JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Electrostatics

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Two uniformly charged spherical conductors A and B of radii 5 mm and 10 mm are separated by a distance of 2 cm. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere A and B will be:

(A) $1 : 2$
(B) $2 : 1$ ✓ Correct
(C) $1 : 1$
(D) $1 : 4$

Answer: (B)

Solution

When connected by a wire, the spheres reach the same potential:

V_{A} = V_{B}

, so

\frac{k Q _{A}}{R _{A}} = \frac{k Q _{B}}{R _{B}}

, giving

\frac{Q _{A}}{Q _{B}} = \frac{R _{A}}{R _{B}} = \frac{5}{10} = \frac{1}{2}

. The electric field at the surface is

E = \frac{k Q}{R ^{2}}

. Therefore,

\frac{E _{A}}{E _{B}} = \frac{Q _{A}}{Q _{B}} \times \frac{R _{B}^{2}}{R _{A}^{2}} = \frac{1}{2} \times \frac{( 10 ) ^{2}}{( 5 ) ^{2}} = \frac{1}{2} \times 4 = 2

, giving ratio

2 : 1

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Electrostatics

Solution

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