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?
Best Answer
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.