Two isolated metallic spheres of radii R and 2 R are charged such that both have the same charge density σ . The spheres are then connected by a thin conducting wire. If the new charge density of the

Question

Two isolated metallic spheres of radii R and 2 R are charged such that both have the same charge density σ . The spheres are then connected by a thin conducting wire. If the new charge density of the larger sphere is σ 1 ​ . The ratio σ 1 ​ to σ is

TestHub · Accepted Answer

Let the initial charge on the smaller sphere be $Q_{1}$ and on the larger sphere be $Q_{2}$ .

$Q_{1} = σ (4 π R^{2})$

$Q_{2} = σ (4 π (2 R)^{2}) = 16 π R^{2} σ$

The total charge of the system is $Q = Q_{1} + Q_{2} = 20 π R^{2} σ$ .

When the spheres are connected by a conducting wire, charge flows until their potentials become equal. Let the new charges on the smaller and larger spheres be $q_{1}$ and $q_{2}$ respectively.

Equating the potentials:

$\frac{1}{4 π ϵ _{0}} \frac{q _{1}}{R} = \frac{1}{4 π ϵ _{0}} \frac{q _{2}}{2 R}$

$q_{2} = 2 q_{1}$

By conservation of charge, $q_{1} + q_{2} = Q$ .

$q_{1} + 2 q_{1} = 20 π R^{2} σ$

$3 q_{1} = 20 π R^{2} σ$

$q_{1} = \frac{20 π R ^{2} σ}{3}$

The new charge on the larger sphere is $q_{2} = 2 q_{1} = \frac{40 π R ^{2} σ}{3}$ .

The new charge density of the larger sphere is $σ_{1} = \frac{q _{2}}{4 π ( 2 R ) ^{2}}$ .

$σ_{1} = \frac{\frac{40 π R ^{2} σ}{3}}{16 π R ^{2}} = \frac{40}{48} σ = \frac{5}{6} σ$

Therefore, the ratio $\frac{σ _{1}}{σ}$ is $\frac{5}{6}$ .

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