# Two charged conducting spheres of radii a and b are connected to each other by a wire.

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.

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 EAbe the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio,

$\frac{E_{A}}{E_{B}}=\frac{Q_{A}}{4 \pi \in_{0} \times a_{2}} \times \frac{b^{2} \times 4 \pi \epsilon_{0}}{Q_{B}}$

$\frac{E_{A}}{E_{B}}=\frac{Q_{A}}{Q_{B}} \times \frac{b^{2}}{a^{2}}$   ...(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

$\therefore \frac{E_{A}}{E_{B}}=\frac{a}{b} \frac{b^{2}}{a^{2}}=\frac{b}{a}$

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.