15.1 Principle of Operation

3.1 When equal and opposite charges placed on plates
3.2 When unequal charges are placed on the two plates

6.1 Working method to find capacitance of any capacitor

7.1 Force between the Plates of a Capacitor
7.2 Energy stored in a charged conductor

8.1 Loss of energy during redistribution of charge

9.1 Effect of Dielectric
9.2 Capacitance of a Capacitor Partially Filled with Dielectric
9.3 Quantities after inserting dielectric in a capacitor (fully)

12.1 Important points in $R$-$C$ circuits

15.1 Principle of Operation
15.2 Construction

To consider principle of operation of van de graaff generator suppose we have a large spherical conducting shell of radius $R$, on which we place a charge $Q$. This charge spreads itself uniformly all over the sphere. The electric field outside the sphere is just that of a point charge $Q$ at the center; while the field inside the sphere vanishes. So the potential outside is that of a point charge; and inside it is constant, namely the value at the radius $R$.

We thus have: Potential inside conducting spherical shell of radius $R$ carrying charge $Q$ = constant. $$ = \frac{1}{{4\pi {\varepsilon _ \circ }}}\frac{Q}{R}$$Now we introduce a small sphere of radius $r$, carrying some charge q, into the large one, and place it at the center.

The potential due to this new charge clearly has the following values at the radii indicated: Potential due to small sphere of radius $r$ carrying charge $q$, at the surface of small sphere $$ = \frac{1}{{4\pi {\varepsilon _ \circ }}}\frac{q}{r}$$at the surface of radius $R$ sphere $$ = \frac{1}{{4\pi {\varepsilon _ \circ }}}\frac{q}{R}$$Taking both charges $q$ and $Q$ into account we have for the total potential $V$ and the potential difference the values$$\begin{equation} \begin{aligned} {V_R} = \frac{1}{{4\pi {\varepsilon _ \circ }}}\left( {\frac{Q}{R} + \frac{q}{R}} \right) \\ {V_r} = \frac{1}{{4\pi {\varepsilon _ \circ }}}\left( {\frac{Q}{R} + \frac{q}{r}} \right) \\ {V_r} - {V_R} = \frac{1}{{4\pi {\varepsilon _ \circ }}}\left( {\frac{1}{r} - \frac{1}{R}} \right) \\\end{aligned} \end{equation} $$Assume now that $q$ is positive. We see that, independent of the amount of charge $Q$ that may have accumulated on the larger sphere and even if it is positive, the inner sphere is always at a higher potential: the difference $V_r$$–$$V_R$ is positive. The potential due to $Q$ is constant up to radius $R$ and so cancels out in the difference.

This means that if we now connect the smaller and larger sphere by a wire, the charge $q$ on the former will immediately flow onto the matter, even though the charge $Q$ may be quite large. The natural tendency is for positive charge to move from higher to lower potential. Thus, provided we are somehow able to introduce the small charged sphere into the larger one, we can in this way keep piling up larger and larger amount of charge on the latter. The potential at the outer sphere would also keep rising, at least until we reach the breakdown field of air.

This is the principle of the van de Graaff generator. It is a machine capable of building up potential difference of a few million volts, and fields close to the breakdown field of air which is about $3 × 10^6 V/m$.