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Two uniformly charged spherical conductors \mathrm{A} and \mathrm{B} of radii \mathrm{5 \mathrm{~mm}} and \mathrm{10 \mathrm{~mm}} are separated by a distance of \mathrm{ 2\mathrm{~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 \mathrm{ A} and \mathrm{ B} will be :

Option: 1

1: 2


Option: 2

2: 1


Option: 3

1: 1


Option: 4

1: 4


Answers (1)

best_answer

When the spheres are connated by conducting wire, their potential will be same

\mathrm{V_1=V_2 }

\mathrm{\frac{K Q_1}{R_1}=\frac{K Q_2}{R_2} }

\mathrm{\frac{Q_1}{Q_2}=\frac{R_1}{R_2}=\frac{1}{2}}

Electic field on the surface of the sphere

\mathrm{ E=\frac{K Q}{R^2} }

\mathrm{ \frac{E_1}{E_2}=\frac{Q_1 R_2^2}{Q_2 R_1^2}=\frac{2}{1} }

Electric field on the surface of the one sphere is independent of other. (as separation is large as compare to size of sphere)

Hence (2) is correct option.

Posted by

Ajit Kumar Dubey

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