JEE Main · 2006 · 30 Apr 2006 · Offline · Electrostatics

Two spherical conductors

A

and

B

of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surface of spheres

A

and

B

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

Answer: (D)

Solution

Joining the spheres with a wire brings them to a common potential

V

. For a sphere of radius

r

holding charge

q

, the surface field is

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

and the potential is

V = \frac{1}{4 π ε _{0}} \frac{q}{r}

, so

E = \frac{V}{r}

. With the same

V

on both,

\frac{E _{A}}{E _{B}} = \frac{r _{B}}{r _{A}} = \frac{2}{1}

. The surface fields are therefore in the ratio

2 : 1

, the smaller sphere carrying the stronger field.

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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JEE Main · 2006 · 30 Apr 2006 · Offline · Electrostatics

Solution

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