Induced Charge on a Conducting Disk

[cardsfan527]

Homework Statement

A large, thin plastic disk with radius R = 1.1 meter carries a uniformly distributed charge of Q = -7e–5 C. A circular piece of aluminum foil is placed d = 3 mm from the disk, parallel to the disk. The foil has a radius of r = 6 cm and a thickness t = 1 millimeter.

a. Find the net electric field at the center of the foil.
b. Calculate the magnitude q of the charge on the left circular face of the foil.

Homework Equations

E = 1/(4*∏*ε)*q/r^2

The Attempt at a Solution

I know that the answer to part (a) is 0, because a conductor will not have an electric field inside of it.

I also know that the charge on one side of the foil disk will be equal and opposite to the charge on the other side, since the disk has a neutral total charge. I'm just unsure of exactly where to go from there. Should I use force equations, since I know that the charges are at rest? Or is there something I know about the electric field that I can use?

 

[Goddar]

Hi.
So you have a zero field inside the conductor, and non-zero right outside. Determine the field outside, remember the formula for induced surface charge density, then integrate over the surface...

 

[cardsfan527]

Thanks, Goddar. I guess my problem is calculating the electric field. Could you send me in the right direction?

 

[Goddar]

Method of images, since the disk is "large"...

 

[paranormal]

I would approach this problem by first considering the properties of conductors and how they interact with electric fields. Conductors are materials that allow charges to move freely, which means that any excess charge placed on the surface of a conductor will distribute itself evenly throughout the surface in order to minimize the repulsion between like charges.

In this case, the plastic disk is a conductor and has a uniform charge distribution, meaning that the charge is evenly spread out over the entire surface. When the aluminum foil is placed near the disk, it will experience a repulsive force from the induced charges on the disk surface. However, since the foil is also a conductor, it will also distribute this induced charge evenly over its surface.

Therefore, the net electric field at the center of the foil will be 0, as you correctly stated. This is because the electric field from the induced charges on the disk will be cancelled out by the induced charges on the foil.

To calculate the magnitude of the charge on the left circular face of the foil, we can use the equation you provided, E = 1/(4*∏*ε)*q/r^2. Since the electric field at the center of the foil is 0, we can rearrange this equation to solve for q:

q = 0 * 4*∏*ε*r^2 = 0

This means that the charge on the left circular face of the foil is 0. However, this does not mean that there is no charge on the foil at all. There will still be an induced charge on the foil, but it will be evenly distributed over the entire surface and cancel out any external electric field.

In conclusion, the net electric field at the center of the foil is 0 and the charge on the left circular face of the foil is 0. This is due to the properties of conductors and how they interact with electric fields.