Q.Two identical conducting balls having positive charges $q_1$ and $q_2$ are separated by a centre to centre distance r. If they are made to touch each other and then separated to the same distance, the force between them will be :

Charge Sharing Between Conductors

Imagine you have two buckets of water at different heights. If you connect them with a pipe at the bottom, water flows from the higher bucket to the lower one until both reach the same water level. That's exactly what happens with charge and conductors — except the "height" is electric potential, and the "water" is charge.

When two conductors are connected by a thin wire, charge flows from the one at higher potential to the one at lower potential. The flow stops the instant both conductors reach the same potential. At that moment, the system is in electrostatic equilibrium.

The Precise Physics

Let conductor 1 have capacitance $C_1$ and initial charge $Q_1$, and conductor 2 have capacitance $C_2$ and initial charge $Q_2$. Before connection, their potentials are:

If $V_1 \neq V_2$, charge flows. After connection, the two conductors become a single conductor (electrically), so they must share a common potential $V_f$. The total charge is conserved:

where $Q_1'$ and $Q_2'$ are the final charges. Since both are now at the same potential $V_f$:

From these two equations, you can solve for the final charges:

And the common potential is:

This is the fundamental result: the final potential is simply the total charge divided by the total capacitance — exactly as if the two conductors had been combined into one.

What Changes and What Doesn't

Conserved: Total charge. Charge is neither created nor destroyed, only redistributed.

Not conserved: Total energy. Some energy is always lost as heat in the connecting wire (or as electromagnetic radiation). You can calculate the energy loss:

This is always positive unless $V_1 = V_2$ initially. So charge sharing is an irreversible process — you cannot get back the original separated charges without doing work.

A Concrete Example

Take a $2\ \mu\text{F}$ capacitor charged to $100\ \text{V}$ and an uncharged $3\ \mu\text{F}$ capacitor. Connect them.

Initial charges: $Q_1 = 200\ \mu\text{C}$, $Q_2 = 0$.

Total capacitance: $C_1 + C_2 = 5\ \mu\text{F}$.

Final potential: $V_f = \frac{200}{5} = 40\ \text{V}$.

Final charges: $Q_1' = 2 \times 40 = 80\ \mu\text{C}$, $Q_2' = 3 \times 40 = 120\ \mu\text{C}$. …