# Questions about parallel plate size

1. May 19, 2013

### Christofferall

Hi everyone, new member here and I apologize in advance if my questions seem somewhat amateur.

Recently I've started learning about capacitors, electrostatics and electric theory in general. I guess I'm having trouble trying to wrap my head around the actual area (size) of the plates of a parallel capacitor. For instance......if I want to size a capacitor that has the charge of 1kwh...(3,600,000 joules), at 50 kv, the equations say I would need a cap size around .00288 farads.

When I try to calculate the area of the plates for a .00288 cap with .5 meter separation, a dielectric permittivity of 3.....I get a plate area size of .00048 meter square.

This seems like an awfully small plate size to hold that amount of energy.

I guess I'm just looking for some validation to my answers, and if not maybe someone could point me in the right direction. Thanks for your replies in advance.

2. May 19, 2013

### Simon Bridge

Welcome to PF;
None of those are units for charge. You want the cap to be able to store 1kWh of energy?
i.e. it should cost you 1kWh to charge it up.

But the plates are a long long way apart - so you don't have to move much charge for that energy.

It is like, if you wanted to store energy in gravity, then you can move a small amount of mass up a long height or a lot of mass up a short height.

So it is with charges - you can store the same energy moving a little charge a long way or a lot of charge a short way.

At 50kV - would you be able to get a spark to jump the 50cm gap in vacuum?

However ... I'd like to see you check the actual calculation step by step.
The energy stored in a parallel-plate capacitor, with a vacuum separation, would be given by:

$$U=\frac{1}{2}\frac{\epsilon_0 A}{d}V^2$$

ref: http://en.wikipedia.org/wiki/Capacitor#Parallel-plate_model

Last edited: May 19, 2013
3. May 19, 2013

### Christofferall

I was using the formula U=1/2cv^2 to find the needed capacitance.

3,600,000=1/2C50,000^2

3,600,000/1/2=>7,200,000

I thought this is telling me the needed capacitor size in farads if I know the amount of energy I want stored in joules (3,600,000) at 50 kv.

From there, knowing the capacitor size in farads, I wanted to calculate the actual plate size needed for a cap rated at .00288 farads.

Not using vacuum in the gap, we'd be using a silicon oil...transformer oil. Say the permittivity is 25, dielectric constant 2.7. This is rated up to 50 kv......probably wouldn't charge the cap to the max but just using it for an example :)
Thanks again.

4. May 21, 2013

### Simon Bridge

That's right - and the capacitance is C=εA/d in the parallel-plate model.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html

Combining these two equations gives you: $$U=\frac{1}{2}\frac{\epsilon_r\epsilon_0 A}{d}V^2$$ ... cutting out the middleman - so to speak ;)

A permittivity of 25 seems rather large but I don't see the units ... perhaps you mean "relative permittivity" $\epsilon_r=25$?

I notice you did not show your calculation step-by-step, so I cannot tell what you did.

You want to find area A given everything else ... which is:

$$A=\frac{2Ud}{\epsilon_r\epsilon_0 V^2}$$

Using:
U=3600000J
V=50000V
d=0.5m
εr=25
ε0=8.85x10-12F/m

... I get a different number to you.
This is why I suggested you go back over your calculation carefully - did you miss something out or use a different formula for capacitance or something?

5. May 23, 2013

### Staff: Mentor

When Tesla needed high voltage condensers for his trans-Atlantic transmitting station, he had long barns constructed, the height of two-storey houses, and filled them with metal panels hanging vertically from the ceiling to form air-spaced condensers. The panels were kept in place with, I think, glass insulators and separators.