Two charged conducting spheres of radii a and b are connected to each other by a conducting wire. The ratio of charges of the two spheres respectively is:

Solution

Physics sir · 3 steps

The wire forces both spheres to one potential, and V = kQ/R then fixes how the charge splits.

1. Same potential. A conducting wire lets charge flow until there is no potential difference left. So in the end both spheres sit at the same potential V.
2. Potential of a sphere. A charged conducting sphere of radius R with charge Q has potential V = kQ/R at its surface. Treat the spheres as far apart, so each one's potential comes from its own charge.

   kQ₁/a = kQ₂/b ⇒ Q₁/Q₂ = a/b

3. So. Charge is shared in proportion to the radius, so the bigger sphere holds more. The ratio is a/b, option (c).

The idea to keep. Connected conductors share one potential. Since V = kQ/R, charge goes in proportion to radius, while surface charge density goes as 1/R.

Answer from the NTA final answer key. Solution by PrepWiser.