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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2022
  • Medium
  • MCQ

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 .

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Answer & Solution

Correct Answer

(B) \(2:1\)

Step-by-step Solution

Detailed explanation

\(V _{ A }= V _{ B }\)

\(\frac{ KQ _{ A }}{ R _{ A }}=\frac{ KQ _{ B }}{ R _{ B }}\)

\(\frac{ Q _{ A }}{ Q _{ B }}=\frac{ R _{ A }}{ R _{ B }}=\frac{1}{2}\)

\(E _{ A }=\frac{ KQ _{ A }}{ R _{ A }^{2}} ; E _{ B }=\frac{ KQ _{ B }}{ R _{ B }^{2}}\)

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

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