Question

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.

Answer 1

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,

$\frac{E_A}{E_B}=\frac{Q_A}{4\pi {\in }_0\times a_2}\times \frac{b^2\times 4\pi {\in }_0}{Q_B}$

$\frac{E_A}{E_B}=\frac{Q_A}{Q_B}\times \frac{b^2}{a^2}$ ...(1)

However, $\frac{Q_A}{Q_B}=\frac{C_{A}V}{C_{B}V}$

And $\frac{C_A}{C_B}=\frac{a}{b}$

$\therefore \frac{Q_A}{Q_B}=\frac{a}{b}$ ...(2)

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

$\therefore \frac{E_A}{E_B}=\frac{a}{b}\frac{b^2}{a^2}=\frac{b}{a}$

Therefore, the ratio of electric fields at the surface is $\frac{b}{a}$.

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.