JEE Main 2025 Apr 8 Shift 2, Physics Q9: Electric Potential

JEE Main2025Apr 8, Shift 2Physics

Two metal spheres of radius $R$ and $3 R$ have same surface charge density $σ$ . If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes $σ_{1}$ and $σ_{2}$ , respectively. The ratio $\frac{σ _{1}}{σ _{2}}$ is:

A
$\frac{1}{9}$
B
$9$
C
$\frac{1}{3}$
D
$3$

Solution

When two conducting spheres are placed in contact, charge redistributes until their potentials are equal.

The potential of an isolated sphere carrying surface charge density $σ$ is $V = \frac{σ r}{ε _{0}}$ .

Equal potential requires $σ_{1} r_{1} = σ_{2} r_{2}$ , so $\frac{σ _{1}}{σ _{2}} = \frac{r _{2}}{r _{1}}$ .

With smaller radius $r_{1} = R$ and larger radius $r_{2} = 3 R$ : $\frac{σ _{1}}{σ _{2}} = \frac{3 R}{R} = 3$ , matching option (4).

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JEE Main 2025 Apr 8 Shift 2, Physics Q9: Electric Potential

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