Question

Two charged conducting spheres of radii $a$ and $b$ are connected to each other by a wire. The ratio of electric fields at the surfaces of two spheres is

• A. $\frac{a}{b}$
• B. $\frac{b}{a}$
• C. $\frac{a^{2}}{b^{2}}$
• D. $\frac{b^{2}}{a^{2}}$

Answer 1

Answer: (B)

$\frac{b}{a}$

Answer 2

Let $q_{1}$ and $q_{2}$ be the charges and $C_{1}$ and $C_{2}$ be the capacitance of two spheres The charge flows from the sphere at higher potential to the other at lower potential, till their potentials becomes equal. After sharing, the charges on two spheres would be $\frac{q_{1}}{q_{2}}=\frac{C_{1}V}{C_{2}V} \ldots\left(i\right)$ Also $\frac{C_{1}}{C_{2}}=\frac{a}{b} \ldots\left(ii\right)$ From $\left(i\right) \frac{q_{1}}{q_{2}}=\frac{a}{b}$ Ratio of surface charge on the two spheres $\frac{\sigma_{1}}{\sigma_{2}}=\frac{q_{1}}{4\pi a^{2}}\cdot\frac{4\pi b^{2}}{q_{2}}$ $=\frac{q_{1}}{q_{2}}\cdot\frac{b^{2}}{a^{2}}=\frac{b}{a} (using \left(ii\right)$) $\therefore$ The ratio of electric fields at the surfaces of two spheres $\frac{E_{1}}{E_{2}}=\frac{\sigma_{1}}{\sigma_{2}}=\frac{b}{a}$