Two charged metallic spheres are joined by a very thin metal wire. If the radius of the larger sphere is four times that of the smaller sphere, the electric field near the larger sphere is (a) twice that near the smaller sphere (b) quarter of that near the smaller sphere (c) same as that near the smaller sphere (d) half of that near smaller sphere

Answer: (b). Joined spheres are at equal potential, so q ∝ R and σ ∝ 1/R. Field just outside a conductor E = (σ)/(ε₀) ∝ 1/R. Rbig = 4Rsmall ⇒ Ebig = 1/4Esmall So the answer is (b) quarter of that near the smaller sphere.

Two charged metallic spheres are joined by a very thin metal wire. If… | MHT CET 2021 (23 Sep, Shift 1)

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Electric Potential & Equipotentials

Two charged metallic spheres are joined by a very thin metal wire. If the radius of the larger sphere is four times that of the smaller sphere, the electric field near the larger sphere is

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Answer: (b)

Joined spheres are at equal potential, so $q \propto R$ and $\sigma \propto \frac{1}{R}$ .
Field just outside a conductor $E = \frac{\sigma}{\varepsilon_0} \propto \frac{1}{R}$ .
$R_{big} = 4R_{small} \Rightarrow E_{big} = \frac{1}{4}E_{small}$
So the answer is (b) quarter of that near the smaller sphere.

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