Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

#### Solution

let a be the radius of a sphere A, Q_{A}_{ }be the charge on the sphere, and C_{A} be the capacitance of the sphere.

Let b be the radius of a sphere B, Q_{B} be the charge on the sphere, and C_{B} be the capacitance of the sphere.

Since the two spheres are connected with a wire, their potential (V) will become equal.

Let E_{A }be the electric field of sphere A and E_{B} be the electric field of sphere B.

Therefore, their ratio,

`"E"_"A"/"E"_"B" = "Q"_"A"/(4piin_0 xx "a"_2) xx ("b"^2 xx 4piin_0)/"Q"_"B"`

`"E"_"A"/"E"_"B" = "Q"_"A"/"Q"_"B" xx ("b"^2)/"a"^2` ......(1)

However, `"Q"_"A"/"Q"_"B" = ("C"_"A""V")/("C"_"B""V")`

And, `("C"_"A")/("C"_"B") = "a"/"b"`

∴ `"Q"_"A"/"Q"_"B" = "a"/"b"` .......(2)

Putting the value of (2) in (1), we obtain

∴ `"E"_"A"/"E"_"B" = "a"/"b""b"^2/"a"^2 = "b"/"a"`

Therefore, the ratio of electric fields at the surface is `"b"/"a"`.

A sharp and pointed end can be treated as a sphere of a very small radius and a flat portion behaves as a sphere of a much larger radius. Therefore, the charge density on sharp and pointed ends of the conductor is much higher than on its flatter portions.