8.0 Redistribution of Charge

8.1 Loss of energy during redistribution of charge

We can understand this by following example. We take two vessels of different size filled with some liquid up to different heights. Initially these are joined by a closed valve. When the valve is open the level in both the vessels become equal but the volume of liquid in the right side vessel is more than the liquid in the left side vessel. This is because the base area (or capacity) of this vessel is greater.

Now, let two conductor of capacitance ${C_1}$ and ${C_2}$ have charges ${q_1}$ and ${q_2}$ respectively when they are joined together by a conducting wire, charge redistributes in these conductors in the ratio of their capacities. Charge redistributes till potential of both the conductors become equal.

Thus, let ${Q_1}$ and ${Q_2} $ be the final charges on the conductors, then $${Q_1} \propto {C_1}$$ or $$\begin{equation} \begin{aligned} {Q_2} \propto {C_2}\\ \frac{{{Q_1}}}{{{Q_2}}} = \frac{{{C_1}}}{{{C_2}}} \\\end{aligned} \end{equation} $$

and if they are spherical conductors,then $$\frac{{{C_1}}}{{{C_2}}} = \frac{{{R_1}}}{{{R_2}}}$$$$\frac{{{Q_1}}}{{{Q_2}}} = \frac{{{C_1}}}{{{C_2}}} = \frac{{{R_1}}}{{{R_2}}}$$

Since, the total charge is ${q_1}+{q_2}$. Therefore,$${Q_1} = \left( {\frac{{{C_1}}}{{{C_1} + {C_2}}}} \right)({q_1} + {q_2}) = \left( {\frac{{{R_1}}}{{{R_1} + {R_2}}}} \right)({q_1} + {q_2})$$

and $${Q_1} = \left( {\frac{{{C_2}}}{{{C_1} + {C_2}}}} \right)({q_1} + {q_2}) = \left( {\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right)({q_1} + {q_2})$$

If we take spherical conductors then potential will be equal after connecting by a wire and can be written as $$\frac{{{q_1}}}{{4\pi {\varepsilon _ \circ }{R_1}}} = \frac{{{q_2}}}{{4\pi {\varepsilon _ \circ }{R_2}}}$$

or $${\sigma _1}{R_1} = {\sigma _2}{R_2}$$

where ${\sigma _1}$ and ${\sigma _2}$ are charge densities on the two spheres. From here we can conclude that the sphere having smaller radius has larger charge density to maintain the same potential.

Now let a single conductor with a non spherical shape. If a charge is given to this conductor, the charge density will not be uniform on the entire surface. A portion where the surface is more "flat" may be considered as part of a sphere of larger radius. The charge density at such portions where the surface is more curved, the charge density will be larger. More precisely, the charge density will be larger where the radius of curvature is small.

If the conductor has a pointed shape like a needle and a charge is given to it, the charge density at the pointed end will be very high. Correspondingly, the electric field near these pointed ends will be very high which may cause dielectric breakdown in air. The charge may jump from the conductor to the air because of increased conductivity of the air. Often this discharge of air is accompanied by a visible glow surrounding the pointed end.

This phenomenon is called Corona Discharge.