Magnitude and direction of an electric field outside of a plate?

In summary, the circular metal plate has 1010 excess electrons uniformly distributed over its surface. The magnitude and direction of the field just outside the plate near its center is 16 C.
  • #1
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Homework Statement



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?

Homework Equations


E = kQ/r2

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.
 
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  • #2
Violagirl said:

Homework Equations


E = kQ/r2

I like gauss law better.:biggrin:

The Attempt at a Solution



Starting with E = kQ/r2,

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

= 16 C
Charge of electron is 1.6x10-19

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?
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
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.
 
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  • #4
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
What is the total surface area of the plate?
And Gauss' law does not look like what you wrote. What is k?
 
  • #6
Violagirl said:
Ok so Gauss's Law is:

E = 2pikσ
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
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...
 
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  • #8
Lets begin at the beginning. What is your Gaussian surface?
 
  • #9
Ok so for a circular plate, will my Gaussian surface be:

E =4pi*r2?
 
  • #10
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
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
Okay, let's 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
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

Is this correct? Am I thinking about this right?
 
  • #14
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
Sorry, its getting a bit late here. I am off to :zzz:
For your problem:
After you have got the charge enclosed, just use the Gauss's law expression. Note that you only need the area of the top face of the cylinder, as h and r are very small, we can assume E to be perpendicular to that face.
:wink:
 

1. What is the formula for calculating the magnitude of an electric field outside of a plate?

The formula for calculating the magnitude of an electric field outside of a plate is E = σ/2ε0, where σ is the surface charge density and ε0 is the permittivity of free space.

2. How does the distance from the plate affect the magnitude of the electric field?

The magnitude of the electric field outside of a plate decreases as the distance from the plate increases. This relationship follows an inverse square law, meaning that the electric field decreases by a factor of four as the distance is doubled.

3. Is the direction of the electric field outside of a plate always perpendicular to the surface of the plate?

Yes, the direction of the electric field outside of a plate is always perpendicular to the surface of the plate. This is because the electric field lines always point away from positive charges and towards negative charges.

4. How does the magnitude of the surface charge density affect the magnitude of the electric field outside of a plate?

The magnitude of the electric field outside of a plate is directly proportional to the surface charge density. This means that as the surface charge density increases, the electric field also increases.

5. Can the direction of the electric field outside of a plate change?

No, the direction of the electric field outside of a plate is always perpendicular to the surface of the plate and does not change. However, the direction of the electric field inside the plate may vary depending on the distribution of charges on the surface.

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