# Does changing the gap between plates change the capacitor voltage?

Consider an ideal capacitor which has a length of $\ell_1$ between its plates. The capacitor terminals are open; they are not connected to any finite valued impedance. Its capacity is $C_1$ and it has an initial voltage of $V_1$.

What happens to the capacitor voltage if we make the gap between the plates $\ell_2=2\ell_1$ without changing the amount of charge on the plates?

My thoughts on this:

Increasing the gap will decrease the capacitance.

$$C_2 = \dfrac{C_1}{2}$$

Since the amount of charge is unchanged, the new capacitor voltage will be

$$V_2 = \dfrac{Q}{C_2} = \dfrac{Q}{\dfrac{C_1}{2}} = 2\dfrac{Q}{C_1} = 2V_1.$$

Is this true? Can we change the capacitor voltage just by moving its plates? For example, suppose that I'm wearing plastic shoes and I have some amount of charge on my body. This will naturally cause a static voltage, since my body and the ground act as capacitor plates. Now, if I climb a perfect insulator building (e.g.; a dry tree), will the static voltage on my body increase?

A Wimshurst machine works by that process.

It puts charge on plates which are close together, then moves the plates apart to generate a high voltage.

When I was at school, in the '70s, a kid made one using PCB material for the disks, and gramophone needles to create the initial charge. The 'work' was done by an electric motor. Based on the length of spark it generated, I think it produced over 200,000V.

His dad took it work, where they designed telephones, and tested early electronic telephones with it.

Yes, the voltage increases. It seems most of us learned of this in school. My Physics professor had a setup with movable plates, and a very sensitive (actually, very high impedance) voltmeter. As the plates were pulled apart, the voltage went up.

This comes from the elemental formula Q=CV. Pulling the plates apart lowers the capacitance. The charge didn't go anywhere, so the voltage must rise. This may seem counterintuitive, but the charge on the plates want to attract each other, and you are doing work by pulling them apart.

You can reproduce the experiment described above if you have a voltmeter with an FET input (or an oscilloscope, if you're that fortunate). Ground the negative lead and hold the other lead in your hand. If your shoes are not conductive, and you don't have any ESD straps connected, you should be able to deflect the meter simply by raising and lowering your foot. By the way, rubbing the carpet creates the charge and picking up your feet and moving away is what raises those static charges to such high voltage levels.

On a practical note, this is how an electret condenser microphone works. As the diaphragm vibrates, the capacitance between it and a fixed plate changes, and the voltage changes with it.

The Voltage definitely increases.

Q = C * U

Since you decrease C by increasing the gap but Q stays the same, U will increase.

In my schooltime I did not want to believe it so my techer sent me into the experiments room with a high voltage power supply, plates, cables, isolators and a galvanometer. I've tested it and it it is true! The voltage increases as you increase the gap.

The electric field between two parallel plates of area $A$ is roughly $E = { Q \over \epsilon A}$, hence the voltage at a distance $x$ apart will be $V(x) = { Q x\over \epsilon A}$.

So, doubling the distance will double the voltage.

The electric field approximation will degrade significantly as $x$ gets larger than some fraction of some characteristic dimension of the plates.

As we know, a capacitor consists of two parallel metallic plates. And the potential between two plates of area A, separation distance d, and with charges +Q and -Q, is given by

$$\Delta V = \frac{Qd}{\varepsilon_0 A}$$

So potential difference is directly proportional to the separation distance.

You're correct. You might notice that while charge is conserved, the energy stored in the capacitor after separating the plates has increased: $$E_1 = \frac{1}{2}C_1V_1^2$$ $$E_2 = \frac{1}{2}C_2V_2^2 = \frac{1}{2}\frac{C_1}{2}(2V_1)^2 = C_1V_1^2 = 2E_1$$ This extra energy comes from the mechanical work that you had to do to move the plates apart against the electrostatic force holding them together.

in the context described with plates not connected, the scenario and formulas indicate that for distance 2l you will require twice the voltage to polarize the same amount of charge.

• This answer could be improved by including the formulas you mention. – Null Sep 30 '14 at 2:07
• "polarize the charge" is not correct technical or scientific jargon. You cannot polarize a charge, you can only polarize an object. – Lorenzo Donati -- Codidact.org Aug 30 '16 at 3:15