Discussion
Two metallic spheres of radii in the ratio 1 : 2 are charged and joined by a connecting wire. The ratio of electric field intensities at the surfaces of the spheres is
Answer: A
The ratio is 2 : 1.
Joined by a wire, the spheres reach the same potential: \(\frac{kq_{1}}{r_{1}} = \frac{kq_{2}}{r_{2}}\), so \(q \propto r\).
\[ \begin{aligned} E &= \frac{kq}{r^{2}} \propto \frac{r}{r^{2}} = \frac{1}{r} \\ \frac{E_{1}}{E_{2}} &= \frac{r_{2}}{r_{1}} = \frac{2}{1} \end{aligned} \]
The smaller sphere has the stronger surface field, which is why charge density is greatest at sharp points of a conductor. The option 1 : 2 is the ratio of the charges, not of the fields.
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