JEE MAIN - Physics (2022 - 26th July Evening Shift - No. 5)
Two uniformly charged spherical conductors $$A$$ and $$B$$ of radii $$5 \mathrm{~mm}$$ and $$10 \mathrm{~mm}$$ are separated by a distance of $$2 \mathrm{~cm}$$. 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 $$A$$ and $$B$$ will be :
1 : 2
2 : 1
1 : 1
1 : 4
Explanation
After connection
$${\sigma _1}{R_1} = {\sigma _2}{R_2}$$
Now $$E = {\sigma \over {{\varepsilon _0}}}$$
$$ \Rightarrow {{{E_1}} \over {{E_2}}} = {{{\sigma _1}} \over {{\sigma _2}}} = {{{R_2}} \over {{R_1}}} = {2 \over 1}$$
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