The discussion centers on applying Gauss' Law to demonstrate that the electric field charge of two infinite, oppositely charged plates is zero. Participants clarify that the total charge is zero when the plates have equal and opposite charges. The formula for the electric field due to a charged plate is derived, showing that the electric field (E) can be expressed as E = Q/(2ε₀A) for a single plate. The conversation emphasizes the importance of understanding continuous charge distributions when analyzing parallel plate capacitors.
PREREQUISITESThis discussion is beneficial for physics students, educators, and anyone interested in understanding electrostatics, particularly those studying electric fields and Gauss' Law in the context of charged plates.
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.Mango12 said: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.
topsquark said: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 said: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.
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?Mango12 said: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?
topsquark said: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 said: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.
Mango12 said:"There are 2 infinite plates side by side. One plate has a negative charge and the other has an equal charge, only positive. Using Gauss' Law or the superimposition of fields, calculate the charge on the OUTSIDE of the plates"
I like Serena said: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)