# Question #df5d0

Jan 19, 2017

Given that the radius of smaller sphere is $R$ and that of larger sphere is $2 R$ and their common surface charge density is $\rho$

So initial charge on smaller sphere ${q}_{1} = 4 \pi {R}^{2} \rho$

And initial charge on larger sphere ${Q}_{1} = 16 \pi {R}^{2} \rho$

Now we know that the capacitance of a sphere is proportional to its radius. So the ratio of their capacitances (smaller:larger) will be $R : 2 R = 1 : 2$

Let the capacitance of smaller one be ${C}_{s} = C$ and that of larger one be ${C}_{l} = 2 C$.

After they are connected by a thin conducting wire remaining at a large distance apart their charge will be redistributed until they gain same potential.

Let this common potential be V.

By principle of conservation of charge we can write

${C}_{s} V + {C}_{l} V = {q}_{1} + {Q}_{1} = 20 \pi {R}^{2} \rho$

$\implies C V + 2 C V = 20 \pi {R}^{2} \rho$

$\implies V = \frac{20 \pi {R}^{2} \rho}{3 C}$

So changed charge on large capacitor $= {C}_{l} V = 2 C \times \frac{20 \pi {R}^{2} \rho}{3 C} = \frac{40 \pi {R}^{2} \rho}{3}$

So the changed surface charge density on larger capacitor
$= \frac{40 \pi {R}^{2} \rho}{3 \times 16 \pi {R}^{2}} = \frac{5}{6} \rho$