# The electric field of a point charge surrounded by a thick spherical shell

In the present post, I will consider the problem of calculating the electric field due to a point charge $$Q$$ surrounded by a conductor which has the form of a thick spherical shell. The conductor has zero net electric charge. The inner radius of the shell is $$R_1$$, and the outer radius is $$R_2$$. I will consider the case when the charge is at the center of symmetry of the spherical shell. I will discuss the induced charges, and also about what happens when the shell is connected to the ground, and what happens when the point charge $$Q$$ is not at the center.

To solve the problem, let’s take a spherical Gaussian surface concentric with the shell. The area of the Gaussian surface is $$A=4\pi r^2$$. For $$r>R_2$$, the Gauss Law $$E A=Q_{enc}/\epsilon_0$$ immediately gives the answer ($$Q_{enc}$$ is the charge enclosed inside the Gaussian surface):

$\vec{E}=\frac{Q}{4\pi\epsilon_0r^2}\hat{r}$

Inside the conductor, i.e., for $$R_1<r<R_2$$, we know that the electric field is zero (This is one of the properties of conductors).

In the hollow part of the system, i.e., for $$0<r<R_1$$, the Gauss Law leads to a similar result as for the field outside the shell:

$\vec{E}=\frac{Q}{4\pi\epsilon_0r^2}\hat{r}$

We summarize the results as follows:

$\vec{E}=\begin{cases} \frac{Q}{4\pi\epsilon_0r^2}\hat{r}, & \text{for r>R_2}\\ \vec{0}, & \text{for R_1<r<R_2}\\ \frac{Q}{4\pi\epsilon_0r^2}\hat{r}, & \text{for 0<r<R_1} \end{cases}$

The solution was easy, but, as we will see in what follows, we can learn some important things from this system.

## The induced charge on the shell

Let’s consider again the region inside the conductor, i.e., the region for $$R_1<r<R_2$$. We know that there is no net charge in the volume occupied by the conductor (This is another property of conductors. Any excedent of charge must reside on the surface of the conductor) and that the electric field is zero in this region. If we take a spherical Gaussian surface with radius $$R_1<r<R_2$$, the Gauss Law implies that the enclosed charge is zero. But the point charge $$Q$$ lies at the center. Therefore, there must be an equal amount of charge, with opposite sign, i.e., a charge $$-Q$$, uniformly (due to the spherical symmetry) distributed on the inner surface of the shell. This charge is induced on the surface of the conductor by the point charge $$Q$$, and has a surface charge density given by

$\sigma_1=-\frac{Q}{4\pi R_1^2}$

But the conductor has zero net charge and there is no net charge in the volume occupied by the conductor. Therefore, there must be an amount of charge $$Q$$ uniformly distributed (again, due to the spherical symmetry) on the outer surface of the shell, with surface charge density

$\sigma_2=\frac{Q}{4\pi R_2^2}$

The electric field outside the shell is due to the surface charge density $$\sigma_2$$ alone. Indeed, for $$r>R_1$$ the electric field of the point charge is canceled by the electric field due to the electric charge distributed on the inner surface of the shell.

If we displace the point charge $$Q$$ from the center (without touching the conductor), the electric field in the hollow region and the surface density $$\sigma_1$$ at the inner radius will change in such a way that the electric field for $$r>R_1$$ is still zero, but the surface charge density $$\sigma_2$$, and the field in the exterior region will remain the same. The surface density $$\sigma_1$$ in the inner surface of the shell can be calculated with help of the method of images. I will explain how to use the method of images to find the electric field, the electric potential, and the charge density on the inner surface of the shell in another post, but before that, I recommend reading the post The method of images in electrostatic explained as never before.

## The electric field after grounding the shell

The previous results have an additional consequence: If we connect the shell to the ground, then the electric field will be zero in the exterior region. I will explain how do we know that in the following section.

If the electric field in the exterior region is zero, then the Gauss Law, applied to a Gaussian surface surrounding the shell, implies that the total enclosed charge is zero. But the point charge $$Q$$ is at the center and an opposite charge $$-Q$$ is distributed on the inner face of the shell. Therefore, the charge at the outer face of the shell has to be zero.

Remember that before grounding the shell, the charge in the outer face was $$Q$$, and not zero. This means that an amount of charge $$-Q$$ was transferred from the ground to the outer face of the shell. Therefore, the conductor is not neutral anymore. It has acquired a net charge $$-Q$$. This charge $$-Q$$ spreads uniformly on the outer face and neutralizes the charge that was present there before the grounding.

What will happen if now we disconnect the shell from the ground? The Electric field inside the conductor is zero all the time. Therefore the situation inside the conductor, at the inner surface, and in the hollow region will remain unchanged. But at the outer surface, the net charge remains zero. Therefore, the Gauss Law, again implies that the electric field in the exterior region continues to be zero after the shell is disconnected from the ground. It is like saying that touching the ground turns off the external electric field.

## The electric field outside a grounded conductor is zero

How do we know that the electric field outside a conductor which is connected to the ground is zero? (Of course, we are assuming that there are no electric charges in the region outside the conductor). The answer is that experimental measurements show that this is so. But we can reach the same conclusion from the electromagnetic theory as we will see immediately.

After grounding the shell, it is easier to calculate first the electric potential $$\Phi$$ in the outer region, and after that, to take the gradient of the potential in order to find the electric field according to the relation $$\vec{E}=-\nabla\Phi$$.

The electric potential is a solution of the Laplace equation $$\nabla^2\Phi=0$$, and in addition, has to satisfy the boundary conditions at infinity and on the outer surface of the shell. The boundary condition at infinity is obviously $$\Phi|_{r=\infty}=0$$.

The physical meaning of grounding a conductor is that we put it at zero electric potential (See the post The physics behind grounding a conductor). Therefore, the boundary condition at the outer face of the shell is $$\Phi_{\text{on the shell}}=0$$.

Now it is easy to guess a solution, namely, $$\Phi=0$$. But there is a mathematical theorem that guarantees that there must be one, and only one solution to the Laplace equation that satisfies the appropriate boundary conditions. Therefore, $$\Phi=0$$ is not just a solution but is the only solution.

Thus, we conclude that both, the electric potential and the electric field, are zero outside of the shell.

## What will happen if the point charge $$Q$$ is displaced from the center?

If the charge $$Q$$ is not at the center, we still can use the Gauss Law to calculate the electric field outside the shell because it remains spherically symmetric there. However, the electric field in the hollow part has not spherical symmetry anymore, and therefore, the Gauss law is not useful to find the field there.

In the hollow region, we have to calculate the electric potential first, and then take its gradient. The electric potential in the hollow can be calculated by using the method of images. I will show how to solve this problem in another post.

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1. Amazing, mindblowing,… mand many more.. no words… You are god… Thanks…. hope you will give such beautiful solutions regularly….. ??❤️

2. Thanks a lot for such an amazing explanation.
I am looking at a problem where I have a charged conducting sphere (radius r1) of certain voltage. For calculating the potential at any point (say, r2>r1) outside the sphere, we take into consideration that the potential at infinity is zero. However, what is the potential at point r2, when the potential is not at infinity, but at a radius, say, r3 (r3>r2>r1) from the centre.

From Comsol, I find that the potential at r2 decreases when r3 in not at infinity, but as r3 is increased, the potential at r2 approaches the value given by Q/(4*pi*epsilon_0*r2). I am unable to find an expression which takes into consideration the dependance of r3 on the potential formula. Please can you help me with this?

3. Sorry for the repost. There was a typo in the problem statement which I have corrected in this post.

I am looking at a problem where I have a charged conducting sphere (radius r1) of certain voltage. For calculating the potential at any point (say, r2>r1) outside the sphere, we take into consideration that the potential at infinity is zero. However, what is the potential at point r2, when the potential is not zero at infinity, but at a radius, say, r3 (r3>r2>r1) from the centre.

From Comsol, I find that the potential at r2 decreases when r3 in not at infinity, but as r3 is increased, the potential at r2 approaches the value given by Q/(4*pi*epsilon_0*r2). I am unable to find an expression which takes into consideration the dependance of r3 on the potential formula. Please can you help me with this?

1. Thanks for your comment and for your question. The answer to your question is as follows: You have to solve the Laplace equation for R1 < r < R3 subject to boundary conditions. Spheric symmetry implies that the solution is just Potential=(A/r)+B, with A and B constants to be determined from the boundary conditions. One of the boundary conditions is that the potential is zero at R3. The other boundary condition is at R1 and there are two possibilities for this, namely, either you fix the potential or you specify the electric field. If you fix the potential V0 at R1 then you will find that A= R1*R3*V0/(R3-R1) and B=-A/R3. If you fix the electric field at R1 you use the relations Sigma=epsilon0*En, where Sigma=Q/(4pi*R1^2), and the normal component of the electric field is En=-(d/dr)Potential. In this case, you get B=-A/R3 as before, and A=Q/(4pi*epsilon0). This gives you the complete solution to your problem.

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