Two charged spheres of radii $R_{1}$ and $R_{2}$ have equal surface charge density. The ratio of their potential at an equidistant external point is

\frac{R_{2}}{R_{1}}

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(\frac{R_{2}}{R_{1}})^{2}

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(\frac{R_{1}}{R_{2}})^{2}

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\frac{R_{1}}{R_{2}}

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Solution

The correct option is D $\frac{R_{1}}{R_{2}}$
Given two spheres of radii $R_{1}$ and $R_{2}$ have equal surface charge density.
To find the ratio of their potentials at an equidistant external point.

Let us suppose that the sphere with radius $R_{1}$ has charge $q_{1}$ on it, and that the sphere with radius $R_{2}$ has a charge $q_{2}$ on it. Now, they have same charge densities, so $\frac{q_{1}}{4 π R_{1}^{2}} = \frac{q_{2}}{4 π R_{2}^{2}}$
So, $\frac{q_{1}}{q_{2}} = {(\frac{R_{1}}{R_{2}})}^{2}$ ...(i)

Potential in first sphere, $V_{1} = \frac{k q_{1}}{R_{1}}$
Potential in second sphere, $V_{2} = \frac{k q_{2}}{R_{2}}$
Thus, $\frac{V_{1}}{V_{2}} = \frac{q_{1}}{R_{1}} . \frac{R_{2}}{q_{2}}$
$= \frac{q_{1}}{q_{2}} {(\frac{R_{1}}{R_{2}})}^{2}$

From (i), $\frac{V_{1}}{V_{2}} = {(\frac{R_{1}}{R_{2}})}^{2} \times \frac{R_{1}}{R_{2}} = \frac{R_{1}}{R_{2}}$

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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

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(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}

r_{1}

r_{2}

r

r = r_{1} + r_{2}

{r_{1}}^{2} + {r_{2}}^{2} = r^{2}

r_{1} + r_{2} < r

{r_{1}}^{2} + {r_{2}}^{2} < r^{2}

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