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Question: Two spheres of radius \(a\) and \(b\) respectively are charged and joined by a wire. The ratio of el...

Two spheres of radius aa and bb respectively are charged and joined by a wire. The ratio of electric field of the spheres is

A

a/ba/b

B

b/ab/a

C

a2/b2a^{2}/b^{2}

D

b2/ab^{2}/a

Answer

b/ab/a

Explanation

Solution

Joined by a wire means they are at the same potential. For same potential

kQ1a1=kQ2a2\frac{kQ_{1}}{a_{1}} = \frac{kQ_{2}}{a_{2}}

⇒ Q1Q2=ab\frac{Q_{1}}{Q_{2}} = \frac{a}{b}

Further, the electric field at the surface of the sphere having radius R and charge Q is kQR2.\frac{kQ}{R^{2}}.

∴ E1E2=kQ1/a2kQ2/b2=Q1Q2×b2a2=ba\frac{E_{1}}{E_{2}} = \frac{kQ_{1}/a^{2}}{kQ_{2}/b_{2}} = \frac{Q_{1}}{Q_{2}} \times \frac{b^{2}}{a^{2}} = \frac{b}{a}