# Homework Help: Magnitude and direction of an electric field outside of a plate?

1. Sep 26, 2013

### Violagirl

1. The problem statement, all variables and given/known data

A circular metal plate of radius 0.2 m has 1010 excess electrons uniformly distributed over its surfaces. What is the magnitude and direction of the field just outside the plate near its center?

2. Relevant equations
E = kQ/r2

3. The attempt at a solution

Starting with E = kQ/r2,

I solved for Q: (1010) (1.60 x 10-9 C)

= 16 C

Plugging into the equation for E:

E = (16 C) (9.0 x 109 N m2/C2) / (.02 m)2

I end up with a really high value of 3.6 x 1012 N/C.

However, I know this is wrong because I should end up with a value of around 720. What am I doing wrong?

For direction, I know the electron will move towards the plate.

Last edited: Sep 26, 2013
2. Sep 26, 2013

### Enigman

I like gauss law better.

Charge of electron is 1.6x10-19

It is because you are assuming all charge to be at the same place. In this case you will either need to integrate (long and tedious) or use gauss' law.

3. Sep 26, 2013

### rude man

Compute the surface charge density σ. Then, if you don't know the formula for the E field just outside the plate, set up a gaussian surface, say a right circular cylinder, with one end-surface inside the plate and the other just outside. Then compute net flux leaving the surfaces. Remember to make the height of the cylinder very short. EDIT: also the area of the end sections.

Last edited: Sep 26, 2013
4. Sep 26, 2013

### Violagirl

Ok so Gauss's Law is:

E = 2pikσ

Q = (1010) (1.6 x 10-19) C

Q = 1.6 x 10-9 C

σ = Q/A = 1.6 x 10-9 C/pi (.02)2 = 2.04 x 10-13 C/m2

E = 2pi (9.0 x 109 N m2/C2) (2.04 x 10-13 C/m2)

E = .011

I'm still not sure where I'm going wrong with this..

5. Sep 26, 2013

### nasu

What is the total surface area of the plate?
And Gauss' law does not look like what you wrote. What is k?

6. Sep 26, 2013

### Enigman

Actually
$\int E.ds=\frac{q}{ε_{o}}$
$E.A=\frac{q}{ε_{o}}$
$E=\frac {σ}{ε_{o}}$ As field is perpendicular to the face of cylinder.
$E=\frac{σ}{1/4 \pi k}$
$E=4 \pi \sigma k$

7. Sep 26, 2013

### Violagirl

Oh whoops, ok, I see I wrote the expression for a conductor, sorry about that! Gauss's Law is otherwise expressed as:

∅E = Q /ε0

Total surface area of the plate, isn't it A = pi*r2?

So finding Q again:

Q = (1010) (1.6 x 10-19 C) = 1.6 x 10-9 C

∅E = (1.6 x 10-9 C) / (8.85 x 10-12 F m-1) = 181?

The final answer should be 720. I'm not sure if I'm missing something to this problem...

Last edited: Sep 26, 2013
8. Sep 26, 2013

### Enigman

Lets begin at the beginning. What is your Gaussian surface?

9. Sep 26, 2013

### Violagirl

Ok so for a circular plate, will my Gaussian surface be:

E =4pi*r2?

10. Sep 26, 2013

### Enigman

That is the equation for a gaussian surface (what gaussian surface I have no idea). Gaussian surface is a 3D closed figure.
See post #3 by rude man.

11. Sep 26, 2013

### Violagirl

I'm not finding much on how to compute finding a value for the Gaussian surface of a circular plate though. Unless I need to integrate the area of A for what a circular plate would normally be? i.e. A of a circle = TTr2?

12. Sep 26, 2013

### Enigman

Okay, lets take RM's gaussian surface.
Let the radius of cylinder be r and height be h.
Now this cylinder will enclose how much charge?

13. Sep 26, 2013

### Violagirl

So for a cylinder:

Won't we have, E = kQ/r-h, so to solve for Q, we'd end up with an equation of:

E(r-h)/k = Q

14. Sep 26, 2013

### Enigman

Noooo....
Lets put that equation aside for this sum.
What is the charge in the cylinder?
You have the charge of the disc and its area (remember a disc has 2 sides) So find the surface charge density and multiply it with area of the disc our gaussian surface i.e.. cylinder encloses.

15. Sep 26, 2013

### Enigman

Sorry, its getting a bit late here. I am off to :zzz: