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

A

a/b

B

b/a

C

a²/b²

D

b²/a²

Step-by-Step Solution

When two charged conductors are connected by a wire, charge flows between them until they acquire the same electric potential (V1=V2V_1 = V_2) [Source 53].

  1. Potential Condition: For spheres of radius aa and bb with charges QaQ_a and QbQ_b: Va=VbkQaa=kQbbQaQb=abV_a = V_b \Rightarrow \frac{kQ_a}{a} = \frac{kQ_b}{b} \Rightarrow \frac{Q_a}{Q_b} = \frac{a}{b} [Source 41].

  2. Electric Field Ratio: The electric field at the surface of a conducting sphere is given by E=kQR2E = \frac{kQ}{R^2} [Source 41]. The ratio of the fields is: EaEb=kQa/a2kQb/b2=(QaQb)×(b2a2)\frac{E_a}{E_b} = \frac{kQ_a / a^2}{kQ_b / b^2} = \left( \frac{Q_a}{Q_b} \right) \times \left( \frac{b^2}{a^2} \right)

  3. Substitution: Substituting the charge ratio derived in step 1: EaEb=(ab)×(b2a2)=ba\frac{E_a}{E_b} = \left( \frac{a}{b} \right) \times \left( \frac{b^2}{a^2} \right) = \frac{b}{a}

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