A Non-Conducting Sphere with a Hole

[jg370]

I have an thin, hollow, non-conducting sphere with charge[tex]\sigma[/tex]. The magnitude of the electric field at the surface is [tex] \sigma[/tex]/[tex]\epsilon[/tex].

I am asked to show that if a tiny hole is made through the sphere, then the magnitude of the electric field in the hole is [tex] \sigma[/tex]/[tex]2\epsilon[/tex].

Here is my trial solution:

When a hole is made in the sphere, we no longer have a close surface. To find the magnitude of the electric field in the hole, let's imagine that we place a "plug" in the hole; then we have a closed surface and,

[tex]\int(sphere) E dS - \int(plug) E dS = \sigma/\epsilon[/tex].

However, this is not getting me the solution sought.

So, after thinking about this some more, I am wordering what happens to the charge on the sphere when a hole is made in it. The sphere is no longer a closed surface. Does the charge redistribute itself on the outside and inside?

However, the sphere is non-conducting, so there will not be any redistribution of charge. The only thing I can think of is that the field in the hole must be provided by the material near the hole.

Any comments that could help solve this problem?

Tks, JG

 

[Doc Al]

superposition principle

Use the superposition for electric fields. See this thread: https://www.physicsforums.com/showthread.php?t=27667

 

[M98Ranger]

370[/quote]

Your approach is on the right track. When a hole is made in the sphere, the electric field inside the hole is provided by the material near the hole. This is because the charge on the sphere is still present and creating an electric field, even though there is a hole. This charge will redistribute itself on the outside and inside of the sphere, but since the sphere is non-conducting, it will not redistribute evenly like it would in a conducting material.

To find the magnitude of the electric field in the hole, you can use Gauss's law. Since the sphere is non-conducting, the electric field will be perpendicular to the surface at all points. Therefore, we can use a Gaussian surface in the shape of a cylinder, with the hole as the base and the top of the cylinder outside the sphere. This will enclose the charge on the sphere and the charge inside the hole.

Applying Gauss's law, we have:

\int E dS = Q_{enc}/\epsilon_0

Since the electric field is constant over the surface of the cylinder, we can pull it out of the integral. Also, the electric field inside the hole will be the same as the electric field on the surface of the sphere (since the material is non-conducting). Therefore, we have:

E \int dS = Q_{enc}/\epsilon_0

Solving for E, we get:

E = Q_{enc}/(\epsilon_0 \int dS)

Since the charge inside the hole is equal to the charge on the sphere, we have:

E = \sigma/(\epsilon_0 \int dS)

The integral of dS over the surface of the cylinder is just the area of the base, which is the area of the hole. Therefore, we have:

E = \sigma/(\epsilon_0 A_{hole})

Since the area of the hole is half the area of the sphere, we have:

E = \sigma/(2\epsilon_0 A_{sphere})

Using the formula for the area of a sphere, we have:

E = \sigma/(2\epsilon_0 (4\pi r^2))

Simplifying, we get:

E = \sigma/(8\pi \epsilon_0 r^2)

This is half of the electric field on the surface of the sphere, which is what we were asked to show. Therefore