Gauss Law and field charge

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Mango12
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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.
 

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topsquark
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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
 
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Mango12
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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?
 
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topsquark
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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
 
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Mango12
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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"
 
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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)
 
  • #7
Mango12
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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 :/
 

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