# Video: Capacitors and Capacitance

What is the capacitance of a large Van de Graaff generator’s terminal given that it stores 5.00 mC of charge at a voltage of 6.0 MV?

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### Video Transcript

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 sixth volts.

This fraction is equal to 0.83 nanofarads. That’s the capacitance of this generator’s terminal.