CBSE Class 12 · 2023 · ElectrostaticsShort Answer

Two charged conducting spheres of radii

a

and

b

are connected to each other by a wire. Find the ratio of the electric fields at their surfaces.

Answer: $\frac{E _{a}}{E _{b}} = \frac{b}{a}$ (the smaller sphere has the larger surface field).

Solution

When joined by a wire the spheres reach a common potential:

\frac{1}{4 π ε _{0}} \frac{q _{a}}{a} = \frac{1}{4 π ε _{0}} \frac{q _{b}}{b}

\frac{q _{a}}{q _{b}} = \frac{a}{b}

The surface field is

E = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}}

, so:

\frac{E _{a}}{E _{b}} = \frac{q _{a}}{q _{b}} \cdot \frac{b ^{2}}{a ^{2}} = \frac{a}{b} \cdot \frac{b ^{2}}{a ^{2}}

\frac{E _{a}}{E _{b}} = \frac{b}{a}

The surface charge density, and hence the field, is larger on the sphere of smaller radius.

CBSE Class 12 Board · 2023 · Subjective

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