# Electrostatics - 3rd plate inserted into parallel plate capacitor

• TH93
In summary: Doing so will change the fields in the regions between the middle and outer plates, and you would need to solve for the new fields.
TH93
Electrostatics -- 3rd plate inserted into parallel plate capacitor

## Homework Statement

This question is from the book by I.S Grant and W.R Phillips on Electromagnetism. Two large parallel plates of area A and distance d apart are maintained at potentials 0 and V. A third similar plate, carrying a charge q, is isolated from the other two and placed midway between them. What is the potential of this plate?

V = -∫E.dl
E =σ/d
σ=q/A

## The Attempt at a Solution

The plate is placed half way between the two plates so I integrated the electric field between 0 and 1/2d. This gave me the potential as σd/2ε$_{0}$. I know that σ is simply the charge on the plate q divided by the area A. Substituting this into my answer gives qd/2ε$_{0}$A. The final answer however, is 1/2V plus the answer which I obtained. I am not sure if what I have done is correct or where the 1/2V came from. Any help would be appreciated.

Once the new plate is added, you have a three plate system. When writing the integral out from 0 to d, you have to break it into two parts, 0 to .5d and .5d to d. You only calculated the lower half of the integral (center plate to 0 potential plate).

The plate held at V is held at V relative to the plate held at 0, and can be expressed as E*d (We will call this V0. Your result, is E*d/2 from the first region (0 to .5d). Call this potential VF. Now the contribution to the center plate from the plate held at V spans the region .5d to d, and therefore can be expressed as VS = V0 - VF (the F and S subscripts correspond to first and second region respectively).

Your expression E*d/2 for the first region is exactly half the original E*d. Therefore, the potential in this region is .5V. Now using VS = V0 - VF, we see that the potential from the second region must be .5V as well.

So now that you have the second region's potential, add that region to your expression for region one, and you have the solution.

Now I challenge you to think about what would happen if you added a fourth plate, or fifth, or Nth plate to the system. What would be the potential between the plate held at V and that held at 0V?

So the potential of the middle plate is equal to the potential difference of the first region plus the potential difference of the second region? I guess this is because both of the plates are contributing to the field and not just the one plate. For the question regarding the adding of n plates, the potential between the plate held at 0 and the plate held at V should still be equal to V due to energy conservation. Adding plates is simply dividing the potential into equal parts.

You can think of these problems in relation to their fields if you want (you wouldn't be wrong), but I have found examining E&M problems by their potentials is much easier (in the case of electrostatics you don't have to worry about vector quantities, just scalar ones). So yes, b/c of conservation laws the plate is still at V, but if you notice, the sum of potentials in each region should equal the total potential of the plate at V relative to the one at 0V (call these the "outer" plates).

Now when adding plates, you aren't dividing the potential into equal parts necessarily. The potential contributions from each of the N plates when summed together must equal the total potential of the "outer" plates, but each individual plate's potential is determined by its distance d from a reference or ground source (in this case the plate at 0V).

If you set up this summation mathematically, you actually wind up deriving the expression for series capacitance. It's worth doing if you have the time, as it's really neat to see that from Gauss' Law, you obtain a real world application to electronic circuitry.

OP: this is a very educational problem!

EDIT: I suggest using two strategically placed Gaussian surfaces to help come up with the 4 inside surface charges. Once you have the surface charges the E fields, and then the potentials, are readily determined.

4 equations, 4 unknowns. Two are from the two Gaussian surfaces, one is the sum of surface charges on both sides of the middle plate, and one is Ed = V where E is the field , d is the distance between plates, and V is your answer.

For extra credit you could position the middle plate asymmetrically with respect to the outer plates.

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## 1. What is a parallel plate capacitor?

A parallel plate capacitor is a device that stores electric charge and energy by creating an electric field between two conducting plates that are parallel to each other.

## 2. How does the third plate affect the capacitance of a parallel plate capacitor?

The third plate inserted into a parallel plate capacitor changes the capacitance by altering the electric field between the two original plates. This change in capacitance can affect the amount of charge and energy that the capacitor can store.

## 3. What is the difference between a parallel plate capacitor with and without a third plate?

A parallel plate capacitor with a third plate inserted has a different capacitance and electric field than one without the third plate. This can result in different charge and energy storage capabilities.

## 4. How does the distance between the plates affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is directly proportional to the distance between the plates. As the distance decreases, the capacitance increases, and vice versa.

## 5. Can a third plate be used to increase the capacitance of a parallel plate capacitor indefinitely?

No, there are physical limitations to the capacitance of a parallel plate capacitor. Eventually, the electric field between the plates will become too strong and cause a breakdown of the insulating material, limiting the maximum capacitance that can be achieved.

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