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 surface of 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 fatter portions ?

Explanation

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))/(4pi in_(0) xx a_(2)) xx (b^(2) xx 4pi in_(0))/(Q_(B))`
`(E_(A))/(E_(B)) = (Q_(A))/(Q_(B)) xx (b^(2))/(a^(2))` ... (i)
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)`.