Parallel Plate Capacitor Spacing Question

jacksonpeeble

If the length and width of each plate of a parallel plate capacitor were doubled, and the spacing between the plates was also doubled, by what factor does the capacitance change?

I know that increasing the length and width of a plate increases its capacitance, and I thought that it was directly proportional, but I'm not sure about spacing the plates... it's an interesting question...

berkeman

Do you know the equation for the capacitance in terms of the area and spacing of the plates? That equation will answer your question.

jacksonpeeble

Is it e*Area/distance?

berkeman

Yes. [tex]C = \frac{\epsilon A}{d}[/tex]

This simple equation is used a lot in capacitor calculations, but keep in mind that it is simplified from the more accurate equation. This equation ignores the fringing field at the edges of the parallel plate capacitor. For a cap with lots of area A compared to the spacing distance d, this is a fine approximation. But if the distance d is, say 1/10 of the length of one of the sides of the cap plate, then you will measure a higher C than you calculated by probably around 10% or so.

But this basic equation should be fine for your conceptual question. So what's the answer?

jacksonpeeble

The capacitance increases by a factor of 2, because 2*2/2=2.

I feel stupid. Thanks for the help.

berkeman

Correct. No reason to feel that way, BTW. We all had to learn to use that equation (and learn its limitations) at some point.

priya amrute

what if the distance between the plates is much larger than the diameter of the parallel plates?? how do we then calculate the capacitance???

berkeman

Not with the simplistic equation we've listed so far, that's for sure.

You tell us. That's how it works around here...

priya amrute

i looked around a lot..all i know is
Decreasing the separation of the plates, decreases the voltage of the capacitor since the electric-field is not affected by the distance between the plates. The voltage on the capacitor is V=Ed. Therefore the voltage increases. For a constant charge, Q, C=Q/V =Q/Ed and vice-versa..then why cant the same formula still remain applicable???

priya amrute

i will be really glad if u can help me with this...we am working on a project where the calculation of resonant frequency is very important for which we require...

berkeman

When the geometry is not simple, you need to calculate the capacitance by integration, typically. C = Q/V, and you perform a surface integration to find Q, and a line integral to find V.