The conducting sphere of radii $r_{1}$ and $r_{2}$ are at the same potential. What is the ratio of their charges?

r_{1} : r_{2}

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r_{2} : r_{1}

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r_{1}^{2} : r_{2}^{2}

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r_{2}^{2} : r_{1}^{2}

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Solution

The correct option is A $r_{1} : r_{2}$
The conducting sphere of radii $= r_{1}$ and $r_{2}$ .
Those have same potential the ratio of their charges $V_{1} = V_{2}$
$V_{1} = \frac{1}{4 π ϵ_{0}} \frac{q_{1}}{r_{1}}$
$V_{2} = \frac{1}{4 π ϵ_{0}} \frac{q_{2}}{r_{2}}$
Since, $V_{1} = V_{2}$
then,
$\frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}}$
or, $\frac{q_{1}}{q_{2}} = \frac{r_{1}}{r_{2}}$
$∴$ $q_{1} : q_{2} = r_{1} : r_{2}$

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Similar questions

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} = R

R_{1}^{2} + R_{2}^{2} = R^{2}

R_{1} + R_{2} < R

R_{1}^{2} + R_{2}^{2} < R_{2}

\frac{{R_{1}}^{2}}{{R_{2}}^{2}} = \frac{9}{576}, \frac{R_{1}}{R_{2}} = ?

r_{1}

r_{2}

R

M

F_{1}

F_{2}

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