Two metal spheres of radii $r_1$ and $r_2\ (>r_1)$ having charges $q_1$ and $q_2$ respectively, kept in air, are brought in contact. Which of the following statements is $not$ correct? (A) The total charge of the two spheres is conserved. (B) Both spheres attain the same potential. (C) The final potential of the system equals $\dfrac{1}{4\pi\varepsilon_0}\dfrac{(q_1+q_2)}{(r_1+r_2)}$. (D) The final potential of the system equals $\dfrac{1}{4\pi\varepsilon_0}\dfrac{(q_1+q_2)(r_1+r_2)}{r_1 r_2}$.

Two metal spheres of radii r₁ and r₂(>r₁) having charges q₁ and q₂ respectively, kept in air, are brought in c

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r_1

CBSECBSE Class XII Board 2026MCQ· 1mImportance★★★★★

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Concept understanding — Common Potential

Common Potential: What Happens When Two Charged Conductors Meet

Imagine you have two water tanks at different heights. One is full, the other half-full. If you connect them with a pipe at the bottom, water flows from the higher tank to the lower one until the water level in both is the same. That final level is not the average of the two starting levels — it depends on the size (cross-section) of each tank.

Charged conductors behave exactly the same way. When you connect two conductors with a thin wire, charge flows from the one at higher potential to the one at lower potential. The flow stops when both are at the same potential — that final potential is called the common potential.

The Precise Statement

When two isolated charged conductors of capacitances $C_1$ and $C_2$ , initially at potentials $V_1$ and $V_2$ , are connected by a conducting wire, charge redistributes between them until they acquire a common potential $V$ given by:

V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}

This is not an average of $V_1$ and $V_2$ — it is a capacitance-weighted average.

Why This Formula? The Physics in Two Steps

Step 1: Total charge is conserved. Before connection, the total charge on the two conductors is:

Q_{\text{total}} = C_1 V_1 + C_2 V_2

After connection, no charge is lost — it just moves around. So the total charge after connection is the same.

Step 2: After connection, both are at the same potential $V$ . So the charge on each becomes:

Q_1' = C_1 V, \quad Q_2' = C_2 V

Conservation of charge gives:

C_1 V + C_2 V = C_1 V_1 + C_2 V_2

Factor $V$ on the left:

V (C_1 + C_2) = C_1 V_1 + C_2 V_2

And you have the formula.

Important

The common potential is not the arithmetic mean $\frac{V_1+V_2}{2}$ . It is closer to the potential of the larger capacitor because a larger capacitance can "absorb" or "supply" more charge with a smaller change in potential.

A Concrete Example

Take $C_1 = 2\ \mu\text{F}$ at $V_1 = 10\ \text{V}$ , and $C_2 = 8\ \mu\text{F}$ at $V_2 = 5\ \text{V}$ .

Total charge: $Q = (2)(10) + (8)(5) = 20 + 40 = 60\ \mu\text{C}$ .

Total capacitance after connection: $C = 2 + 8 = 10\ \mu\text{F}$ .

Common potential: $V = \frac{60}{10} = 6\ \text{V}$ .

Notice: $6\ \text{V}$ is much closer to $5\ \text{V}$ (the potential of the larger $8\ \mu\text{F}$ capacitor) than to $10\ \text{V}$ . The larger capacitor dominates the final potential — just like a wider water tank dominates the final water level.

Watch out

A common mistake is to think the common potential is the average of the two potentials. That would give $7.5\ \text{V}$ here — which is wrong. Always use the capacitance-weighted formula.

What Happens to the Charge on Each?

After connection, the charge on each conductor is:

Q_1' = C_1 V, \quad Q_2' = C_2 V

In the example above: $Q_1' = (2)(6) = 12\ \mu\text{C}$ (it lost $8\ \mu\text{C}$ ), and $Q_2' = (8)(6) = 48\ \mu\text{C}$ (it gained $8\ \mu\text{C}$ ). Charge flows from the smaller capacitor (higher initial potential) to the larger one (lower initial potential) until balance is reached.

The Big Picture

Common potential is simply electrostatic equilibrium — the condition that no net electric field exists inside the connecting wire, which happens only when both conductors are at the same potential. The formula is a direct consequence of charge conservation and the definition of capacitance $Q = CV$ .

Common potential is a closely related, frequently tested formula in the NCERT Class 12 Physics Electrostatics chapter, and 'common potential formula two capacitors' or 'common potential important questions class 12 physics' are typical exam-prep searches. This capacitance-weighted-average result is a classic JEE Main and NEET numerical, often confused with a simple average of the two potentials.

When two conducting spheres are brought into contact, charge redistributes until both reach the same potential. The total charge is conserved, and the common potential is $V = \frac{1}{4\pi\varepsilon_0} \frac{q_1+q_2}{r_1+r_2}$ . Option (D) is the incorrect statement.

The key idea here is that conductors in contact form a single conductor — charge flows until the electric potential is the same everywhere on the combined surface. For two isolated spheres far apart, each has its own potential. When they touch, they become one equipotential system.

Let’s recall the potential of an isolated conducting sphere of radius $r$ carrying charge $q$ :

V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}

This formula holds because the sphere’s charge resides on its surface, and for points outside (or on the surface), the sphere behaves like a point charge at its centre.

Now, when the two spheres are brought into contact, charge flows between them until both spheres are at the same potential $V_f$ . The total charge is conserved — no charge is created or destroyed, only redistributed.

Charge conservation If the final charges on the spheres are $q_1'$ and $q_2'$ , then

q_1' + q_2' = q_1 + q_2

This is always true. So statement (A) is correct.

Equal potentials after contact Since they are conductors in contact, the final potential of each sphere must be the same:

\frac{1}{4\pi\varepsilon_0} \frac{q_1'}{r_1} = \frac{1}{4\pi\varepsilon_0} \frac{q_2'}{r_2}

Hence $q_1'/r_1 = q_2'/r_2$ . Statement (B) is correct.

Finding the common potential From the equal-potential condition: $q_1' = \frac{r_1}{r_2} q_2'$ . Using charge conservation:

\frac{r_1}{r_2} q_2' + q_2' = q_1 + q_2 \quad\Rightarrow\quad q_2' \left( \frac{r_1 + r_2}{r_2} \right) = q_1 + q_2

So $q_2' = \frac{r_2}{r_1+r_2}(q_1+q_2)$ and similarly $q_1' = \frac{r_1}{r_1+r_2}(q_1+q_2)$ .

The common potential is then:

V_f = \frac{1}{4\pi\varepsilon_0} \frac{q_1'}{r_1} = \frac{1}{4\pi\varepsilon_0} \frac{1}{r_1} \cdot \frac{r_1}{r_1+r_2}(q_1+q_2) = \frac{1}{4\pi\varepsilon_0} \frac{q_1+q_2}{r_1+r_2}

This matches statement (C). So (C) is correct.

Checking statement (D) Statement (D) claims:

V_f = \frac{1}{4\pi\varepsilon_0} \frac{(q_1+q_2)(r_1+r_2)}{r_1 r_2}

This is clearly different from the expression we derived. It has $(r_1+r_2)$ in the numerator instead of the denominator — a factor of $(r_1+r_2)^2$ off. So (D) is false.

Watch out

A common mistake is to think the final potential is the average of the initial potentials, or to incorrectly combine radii. The correct formula has $(r_1+r_2)$ in the denominator, not the numerator.

Tip

Notice that the final potential is the same as if the total charge were placed on a single sphere of radius $r_1+r_2$ . This makes physical sense: when two spheres touch, they behave like one larger conductor whose effective radius is the sum of the individual radii (for the purpose of potential calculation, assuming they are far apart initially).

✓Final answer

The statement that is not correct is option (D) .

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

CBSE 2026Set 55/2/11 markMCQ
$r_1$
›Reveal solutionSolution
When two conducting spheres are brought into contact, charge redistributes until both reach the same potential. The total charge is conserved, and the common potential is $V = \frac{1}{4\pi\varepsilon_0} \frac{q_1+q_2}{r_1+r_2}$ . Option (D) is the incorrect statement.

The key idea here is that conductors in contact form a single conductor — charge flows until the electric potential is the same everywhere on the combined surface. For two isolated spheres far apart, each has its own potential. When they touch, they become one equipotential system.

Let’s recall the potential of an isolated conducting sphere of radius $r$ carrying charge $q$ :
$V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$
This formula holds because the sphere’s charge resides on its surface, and for points outside (or on the surface), the sphere behaves like a point charge at its centre.

Now, when the two spheres are brought into contact, charge flows between them until both spheres are at the same potential $V_f$ . The total charge is conserved — no charge is created or destroyed, only redistributed.

Charge conservation If the final charges on the spheres are $q_1'$ and $q_2'$ , then

$q_1' + q_2' = q_1 + q_2$
This is always true. So statement (A) is correct.

Equal potentials after contact Since they are conductors in contact, the final potential of each sphere must be the same:

$\frac{1}{4\pi\varepsilon_0} \frac{q_1'}{r_1} = \frac{1}{4\pi\varepsilon_0} \frac{q_2'}{r_2}$
Hence $q_1'/r_1 = q_2'/r_2$ . Statement (B) is correct.

Finding the common potential From the equal-potential condition: $q_1' = \frac{r_1}{r_2} q_2'$ . Using charge conservation:

$\frac{r_1}{r_2} q_2' + q_2' = q_1 + q_2 \quad\Rightarrow\quad q_2' \left( \frac{r_1 + r_2}{r_2} \right) = q_1 + q_2$
So $q_2' = \frac{r_2}{r_1+r_2}(q_1+q_2)$ and similarly $q_1' = \frac{r_1}{r_1+r_2}(q_1+q_2)$ .

The common potential is then:
$V_f = \frac{1}{4\pi\varepsilon_0} \frac{q_1'}{r_1} = \frac{1}{4\pi\varepsilon_0} \frac{1}{r_1} \cdot \frac{r_1}{r_1+r_2}(q_1+q_2) = \frac{1}{4\pi\varepsilon_0} \frac{q_1+q_2}{r_1+r_2}$
This matches statement (C). So (C) is correct.

Checking statement (D) Statement (D) claims:

$V_f = \frac{1}{4\pi\varepsilon_0} \frac{(q_1+q_2)(r_1+r_2)}{r_1 r_2}$
This is clearly different from the expression we derived. It has $(r_1+r_2)$ in the numerator instead of the denominator — a factor of $(r_1+r_2)^2$ off. So (D) is false.

Watch out

A common mistake is to think the final potential is the average of the initial potentials, or to incorrectly combine radii. The correct formula has $(r_1+r_2)$ in the denominator, not the numerator.

Tip

Notice that the final potential is the same as if the total charge were placed on a single sphere of radius $r_1+r_2$ . This makes physical sense: when two spheres touch, they behave like one larger conductor whose effective radius is the sum of the individual radii (for the purpose of potential calculation, assuming they are far apart initially).

✓Final answer
$The statement that is not correct is option (D) .$

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