8.1 Loss of energy during redistribution of charge

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

From the above example, we are going to show that the redistribution of charge is always loss of the energy.

Initial electic potential energy, $${U_i} = \frac{1}{2}\frac{{{q_1}^2}}{{{C_1}}} + \frac{1}{2}\frac{{{q_2}^2}}{{{C_2}}}$$

Final electric potential energy, $${U_f} = \frac{1}{2}\frac{{{{({q_1} + {q_2})}^2}}}{{{C_1} + {C_2}}}$$ $$\Delta U = {U_i} - {U_f} = \frac{1}{2}\left( {\frac{{{q_1}^2}}{{{C_1}}} + \frac{{{q_2}^2}}{{{C_2}}} - \frac{{{{({q_1} + {q_2})}^2}}}{{{C_1} + {C_2}}}} \right)$$

or $$\begin{equation} \begin{aligned} \Delta U = \frac{1}{{2{C_1}{C_2}({C_1} + {C_2})}}\left[ {{q_1}^2{C_1}{C_2} + {q_1}^2{C_2}^2 + {q_2}^2{C_1}^2 + {q_2}^2{C_1}{C_2} - {q_1}^2{C_1}{C_2} - {q_2}^2{C_1}{C_2} - 2{q_1}{q_2}{C_1}{C_2}} \right] \\ = \frac{{{C_1}^2{C_2}^2}}{{2{C_1}{C_2}({C_1} + {C_2})}}\left[ {\frac{{{q_1}^2}}{{{C_1}^2}} + \frac{{{q_2}^2}}{{{C_2}^2}} - \frac{{2{q_1}{q_2}}}{{{C_1}{C_2}}}} \right] \\ = \frac{{{C_1}{C_2}}}{{2({C_1} + {C_2})}}\left[ {{V_1}^2 + {V_2}^2 - 2{V_1}{V_2}} \right] \\\end{aligned} \end{equation} $$

or$$\Delta U = \frac{1}{{2{C_1}{C_2}({C_1} + {C_2})}}{({V_1} - {V_2})^2}$$

As we know $C_1$, $ C_2$ and ${({V_1} - {V_2})^2}$ are always, positive. Since ${U_i} > {U_f}$, $i.e.$,So energy decreases. Hence energy is always lost in redistribution of charge. Now if $${V_1}={V_2}$$ then $$\Delta U = 0$$

Example: An insulated conductor initially uncharged is charged by repeated contacts with a plate which after each contact is replenished to a charge $Q$. If$ q$ is the charge on the conductor after first operation. Find the maximum charge which can be given to the conductor in this way.

Solution: Assume $C_1$, be the capacitance of plate and $C_2$ that of the conductor. After first contact charge on conductor is $q$. Therefore, charge on plate will remain $Q-q$. As we know the charge redistributes in the ratio of capacities. $\frac{{Q - q}}{q} = \frac{{{C_1}}}{{{C_2}}}$. . .(3)

Let $ q_m$ be the maximum charge which can be given to the conductor. Then flow of charge from the plate to the conductor will stop when,$${V_{conductor}}={V_{plate}}$$ $$ \frac{{{q_m}}}{{{C_2}}} = \frac{Q}{{{C_1}}}$$ $${q_m} = \left( {\frac{{{C_2}}}{{{C_1}}}} \right)Q$$

Substituting ${\frac{{{C_2}}}{{{C_1}}}}$ from Eq.(3), we get$${q_m} = \frac{{Qq}}{{Q - q}}$$

Example: Two isolated spherical conductors have radius 10 $cm $and 20 $cm $ respectively. They have charges of 16 $\mu C$ and -4 $\mu C $. Find the charges after they are connected by a conducting wire. Also find the common potential after redistribution.

Solution: Net charge $= (16-4)\mu C =12 \mu C$

Charge is distributed in the ratio of their capacities (or radius in case of spherical conductors), $i.e.$,$$\frac{{{Q_1}}}{{{Q_2}}} = \frac{{{R_1}}}{{{R_2}}} = \frac{{10}}{{20}} = \frac{1}{2}$$

By partial distribution law$$\begin{equation} \begin{aligned} {{\text{Q}}_1} = \left( {\frac{1}{{1 + 2}}} \right)(12) = 4\mu C \\ \end{aligned} \end{equation} $$

Common potential$$\begin{equation} \begin{aligned} V = \frac{{{q_1} + {q_2}}}{{{C_1} + {C_2}}} = \frac{{(12 \times {{10}^{ - 6}})}}{{4\pi {\varepsilon _ \circ }\left( {{R_1} + {R_2}} \right)}} \\ {\text{ = }}\frac{{(12 \times {{10}^{ - 6}})(9 \times {{10}^9})}}{{(30 \times {{10}^{ - 2}})}} \\ {\text{ = 3}}{\text{.6}} \times {\text{1}}{{\text{0}}^5}{\text{ volt}} \\\end{aligned} \end{equation} $$