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_0xxa_2)xx(b^2xx4piin_0)/Q_B`

`E_A/E_B=Q_A/(a_2)xx(b^2)/Q_B`

However, `Q_A/Q_B=(C_AV)/(C_BV)`

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

`E_A/E_B=(ab^2)/(ba^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 very small radius and a flat portion behaves as a sphere of much larger radius.Therefore, charge density on sharp and pointed ends of the conductor is much higher than on its flatter portions.