2 Charged thick plates placed next to each other.

In summary: Then use Gauss's Law to find E for x > 0 . Finally, use the integral sign to combine the two results to get E = Rho*x/ε.
  • #1
Ghastn
4
0

Homework Statement



Two thick, parallel plates of thickness d and uniform charge densities
(Coulombs per unit volume) ρ and −ρ are placed next to each other, as
shown in the figure. The negatively charged plate is located between –d
and 0 on the x-axis and the positively charged plate between 0 and d on
the x-axis. The z-axis points out of the page. Assume that both plates
are infinite in y and z. Find the expressions for the x-, y-, and zcomponents
of the electric field E(x) as a function of the x-coordinate.
Write your solution in terms of ε0, not ke.
Is the net force per unit area between the plates: attractive, repulsive or zero.
Setup the integral to compute the pressure on the plates


Homework Equations



I believe Gauss Law could be used: Flux = Qin/ε.

The Attempt at a Solution


I would imagine a cylinder crossing the 2 plates and measure the field on each surface: they are going to be in the same direction, I think. The problem is that I cannot fully imagine the situation: Is the field w on the left and equal to the sum on the right?
 
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  • #3
I don't think that would work as we are not looking at the interaction between the two plates but rather the superposition of the two fields: Here is the image:
http://img163.imageshack.us/img163/5497/phys.jpg
Any guesses?
 
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  • #4
I'm suggesting that you can divide each think plate into a stack of parallel thin plates. The field at some point within a thick plate then translates to the field between thin plates (with different total charges each side). As for the field beyond the thick plates.. you know what that will be, yes?
 
  • #5
Will that be a ZERO?
I mean, the question asks for the field everywhere: that excludes the plate, right?
SO they are asking about the field outside of it! = ZERO?
 
  • #6
I believe it will be zero outside the plates, but I also think the question is asking for the field inside the plates as well.
 
  • #7
Ok, so I tried this problem and I found that inside the sheets, E = Rho*x/ε.
Would that be correct?
 
  • #8
Ghastn said:
Ok, so I tried this problem and I found that inside the sheets, E = Rho*x/ε.
Would that be correct?
No. That does not approach zero as x → d .

How did you arrive at that answer.

You know that E = 0 for x < -d and E = 0 for x > d .

Start at x = -d and use Gauss's Law to find E for -d ≤ x ≤ 0 .
 

Related to 2 Charged thick plates placed next to each other.

1. What is the purpose of placing two charged thick plates next to each other?

The purpose of placing two charged thick plates next to each other is to create an electric field between the plates. This electric field can be used for various applications such as capacitors, particle accelerators, and electrostatic precipitators.

2. How do the charges on the plates affect the electric field between them?

The charges on the plates create an electric field between them that is directly proportional to the magnitude of the charges and inversely proportional to the distance between the plates. The field lines will be directed from the positive plate to the negative plate.

3. What happens if the plates have opposite charges?

If the plates have opposite charges, they will attract each other and the electric field between them will be stronger. The positive charges on one plate will be attracted to the negative charges on the other plate, creating a strong electric force between the plates.

4. Can the strength of the electric field between the plates be changed?

Yes, the strength of the electric field between the plates can be changed by altering the distance between the plates or by changing the amount of charge on the plates. Increasing the distance between the plates will weaken the electric field, while increasing the charge on the plates will strengthen it.

5. What happens if the plates have the same charge?

If the plates have the same charge, they will repel each other and the electric field between them will be weaker. This is because the like charges on each plate will repel each other, resulting in a weaker overall electric force between the plates.

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