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Properties of Field Lines

Two conductin...

Question

Two conducting spheres of radii $r_{!}$ and $r_{2}$ have same electric field near their surfaces. The ratio of their electric potentials is :-

r_{1}^{2} / r_{2}^{2}

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r_{2}^{2} / r_{1}^{2}

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r_{1} / r_{2}

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r_{2} / r_{1}

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Solution

The correct option is D $r_{2} / r_{1}$
Two conducting spheres of radii $r_{1}$ and $r_{2}$
$E$ $α$ $\frac{1}{r^{3}} ˆ r$
$\frac{E_{1}}{E_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}} \times \frac{{ˆ r}_{1}}{{ˆ r}_{2}}$
or, $\frac{E_{1}}{E_{2}} = \frac{r_{2}}{r_{1}}$

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Similar questions

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} = 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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