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.

(B) Let $a$ and $b$ be the radii of two conducting spheres $A$ and $B$,respectively. Let $Q_A$ and $Q_B$ be the charges on them,and $V_A$ and $V_B$ be their potentials.
Since the spheres are connected by a wire,they reach the same potential,so $V_A = V_B = V$.
The potential of a conducting sphere is given by $V = \frac{Q}{4 \pi \epsilon_0 R}$.
Thus,$V = \frac{Q_A}{4 \pi \epsilon_0 a} = \frac{Q_B}{4 \pi \epsilon_0 b}$,which implies $\frac{Q_A}{Q_B} = \frac{a}{b}$.
The electric field at the surface of a sphere is $E = \frac{Q}{4 \pi \epsilon_0 R^2}$.
The ratio of the electric fields is $\frac{E_A}{E_B} = \frac{Q_A / (4 \pi \epsilon_0 a^2)}{Q_B / (4 \pi \epsilon_0 b^2)} = \frac{Q_A}{Q_B} \times \frac{b^2}{a^2}$.
Substituting $\frac{Q_A}{Q_B} = \frac{a}{b}$,we get $\frac{E_A}{E_B} = \frac{a}{b} \times \frac{b^2}{a^2} = \frac{b}{a}$.
Since surface charge density $\sigma = \frac{Q}{4 \pi R^2} = \epsilon_0 E$,we have $\sigma \propto E$. For a conductor,the potential is constant throughout. $A$ sharp point has a very small radius of curvature $(R \to 0)$,which implies a very high electric field $(E \propto 1/R)$ and consequently a very high surface charge density $\sigma$ at that point.