Question

Two charged conducting spheres of radii a and b are connected toeach other by a wire. What is the ratio of electric fields at the surfacesof the two spheres? Use the result obtained to explain why chargedensity on the sharp and pointed ends of a conductor is higherthan on its flatter portions.

Answer 1

Given: The radii of the two charged conducting spheres are a and b.

The ratio of electric fields of sphere A and B is given as,

E A E B = Q A 4π ε 0 a 2 × 4π ε 0 b 2 Q B

Where, the charge on sphere A is Q A , the charge on sphere B is Q B the electric field of sphere A is E A and the electric field of sphere B is E B .

Further simplify the above equation as,

E A E B = b 2 a 2 × Q A Q B (1)

The ratio of charges is given as,

Q A Q B = C A V C B V

Where, the capacitance of sphere A is C A , the capacitance of sphere B is C B and the potential is V.

Further simplify the above equation as,

Q A Q B = C A C B (2)

Since, C A C B = a b .

By substituting the given value in the equation (2), we get

Q A Q B = a b (3)

By substituting the result of equation (3) in equation (1), we get

E A E B = b 2 a 2 × a b = b a

Thus, the ratio of electric fields at the surface of s is b a and the sharp and pointed end of a conductor can be considered as a sphere of very small radius. The flat portion of the conductor behaves as a sphere of much larger radius so the charge density on the sharp and pointed ends of a conductor is much higher than on the flatter portions of the conductors.