0
$\begingroup$

My textbook has the following question:

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $ \sigma/2\epsilon_0$ $\hat{\mathbf{n}}$, where $\hat{\mathbf{n}}$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.

I was puzzled how to solve this, so I looked up the hints given in the Answers section, and here's what it said:

Consider the conductor with the hole filled up. Then the field just outside is $ \sigma/\epsilon_0$ $\hat{\mathbf{n}}$ and is zero inside. View this field as a superposition of the field due to the filled up hole plus the field due to the rest of the charged conductor. Inside the conductor, these fields are equal and opposite. Outside they are equal both in magnitude and direction. Hence, the field due to the rest of the conductor is $ \sigma/2\epsilon_0$ $\hat{\mathbf{n}}$.

Is there any alternative way to prove the result that doesn't use a combination of superposition principle and Gauss's law? I find this proof to be a bit counterintuitive.

$\endgroup$

1 Answer 1

1
$\begingroup$

One of the standard ways we deal with holes in charged objects is to start with the complete object then superimpose an object with the opposite charge density where we want the hole. Then in the hole the opposite signs cancel and we get a zero net charge where the hole is.

In this case we start with a complete spherical shell with charge density $\sigma$ so the total charge is $4\pi r^2\sigma$, and the field just outside the surface of the sphere is the same as a point charge at the centre i.e.

$$\begin{align} E_1 &= \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2} \\ &= \frac{1}{4\pi\epsilon_0}\frac{4\pi r^2\sigma}{r^2} \\ &= +\frac{\sigma}{\epsilon_0} \end{align}$$

where the plus sign means the field is directed radially outwards.

Now take a small disk with a charge density $-\sigma$, and place it on the surface to create the hole where the $+\sigma$ and $-\sigma$ cancel to zero. The field due to the disk will be complicated, but right at the centre of the disk the field will be the same as the field from an infinite plane i.e.

$$ E_2 = -\frac{\sigma}{2\epsilon_0} $$

where the minus sign means the field is directed inwards towards the disk.

And when we superimpose the disk onto the sphere we just add the fields, so the field at the centre of the disk is:

$$ E_3 = E_1 + E_2 = +\frac{\sigma}{\epsilon_0} -\frac{\sigma}{2\epsilon_0} = +\frac{\sigma}{2\epsilon_0} $$

That is, the field has a magnitude of $\sigma/2\epsilon_0$ and is directed radially outwards.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.