Two metallic spheres of radii $1\ \text{cm}$ and $3\ \text{cm}$ are given charges of $-1\times10^{-2}\ \text{C}$ and $5\times10^{-2}\ \text{C}$ respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is (AIIPMT 2012) (a) $3\times10^{-2}\ \text{C}$ (b) $4\times10^{-2}\ \text{C}$ (c) $1\times10^{-2}\ \text{C}$ (d) $2\times10^{-2}\ \text{C}$

Two metallic spheres of radii 1cm and 3cm are given charges of -1×10⁻²C and 5×10⁻²C respectively. If these are

I. Multiple Choice Questions · Q15

1\ \text{cm}

(a)

3\times10^{-2}\ \text{C}

(b)

4\times10^{-2}\ \text{C}

(c)

1\times10^{-2}\ \text{C}

(d)

2\times10^{-2}\ \text{C}

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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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Step 1. Total charge before connecting: $Q_{\text{total}} = -1\times10^{-2} + 5\times10^{-2} = 4\times10^{-2}\ \text{C}$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$q_1$
(a) zero
(b) less than before
(c) more than before
(d) same as before
›Reveal solutionSolution

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

Working

Original force: $F_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ .

Since the balls are identical conductors, on touching, the total charge redistributes equally: each now carries $q' = \dfrac{q_1+q_2}{2}$ .

After separating back to $r$ : $F_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q'^2}{r^2}$ .

Comparing:

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

…

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CBSE 2024Set ANNUAL1 markMCQ
$q_1$
(a) more than before
(b) less than before
(c) zero
(d) same as before
›Reveal solutionSolution

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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CBSE 2022Set ANNUAL1 markMCQ
$-1 \times 10^{-2}$
(a) $1 \times 10^{-2}$ C
(b) $3 \times 10^{-2}$ C
(c) $2 \times 10^{-2}$ C
(d) $4 \times 10^{-2}$ C
›Reveal solutionSolution

Total charge is conserved when the spheres are connected; since they reach a common potential, the final charge on each divides in the ratio of their radii, giving $3\times10^{-2}$ C on the bigger (3 cm) sphere.

Working

Total charge before connection (conserved):

$Q=Q_1+Q_2=(-1\times10^{-2})+(5\times10^{-2})=4\times10^{-2}\ \text{C}$

When connected by a conducting wire, both spheres reach a common potential $V$ . For an isolated sphere, $V=\dfrac{kQ}{r}$ , so at common potential:

$\dfrac{Q_1'}{r_1}=\dfrac{Q_2'}{r_2}\ \Rightarrow\ \dfrac{Q_1'}{1}=\dfrac{Q_2'}{3}\ \Rightarrow\ Q_2'=3Q_1'$

Using charge conservation: …

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