Question

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 )
( Rightarrow C V+2 C V=20 pi R^{2} rho )
( Rightarrow 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 )

# Two metallic solid spheres of radii Rand 2R are charged such that both of them have same charge density o. If the spheres are located far away from each other and connected by a thin conducting wire, find the new charge density on bigger sphere. A) (2/3) B) (1/3) C) (5/6) D) (5/3)

Solution