Two conducting spheres of radii $r_{1}$ and $r_{2}$ are at the same potential. What is the ratio of the charges on them?

\sqrt{\frac{r_{1}}{r_{2}}}

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\frac{r_{1}}{r_{2}}

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\frac{r_{1}^{2}}{r_{2}^{2}}

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\sqrt{r_{1} r_{2}}

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Solution

The correct option is B $\frac{r_{1}}{r_{2}}$
The potential on the surface of a conducting sphere having charge q and radius R is
$V = \frac{q}{4 π ε_{0} R}$
Let the charges on the surface of radii $R_{1}$ and $R_{2}$ be $q_{1}$ and $q_{2}$
$⟹ \frac{q_{1}}{4 π ε_{0} R_{1}} = \frac{q_{2}}{4 π ε_{0} R_{2}}$
$⟹ \frac{q_{1}}{R_{1}} = \frac{q_{2}}{R_{2}}$
$⟹ \frac{q_{1}}{q_{2}} = \frac{R_{1}}{R_{2}}$

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Similar questions

Two conducting spheres of radii $r_{1}$ and r₂ are at the same potential. Their charges are in the ratio of

r_{1}

r_{2}

If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

If the sum of the areas of two circles with radii $R_{1}$ and $R_{2}$ is equal to the area of a circle of radius R then

(a) $R_{1} + R_{2} = R$

(b) $R_{1} + R_{2} > R$

(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$

R_{1}

R_{2}

R_{1} > R_{2}

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