As the two spheres are connected to each other by wire, so they have same electric potential i.e., $V_a$ = $V_b$ or $\frac{1}{4\pi\varepsilon_0} \frac{q_q}{q}$ = $\frac{1}{4\pi\varepsilon_0} \frac{q_b}{b}$ or $\frac{q_{qa}}{q_b}$ = $\frac{a}{b}$ ... (i) So, the ratio of electric fields at the surfaces of the two spheres is $\frac{E_a}{E_b}$ = $\frac{\frac{1}{4\pi\varepsilon_0} \cdot \frac{q_a}{a^2}}{\frac{1}{4\pi\varepsilon_0} \cdot \frac{q_b}{b^2}}$ = $\frac{q_a}{q_b}$ × $\frac{b^2}{a^2}$ = $\frac{a}{b} \times \frac{b^2}{a^2}$ or $\frac{E_a}{E_b}$ = $\frac{b}{a}$ If b > a, then $E_a$ > $E_b$ i.e. sphere with smaller radius produces more electric field on its surface. Hence, the charge density on the sharp and pointed ends of conductor is higher than on its flatter portions.