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 r

Question

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

TestHub · Accepted Answer

When two conductors are connected by a conducting wire, their potentials become equal. Therefore, $V_{A} = V_{B}$
The potential of a spherical conductor is given by, $V = \frac{k Q}{R}$
$\Rightarrow \frac{k Q_{A}}{R_{A}} = \frac{k Q_{B}}{R_{B}}$
$\Rightarrow Q \propto R$
Now the electric field near the surface of the conductor will be, $E = \frac{k Q}{R^{2}}$
$\Rightarrow \frac{E_{A}}{E_{B}} = \frac{\frac{k Q_{A}}{R_{A}^{2}}}{\frac{k Q_{B}}{R_{B}^{2}}} = \frac{R_{B}}{R_{A}}$
$\Rightarrow \frac{E_{A}}{E_{B}} = \frac{10}{5} = \frac{2}{1}$

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