Solving an Electric Field in a Grounded Conducting Sphere

[Dustinsfl]

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
$$
u(r,\theta) = Er\cos\theta\left(1 - \frac{a^3}{r^3}\right).
$$

 

[jvicens]

Thank you for your question. It seems like you are trying to solve for the perturbation potential in a grounded conducting sphere that is placed in a uniform electric field in the $z$-direction. In order to do this, we can use a perturbation scheme for the total potential $u(r,\theta) = Er\cos\theta + u'$. This means that we can write the perturbation potential as $u'(r,\theta) = \sum_{n = 0}^{\infty}\frac{A_n}{r^{n + 1}}P_n(\cos\theta)$, where $A_n$ are coefficients that we need to solve for.

We are given the boundary condition $u(a,\theta) = 0$, which means that at the surface of the grounded sphere, the total potential is equal to the potential of the uniform electric field $Er\cos\theta$ plus the perturbation potential $u'$. Using this condition, we can solve for the coefficient $A_1$ and obtain the perturbation potential $u'(r,\theta) = -Ea^3\frac{\cos\theta}{r^2}$.

Finally, we can use this perturbation potential to solve for the total potential $u(r,\theta) = Er\cos\theta + u'$. This gives us a total potential of $u(r,\theta) = Er\cos\theta\left(1 - \frac{a^3}{r^3}\right)$. I hope this helps you understand how to incorporate the given electric field into your problem. Let me know if you have any further questions.