# Charged Sphere with a Hole - Check my work?

• Raen
In summary: The Attempt at a Solution Both equations (for disk+sphere) are true in a specific range of z (inside or outside?). Only.
Raen
Charged Sphere with a Hole -- Check my work?

## Homework Statement

You have a spherical shell of radius a and charge Q. Your sphere is uniformly charged except for the region where θ<= 1° (which has σ = 0).
Imagine that your fi eld point is somewhere on the positive z-axis (so z could be larger or smaller than a). Determine E as a function of z.

I believe I can represent this as a uniformly charged sphere without a hole and a thin disk with a charge density of -σ. Then the law of superposition let's me add the two together. I think I did it right, but before I go on to the computer program portion of the assignment, I'd love if somebody would double-check my logic and work. If you see an error, please let me know. If you think it's correct, let me know that, too.

## Homework Equations

sin(1°) = r/a Where r is the radius of the disk. r = sin(1°)a = 0.017a

E field of a sphere: E(r) = Q/(4∏r2ε0) = σa2/(r2ε0)

E field of a disk: E(z) = q/(2∏ε0(0.017a2)*(1-z2/(√z2+(0.017a)2))

## The Attempt at a Solution

Along the z-axis, the E field of the sphere can be written as E(z) = σa2/(z2ε0)

Therefore, Etotal = σa2/(z2ε0) + q/(2∏ε0(0.017a2)*(1-z2/(√z2+(0.017a)2))

or

Etotal = σ/ε0*(a2/z2 - 1/2 + z/√(z2 + (0.017a)2)

Is this correct? I'm sorry if it's messy and thank you, thank you in advance.

Both equations (for disk+sphere) are true in a specific range of z (inside or outside?) only.
It might be useful to calculate the field for both cases individually.

I know the equation I have for the sphere contribution only works outside of the sphere, because the electric field inside a (normal) uniformly charged sphere is zero. The field around the disk, though, I thought would look the same in both directions. That would make the Etotal in my initial post the Etotal outside the sphere, while inside the sphere it would be

Etotal = -σ/(2ε0)*(1-z/√(z2+(0.017a)2))

The magnitude is the same, but the direction is not.

Right, it would be in the opposite direction...but now I'm confused. Going by Gauss' Law, inside the sphere the E field will still be zero, won't it? Even with the little disk there's still no enclosed charge inside the sphere and therefore no E field...

Raen said:

E field of a disk: E(z) = q/(2∏ε0(0.017a2)*(1-z2/(√z2+(0.017a)2))

This holds if the disk's center is at the origin, at z=0. But this disk is at z = a , so in your equation make the substitution z → z-a . I think that the rest are correct...

## 1. What is a charged sphere with a hole?

A charged sphere with a hole is a physical system where a conducting sphere has a small hole in it, causing a non-uniform distribution of charge on its surface. This can occur when a charged object is placed inside a larger, grounded conducting sphere.

## 2. How is the charge distributed on a charged sphere with a hole?

The charge distribution on a charged sphere with a hole is non-uniform, with a higher concentration of charge near the edges of the hole. This is due to the electric field lines being compressed at the edges of the hole, resulting in a higher charge density.

## 3. What is the electric potential of a charged sphere with a hole?

The electric potential of a charged sphere with a hole can be calculated using the formula V = kQ/R, where k is the Coulomb's constant, Q is the total charge on the sphere, and R is the radius of the sphere. However, this formula only applies for points outside of the hole. Inside the hole, the electric potential is constant and equal to the potential of the charged object inside the sphere.

## 4. How does the presence of a hole affect the capacitance of a charged sphere?

The presence of a hole in a charged sphere affects its capacitance by decreasing it. This is because the electric field lines are compressed near the edges of the hole, reducing the distance between the charges and thus reducing the capacitance. The smaller the hole, the greater the decrease in capacitance.

## 5. How does the charge distribution on a charged sphere with a hole change when a charged object is placed inside the hole?

When a charged object is placed inside the hole of a charged sphere, the charge distribution on the sphere's surface becomes more uniform. This is because the electric field lines from the charged object cancel out the compressed field lines near the edges of the hole, resulting in a more even charge distribution on the sphere's surface.

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