$(a)\;\large\frac{a}{b}\qquad(b)\;\large\frac{b}{a}\qquad(c)\;\large\frac{a^2}{b^2}\qquad(d)\;\large\frac{b^2}{a^2}$

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Answer : (a) $\;\large\frac{a}{b}$

Explanation : Since electric fields near surfaces are equal

$\large\frac{k\;q_{1}}{a^2}=\large\frac{k\;q^2}{b^2}$

$\large\frac{q_{1}}{q_{2}}=\large\frac{a^2}{b^2}$

The ratio of potential on their surfaces

$\large\frac{k\;q_{1}}{a}\;\large\frac{b}{k\;q_{2}}=\large\frac{q_{1}}{q_{2}}\;\large\frac{b}{a}$

$ratio=\large\frac{a^2}{b^2}\;\large\frac{b}{a}=\large\frac{a}{b}\;.$

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