Understanding Charge Distribution in Capacitors: One Plate or Both?

[chopnhack]

Hello all,

I have a simple question about capacitors. We are learning about them in class and in some questions, I seem to come up with different answers than the solution sets, always by 2. I apparently am using the area of each plate in a capacitor to derive my results.

My question is when working with capacitors, does one use the area of one electrode or both?

I expected to use both as I believe there would be charge density on both electrodes.

Thanks in advance.

 

[Vibhor]

My question is when working with capacitors, does one use the area of one electrode or both?

Area of one plate .

 

[chopnhack]

Area of one plate .

Can anyone explain why? Is it possibly because charge migrates to one side with the other side simply being polarized, but not conducting? I seem to recall that there is no flow through the capacitor, rather a push and pull if you will, on either side.

 

[Vibhor]

Area of a parallel plate capacitor comes in picture while deriving its capacitance . Charge density is given by Charge on one plate divided by area of that plate .

 

[chopnhack]

Area of a parallel plate capacitor comes in picture while deriving its capacitance . Charge density is given by Charge on one plate divided by area of that plate .

It's a bit tricky. So when calculating capacitance, if we are offered the total area of both plates, we would need to divide by 2?

Here are two sample questions that I have answered correctly according to the answer guide:

A 15-pF capacitor has a potential difference of 1.50 V between its plates. What is the magnitude of charge on each plate?
A: 2.25x10^-11C

But the question asks for the charge of each plate, the above was calculated using Q=C*V - is this not the total charge of the capacitor? Shouldn't the answer then be half of this? or does this go back to what you were saying, use one plate?

An electric field of 2.8 × 105 V.m-1 exists between two parallel plates each of area 21.0 cm2 and separated by 0.25 cm of air. What is the charge on each plate?
A: 5x10^-9C

In this case I used σ = E ⋅ ε_o and took the surface charge density and multiplied it by the air gap to get the total charge.

Sorry, if I am being thick, I just want to be clear on this before proceeding in my studies.
Thanks!

 

[Vibhor]

But the question asks for the charge of each plate, the above was calculated using Q=C*V - is this not the total charge of the capacitor? Shouldn't the answer then be half of this? or does this go back to what you were saying, use one plate?

Charge means magnitude of charge on one plate .

In a parallel plate capacitor the sum of charges on inside faces sums up to zero .