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Electrostatics

JEE Main level, single correct, medium

150 of 311

Question 150 of 311

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 _______.

Show answer and solution

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

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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