Question

Two spherical conductors of radii $R_{1}$ and $R_{2}$ are separated by a distance much larger than the radius of either sphere. The spheres are connected by a conducting wire as shown in figure. If the charges on the spheres in equilibrium are $q_{1}$ and $q_{2}$, respectively, what is the ratio of the field strength at the surfaces of the spheres?

• A. $R_{2} / R_{1}$
• B. $R_{2}^{2} / R_{1}^{2}$
• C. $R_{1} / R_{2}$
• D. $R_{1}^{2} / R_{2}^{2}$

Answer 1

Answer: (A)

Their potential will be same,
i.e., $V_{1}=V_{2}$ or $\frac{k q_{1}}{R_{1}}=\frac{k q_{2}}{R_{2}}$
or $\frac{q_{1}}{q_{2}}=\frac{R_{1}}{R_{2}}$
or $\frac{E_{1}}{E_{2}}=\frac{k q_{1} / R_{1}^{2}}{k q_{2} / R_{2}^{2}}=\frac{q_{1}}{q_{2}}\left(\frac{R_{2}}{R_{1}}\right)^{2} $
$=\frac{R_{1}}{R_{2}}\left(\frac{R_{2}}{R_{1}}\right)^{2}=\frac{R_{2}}{R_{1}}$