Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres a and B is?

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Solution

Spherical conductors
A spherical conductor's surface is evenly distributed by the mutual repulsion of its excess positive charges.
The resulting electric field is zero inside and perpendicular to the surface.
The field outside the conductor is the same as a point charge with an extra charge at its center.
Step 1: Given Data
Two evenly charged spherical conductors A and B, with radii of 1 mm and 2 mm, are spaced 5 cm apart.
After connection, $V 1 = V 2 $
$\Rightarrow K \frac{Q_{1}}{r_{1}} = K \frac{Q_{2}}{r_{2}}$
$\Rightarrow \frac{Q_{1}}{r_{1}} = \frac{Q_{2}}{r_{2}}$
Step 2: The ratio of electric fields
$\frac{E_{1}}{E_{2}} = \frac{K \frac{Q_{1}}{r_{1}}}{K \frac{Q_{2}}{r_{2}}}$
$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{Q_{1}}{{r_{1}}^{2}} \times \frac{{r_{2}}^{2}}{Q_{2}}$
$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{r_{1}}{{r_{1}}^{2}} \times \frac{{r_{2}}^{2}}{r_{2}} \Rightarrow \frac{E_{1}}{E_{2}} = \frac{r_{2}}{r_{1}} = \frac{2}{1}$
The produced effects may be disregarded because the gap between the spheres is wide compared to their diameters.
Therefore, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is 2:1

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