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 _______.
Pick an option to check your answer.
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: \(\dfrac{9}{4}\)
\(V_1 = V_2\)
\(\dfrac{k Q_1}{R_1} = \dfrac{k Q_2}{R_2}\)
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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