Two identical conducting balls having positive charges $q_1$ and $q_2$ are separated by a center to center distance 'r'. If they are made to touch each other and then separated to the same distance, the force between them will be : (a) more than before (b) less than before (c) zero (d) same as before

Two identical conducting balls having positive charges q₁ and q₂ are separated by a center to center distance

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q_1

(a) more than before

(b) less than before

(d) same as before

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Concept understanding — Charge Sharing Between Conductors

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.

Note

The connecting wire is assumed to have negligible capacitance, so it doesn't store any charge itself — it's just a path for charge to move.

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:

V_1 = \frac{Q_1}{C_1}, \quad V_2 = \frac{Q_2}{C_2}

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:

Q_1 + Q_2 = Q_1' + Q_2'

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

V_f = \frac{Q_1'}{C_1} = \frac{Q_2'}{C_2}

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

Q_1' = \frac{C_1}{C_1 + C_2} (Q_1 + Q_2), \quad Q_2' = \frac{C_2}{C_1 + C_2} (Q_1 + Q_2)

And the common potential is:

V_f = \frac{Q_1 + Q_2}{C_1 + C_2}

V_f = \frac{Q_{\text{total}}}{C_{\text{total}}}

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:

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

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.

Watch out

A common mistake: assuming total energy is conserved. It is not. Only charge is conserved. The lost energy goes into heating the wire or radiating.

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

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After touching, the charge redistributes equally between the identical balls; since the AM-GM inequality gives $\left(\dfrac{q_1+q_2}{2}\right)^2 \ge q_1q_2$ , the resulting force is generally larger than before (equal only when $q_1=q_2$ ).

Working

Original force between the two balls, separated by $r$ :

$F_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$

Since the balls are identical conducting spheres, when brought into contact the total charge $q_1+q_2$ redistributes equally between them (identical spheres share charge equally regardless of the individual initial values). Each now carries

$q' = \dfrac{q_1+q_2}{2}$

After separating back to distance $r$ , the new force is

$F_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q'^2}{r^2} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(q_1+q_2)^2}{4r^2}$

Comparing $q'^2$ to $q_1q_2$ :

$q'^2 - q_1q_2 = \dfrac{(q_1+q_2)^2}{4} - q_1q_2 = \dfrac{(q_1-q_2)^2}{4} \ge 0$ …

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Q.Two identical conducting balls having positive charges q1q_1q1​ and q2q_2q2​ are separated by a center to center distance 'r'. If they are made to touch each other and then separated to the same distance, the force between them will be :

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

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