What is the capacitance of a large
Van de Graaff generator terminal given that it stores 5.00 millicoulombs of charge at a voltage
of 6.0 megavolts?
To start off, let’s show a Van de
Graaff generator sketch and see what its terminal is. When a Van de Graaff generator is
running, positive charge is distributed across a metal sphere at the top or terminal
of the generator.
We’re told that this charge which
we can call 𝑄 is equal to 5.00 times 10 to the negative third coulombs. We’re told, moreover, that this
charge is stored at a potential difference we can call 𝑉 of 6.0 times 10 to the
sixth volts. That potential difference tells us
about the field that these charges find themselves in.
Knowing all this, we want to solve
for the capacitance of the terminal, this big spherical ball, of the Van de Graaff
generator. To do that, we can recall a
relation between capacitance, charge 𝑄, and voltage 𝑉. For a capacitor the capacitance,
𝑐, is equal to the charge stored on the capacitor divided by the potential
difference, which separates the positive from the negative charge.
In the case of our generator, the
negative charge is directed downwards towards the base. For our purposes, we only need to
know that the potential difference separating these two charge types is given to
us. We can say that the capacitance is
equal to 5.00 times 10 to the negative third coulombs divided by 6.0 times 10 to the
This fraction is equal to 0.83
nanofarads. That’s the capacitance of this