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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2026
  • Medium
  • MCQ

Two charged conducting spheres \(S_1\) and \(S_2\) of radii \(8\) cm and \(18\) cm are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on \(S_1\) and \(S_2\) spheres are \(E_{S_1}\) and \(E_{S_2}\) respectively. The value of \(\dfrac{E_{S_1}}{E_{S_2}}\) is _______.

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Answer & Solution

Correct Answer

(D) \(\dfrac{9}{4}\)

Step-by-step Solution

Detailed explanation

When two conducting spheres are connected by a wire, charge flows until their potentials become equal.

\(V_1 = V_2\)

\(\dfrac{k Q_1}{R_1} = \dfrac{k Q_2}{R_2}\)

The electric field on the surface of a conducting sphere is given by \(E = \dfrac{k Q}{R^2}\), which can be written as \(E = \dfrac{V}{R}\).

Since \(V_1 = V_2\), we have:

\(E_{S_1} R_1 = E_{S_2} R_2\)

\(\dfrac{E_{S_1}}{E_{S_2}} = \dfrac{R_2}{R_1}\)

Given \(R_1 = 8\) cm and \(R_2 = 18\) cm:

\(\dfrac{E_{S_1}}{E_{S_2}} = \dfrac{18}{8} = \dfrac{9}{4}\)

Answer: \(\dfrac{9}{4}\)
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