JEE Main & Advanced Physics Electrostatics & Capacitance Redistribution of Charges and Loss of Energy

When two charged conductors joined together through a conducting wire, charge begins to flow from one conductor to another from higher potential to lower potential.

This flow of charge stops when they attain the same potential.

Due to flow of charge, loss of energy also takes place in the form of heat through the connecting wire.

Suppose there are two spherical conductors of radii \[{{r}_{1}}\] and \[{{r}_{2}},\]having charge \[{{Q}_{1}}\] and \[{{Q}_{2}},\] potential \[{{V}_{1}}\] and \[{{V}_{2}},\]energies \[{{U}_{1}}\] and \[{{U}_{2}}\] and capacitance \[{{C}_{1}}\] and \[{{C}_{2}}\] respectively.

If these two spheres are connected through a conducting wire, then alteration of charge, potential and energy takes place.

(1) New charge : According to the conservation of charge

\[{{Q}_{1}}+{{Q}_{2}}=Q_{1}^{'}+Q_{2}^{'}=Q\] (say), also  \[\frac{Q_{1}^{'}}{Q_{2}^{'}}=\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\]

\[\Rightarrow \]\[Q_{2}^{'}=Q\,\left[ \frac{{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}} \right]\] and similarly \[Q_{1}^{'}=Q\,\left[ \frac{{{r}_{1}}}{{{r}_{1}}+{{r}_{2}}} \right]\]

(2) Common potential : Common potential

\[(V)=\frac{Total charge}{Total capacity}\]\[=\frac{{{Q}_{1}}+{{Q}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{Q_{1}^{'}+Q_{2}^{'}}{{{C}_{1}}+{{C}_{2}}}\]\[=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]

(3) Energy loss : The loss of energy due to redistribution of charge is given by

\[\Delta \,U={{U}_{i}}-{{U}_{f}}=\frac{{{C}_{1}}{{C}_{2}}}{2({{C}_{1}}+{{C}_{2}})}\,{{({{V}_{1}}-{{V}_{2}})}^{2}}\]