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 Guass' 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)
Gauss Law is a fundamental law in electromagnetism that relates the electric field at a point to the amount of electric charge enclosed by a surface surrounding that point.
Gauss Law is used to calculate the electric field at a point due to a distribution of electric charges. It is also used to determine the total charge enclosed by a surface if the electric field is known.
A Gaussian surface is an imaginary surface that is used in Gauss Law calculations. It is a closed surface that encloses a certain amount of charge and is chosen to simplify the calculation of the electric field.
Gauss Law is a more general form of Coulomb's Law. Coulomb's Law only applies to point charges, while Gauss Law can be used for any distribution of charges. In certain cases, Gauss Law can lead to a simpler and more elegant solution than Coulomb's Law.
Gauss Law is one of the four Maxwell's equations that form the basis of classical electromagnetism. It is used in many areas of physics, such as optics, electronics, and high energy physics, to understand and predict the behavior of electric fields and charges.