# Gauss Law and field charge

• MHB
Mango12
I need to prove that the field charge of 2 infinite, oppositely charged plates is 0 using Gauss' law. I know that the sum of E=4PiK(Sum of q) but I don't know how to prove the charge is 0.

Gold Member
MHB
I need to prove that the field charge of 2 infinite, oppositely charged plates is 0 using Gauss' law. I know that the sum of E=4PiK(Sum of q) but I don't know how to prove the charge is 0.
What is a "field charge?" Generally you only have the total charge being 0 when you have equal and oppositely charged plates, so that will be a given when the problem is posed.

-Dan

Mango12
What is a "field charge?" Generally you only have the total charge being 0 when you have equal and oppositely charged plates, so that will be a given when the problem is posed.

-Dan

Field charge is the total charge. And I need to find the charge on the outside of the plates. I know it would be 0, so maybe I have to set 2 equations equal to each other and show that they cancel to equal 0?

Gold Member
MHB
I need to prove that the field charge of 2 infinite, oppositely charged plates is 0 using Gauss' law. I know that the sum of E=4PiK(Sum of q) but I don't know how to prove the charge is 0.

Field charge is the total charge. And I need to find the charge on the outside of the plates. I know it would be 0, so maybe I have to set 2 equations equal to each other and show that they cancel to equal 0?
Again, the total charge of the two plates is 0 if and only if they both have the same charge on them. This makes no sense. Can you please post the whole question?

Note: In your first post above you mention the E field due to a distribution of charged particles. When dealing with a parallel plate capacitor you are dealing with a continuous charge distribution so that formula cannot be used.

-Dan

Mango12
Again, the total charge of the two plates is 0 if and only if they both have the same charge on them. This makes no sense. Can you please post the whole question?

Note: In your first post above you mention the E field due to a distribution of charged particles. When dealing with a parallel plate capacitor you are dealing with a continuous charge distribution so that formula cannot be used.

-Dan

"There are 2 infinite plates side by side. One plate has a negative charge and the other has an equal charge, only positive. Using Guass' Law or the superimposition of fields, calculate the charge on the OUTSIDE of the plates"

Homework Helper
MHB
I need to prove that the field charge of 2 infinite, oppositely charged plates is 0 using Gauss' law. I know that the sum of E=4PiK(Sum of q) but I don't know how to prove the charge is 0.

What do you mean exactly by Gauss's law?

For a point charge $q$, we have $E=\frac{q}{4\pi\epsilon_0 r^2}$.
More generally, Gauss's law says that the surface integral of $E$ is equal to the enclosed charge $Q$ divided by $\epsilon_0$:
$$\bigcirc\!\!\!\!\!\!\!\!\iint E\cdot dS = \iiint \frac\rho{\epsilon_0} dV = \frac Q{\epsilon_0}$$

"There are 2 infinite plates side by side. One plate has a negative charge and the other has an equal charge, only positive. Using Guass' Law or the superimposition of fields, calculate the charge on the OUTSIDE of the plates"

I may be going on a rampage, but suppose we look at just one plate with charge $Q$ in a cross section of area $A$.
And suppose we define a cylinder with the plate in the middle, a cross section with area $A$, and a distance $d$ to each end of the cylinder.
Then due to symmetry, we have that the surface integral is $2\cdot E \cdot A$.
Gauss's law says that this is equal to $\frac {Q}{\epsilon_0}$.
So:
$$2EA = \frac {Q}{\epsilon_0} \quad\Rightarrow\quad E = \frac{Q}{2\epsilon_0 A}$$

Are you still with me?
Or am I going on a rampage? (Wondering)

Mango12
What do you mean exactly by Gauss's law?

For a point charge $q$, we have $E=\frac{q}{4\pi\epsilon_0 r^2}$.
More generally, Gauss's law says that the surface integral of $E$ is equal to the enclosed charge $Q$ divided by $\epsilon_0$:
$$\bigcirc\!\!\!\!\!\!\!\!\iint E\cdot dS = \iiint \frac\rho{\epsilon_0} dV= \frac Q{\epsilon_0}$$

I may be going on a rampage, but suppose we look at just one plate with charge $Q$ in a cross section of area $A$.
And suppose we define a cylinder with the plate in the middle, a cross section with area $A$, and a distance $d$ to each end of the cylinder.
Then due to symmetry, we have that the surface integral is $2\cdot E \cdot A$.
Gauss's law says that this is equal to $\frac {Q}{\epsilon_0}$.
So:
$$2EA = \frac {Q}{\epsilon_0} \quad\Rightarrow\quad E = \frac{Q}{2\epsilon_0 A}$$

Are you still with me?
Or am I going on a rampage? (Wondering)

I kind of get it. I think I figured it out a different way though, because we never use integrals in class so he wouldn't want me to solve it that way outside of class :/