- #1
Dustinsfl
- 2,281
- 5
I feel like I need to incorporate ##\mathbf{E} = E\hat{\mathbf{z}}## but I don't know what to do with it.
A grounded conducting sphere of radius ##a## is placed in an (effectively) infinite uniform electric field ##\mathbf{E} = E\hat{\mathbf{z}}##. The potential for a uniform electric field in the ##z##-direction is given by ##Er\cos\theta##. The boundary condition at the surface of the grounded sphere is that
$$
u(a,\theta) = 0.
$$
Use a perturbation scheme for the total potential
$$
u(r,\theta) = Er\cos\theta + u'
$$
to solve for the perturbation potential ##u'##.
Using our giving condition, we have ##u(a,\theta) = Ea\cos\theta + u'(a,\theta) = 0##. That is,
$$
u'(a,\theta) = -Ea\cos\theta = \sum_{n = 0}^{\infty}\frac{A_n}{a^{n + 1}}P_n(\cos\theta).
$$
From our previous work, we know that we only need the ##n = 1## term. Therefore, ##-Ea\cos\theta = \frac{A_1}{a^2} P_1(\cos\theta)\Rightarrow -Ea^3 = A_1##. We have that the perturbation potential is
$$
u'(r,\theta) = -Ea^3\frac{\cos\theta}{r^2}
$$
and that the total potential is
$$
u(r,\theta) = Er\cos\theta\left(1 - \frac{a^3}{r^3}\right).
$$
##P_n## is the nth Legendre polynomial.
A grounded conducting sphere of radius ##a## is placed in an (effectively) infinite uniform electric field ##\mathbf{E} = E\hat{\mathbf{z}}##. The potential for a uniform electric field in the ##z##-direction is given by ##Er\cos\theta##. The boundary condition at the surface of the grounded sphere is that
$$
u(a,\theta) = 0.
$$
Use a perturbation scheme for the total potential
$$
u(r,\theta) = Er\cos\theta + u'
$$
to solve for the perturbation potential ##u'##.
Using our giving condition, we have ##u(a,\theta) = Ea\cos\theta + u'(a,\theta) = 0##. That is,
$$
u'(a,\theta) = -Ea\cos\theta = \sum_{n = 0}^{\infty}\frac{A_n}{a^{n + 1}}P_n(\cos\theta).
$$
From our previous work, we know that we only need the ##n = 1## term. Therefore, ##-Ea\cos\theta = \frac{A_1}{a^2} P_1(\cos\theta)\Rightarrow -Ea^3 = A_1##. We have that the perturbation potential is
$$
u'(r,\theta) = -Ea^3\frac{\cos\theta}{r^2}
$$
and that the total potential is
$$
u(r,\theta) = Er\cos\theta\left(1 - \frac{a^3}{r^3}\right).
$$
##P_n## is the nth Legendre polynomial.