# Calculating Charge created by Van De Graff Generator

So I have a few questions regarding a Van De Graff Generator. With the Equation: V=kQ/r where V is the Voltage Potential, k is the constant, Q is the charge in coulombs and r is the radius of the sphere. 1.) Do I have to use a sphere? Can I use, say a cylinder that is metal?

2.) How does the Charge concept work? I am trying to understand the concept of this equation since I didn't understand it well the first time in class

Danke!

There is a related answer which gives an estimate of the charge which is stored on a Van de Graaff generator dome.

The shape of the dome is important particularly as it must have no sharp edges or points.
This means that if you were to use a cylindrical shape you would need to round the edge.

The reason for this is that as the radius of curvature of a conductor decrease the charge density in that area increase which in turn increases the electric field in that region.

If the electric field is too high the air breaks down and becomes a conductor resulting in charge being lost from the dome thus reducing the maximum potential of the dome.

The left hand diagram is of a dome with a sharp edges (left) with the sharpness of the edge reduced by placing "insulating" tape over the sharp edge (right).

A better solution is to fold the sharp edge inside the dome.

Dust on the dome will act as points and dents will have edges which will reduce the effectiveness of a dome to store charge at high voltages.

Any object has a self capacitance $$C$$. This relates the charge on the body to the potential at its surface relative to infinity via the usual capacitor equation:

$$Q = CV \tag{1}$$

For a spherical object the self capacitance is easily calculated since Gauss' law tells us that the field near the surface of the object is the same as the field from a point mass of the same charge at the centre of the sphere. That is, if $$R$$ is the radius of the sphere the potential is just:

$$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R} \tag{2}$$

And comparing this with equation (1) we see this means:

$$C = 4\pi\epsilon_0 R$$

This is why when we make an approximate calculation of the charge on a Van de Graaff generator we treat it as a sphere. In principle it is possible to do the calculation for other shapes but this gets complicated because we no longer have the spherical symmetry to help us.

• Okay, but how do the charges differ from the voltage potential? Doesn't the voltage potential come from the charges? – Captain Caboose May 8 '19 at 7:39
• @CaptainCaboose yes, the voltage is proportional to the charge. With no charge the potential of the sphere is obviously zero. As we add charge to the sphere the potential increases. Specifically the voltage is related to the charge by $V = Q/C$, where $C$ is the self capacitance. – John Rennie May 8 '19 at 7:43