# Potential from point charge at distance $l$ from conducting sphere

#### migueldbg

Problem Statement
A point charge $q$ is located at a distance $l$ from the center of a conducting grounded sphere with radius $a < l$. Find the potential on the points outside the sphere and determine the charge distribution on the surface of the sphere (Hint: use a Legendre polynomial expansion with the origin at the center of the sphere, choosing the polar axis to be along the direction of the point charge.)
Relevant Equations
$\phi(\mathbf{r}) = \frac{q}{|\mathbf{r} - \mathbf{r'}|}$
After looking around a bit, I found that, considering the polar axis to be along the direction of the point charge as suggested by the exercise, the following Legendre polynomial expansion is true:
$$$$\frac{1}{|\mathbf{r} - \mathbf{r'}|} = \sum_{n=0}^\infty \frac{r^{n}_{<}}{r^{n+1}_{>}}$$P_n(\cos{\theta}).$$
where $P_l$ are the Legendre polynomial and $r_>$ is the larger distance between $\mathbf{r}$ (observation point) and $\mathbf{r'}$ (charge position), while $r_<$ is the smaller distance between the two. In that case, the potential due to a point charge, for the case where we are considering an observation point at a distance $r < l$, would be:
$$$$\phi(r, \theta) = q \sum_{n=0}^\infty \frac{r^n}{l^{n+1}}P_n(\cos{\theta})$$.$$
Since the conducting sphere is grounded, we have that its surface will be an equipotential surface where $\phi = 0$. This means that, at r = a, where $a < l$, we would have:
$$$$\phi(a, \theta) = q \sum_{n=0}^\infty \frac{a^n}{l^{n+1}}P_n(\cos{\theta}) = 0$$$$
for every $\theta$. This is where I'm stuck. I don't see how Equation 3 could be satisfied. I'm aware that this problem can be and has been solved using the method of image charges, but I was wondering if anyone had any insight as to how to find the solution using the hint given by the exercise. Thanks in advance!

Related Advanced Physics Homework News on Phys.org

#### PeroK

Homework Helper
Gold Member
2018 Award
Problem Statement: A point charge $q$ is located at a distance $l$ from the center of a conducting grounded sphere with radius $a < l$. Find the potential on the points outside the sphere and determine the charge distribution on the surface of the sphere (Hint: use a Legendre polynomial expansion with the origin at the center of the sphere, choosing the polar axis to be along the direction of the point charge.)
Relevant Equations: $\phi(\mathbf{r}) = \frac{q}{|\mathbf{r} - \mathbf{r'}|}$

After looking around a bit, I found that, considering the polar axis to be along the direction of the point charge as suggested by the exercise, the following Legendre polynomial expansion is true:
$$$$\frac{1}{|\mathbf{r} - \mathbf{r'}|} = \sum_{n=0}^\infty \frac{r^{n}_{<}}{r^{n+1}_{>}}$$P_n(\cos{\theta}).$$
where $P_l$ are the Legendre polynomial and $r_>$ is the larger distance between $\mathbf{r}$ (observation point) and $\mathbf{r'}$ (charge position), while $r_<$ is the smaller distance between the two. In that case, the potential due to a point charge, for the case where we are considering an observation point at a distance $r < l$, would be:
$$$$\phi(r, \theta) = q \sum_{n=0}^\infty \frac{r^n}{l^{n+1}}P_n(\cos{\theta})$$.$$
Since the conducting sphere is grounded, we have that its surface will be an equipotential surface where $\phi = 0$. This means that, at r = a, where $a < l$, we would have:
$$$$\phi(a, \theta) = q \sum_{n=0}^\infty \frac{a^n}{l^{n+1}}P_n(\cos{\theta}) = 0$$$$
for every $\theta$. This is where I'm stuck. I don't see how Equation 3 could be satisfied. I'm aware that this problem can be and has been solved using the method of image charges, but I was wondering if anyone had any insight as to how to find the solution using the hint given by the exercise. Thanks in advance!
You have a point charge $q$ outside the sphere. At each point on the sphere, there will be a potential due to this charge. The distance to the charge depends on $\theta$. So, you can calculate the potential due to the charge.

The unknown charge distribution on the sphere must cancel this potential. If you can express the potential due to the point charge (on the surface of the sphere) in terms of Legendre polynomials, then you have the potential of the charge distribution in the same form.

This is what you have with equation (3). But, instead of setting this equation to zero, this must be the (negative) potential of your charge distribution on the surface of the sphere.

#### migueldbg

You have a point charge $q$ outside the sphere. At each point on the sphere, there will be a potential due to this charge. The distance to the charge depends on $\theta$. So, you can calculate the potential due to the charge.

The unknown charge distribution on the sphere must cancel this potential. If you can express the potential due to the point charge (on the surface of the sphere) in terms of Legendre polynomials, then you have the potential of the charge distribution in the same form.

This is what you have with equation (3). But, instead of setting this equation to zero, this must be the (negative) potential of your charge distribution on the surface of the sphere.
I see. So the potential due to the induced charge $\phi_\sigma$ at $r = a$ must be equal to the negative of the potential due to the point charge at $r = a$, such that the sum of the two equals zero. That means then that we would have
$$$$\phi_\sigma(a, \theta) = -q \sum_{n=0}^\infty \frac{a^n}{l^{n+1}}P_n(\cos{\theta})$$$$
From this equation, can we then determine what would be the potential $\phi_\sigma(r, \theta)$? I thought about making use of the solution to the Laplace equation in spherical coordinates, but they should not hold at $r = a$, since at that point there would be a charge density. I actually tried to do that, but it resulted in a total potential that would be zero at points other than the surface of the sphere, which shouldn't be the case.

#### PeroK

Homework Helper
Gold Member
2018 Award
I see. So the potential due to the induced charge $\phi_\sigma$ at $r = a$ must be equal to the negative of the potential due to the point charge at $r = a$, such that the sum of the two equals zero. That means then that we would have
$$$$\phi_\sigma(a, \theta) = -q \sum_{n=0}^\infty \frac{a^n}{l^{n+1}}P_n(\cos{\theta})$$$$
From this equation, can we then determine what would be the potential $\phi_\sigma(r, \theta)$? I thought about making use of the solution to the Laplace equation in spherical coordinates, but they should not hold at $r = a$, since at that point there would be a charge density. I actually tried to do that, but it resulted in a total potential that would be zero at points other than the surface of the sphere, which shouldn't be the case.
You can use the result of the separation of variables to express the potential iun terms of Legendre polynomials. Then, equate that with what you have at $r = a$.

This applies for the potential inside and outside the sphere.

### Want to reply to this thread?

"Potential from point charge at distance $l$ from conducting sphere"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving