3

1 4π

ϕ(x )

∂½R3

∂G(x, x ) da . ∂n

Because the space is empty of charges, the ﬁrst integral identically vanishes and we must only consider the boundary integral. Up to a sign which we will set a posteriori, we see that ∂G(x, x ) ∂x3 =− x3 =0

x3 − x3 ((x1 − x1 )2 + (x2 − x2 2x3 ((x1 − x1 )2 + (x2 − x2 )2 + x2 ) 3

3/2

)2

+ (x3 − x3 .

3/2 )2 )

+

x3 + x3 ((x1 − x1 )2 + (x2 − x2 )2 + (x3 + x3 )2 )

3/2 x3 =0

,

=−

Later, we will see that the n -direction should coincide with −x3 so that the potential at the surface is positive; this is identical to inserting the seemingly spurious minus sign in the above calculation.

4

JACOB LEWIS BOURJAILY

We can now compute the potential. Using our work above, it is clear that 1 ∂G(x, x ) ϕ(ρ, φ, z) = − ϕ(x ) da , 4π ∂½R3 ∂n = 1 4π

2π 0 0 a

V

2zρ dρ dφ ((ρ cos φ − ρ cos φ

2π 0 a 0 (ρ2 2

)2

+ (ρ sin φ − ρ sin φ )2 + z 2 ) ρ dρ dφ

3/2

3/2

,…