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Question

# Consider the force $F$ on a charge $\text{'}q\text{'}$ due to a uniformly charged spherical shell of radius $R$ carrying charge $Q$ distributed uniformly over it. Which one of the following statements is true for $F$, if $\text{'}q\text{'}$ is placed at distance $r$ from the center of the shell?

A

$F=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{qQ}{{R}^{2}}>F>0forr

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B

$F=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{qQ}{{r}^{2}}forr>R$

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C

$F=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{qQ}{{r}^{2}}forallr$

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D

$F=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{qQ}{{R}^{2}}forr

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Solution

## The correct option is B $F=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{qQ}{{r}^{2}}forr>R$Step 1. Finding the Electric field E at $r.Consider a gaussian surface of radius $r$ inside the Spherical shell of radius $R$Using the integral form of the gauss law of electrostatic $\oint E.ds=\frac{{Q}_{en}}{{\epsilon }_{0}}$, [Where ${Q}_{en}$is the enclosed charge and ${\epsilon }_{0}$ is the permittivity of the free space, and $ds$ is the area enclosed.] In this case, ${Q}_{en}=0$ $\oint E.ds=0$ $E=0$Step 2. Finding the electric field E at $r>R$Consider a gaussian surface of radius $r$ outside the spherical shell of radius $R$Using an integral form of gauss law $\oint E.ds=\frac{{Q}_{en}}{{\epsilon }_{0}}$ $E\oint ds=\frac{Q}{{\epsilon }_{0}}$ $E\left(4{\mathrm{\pi r}}^{2}\right)=\frac{Q}{{\epsilon }_{0}}$ [Surface area of the sphere$=4{\mathrm{\pi r}}^{2}$] $E=\frac{Q}{4{\mathrm{\pi \epsilon }}_{0}{\mathrm{r}}^{2}}$Step 3. Finding the Force, $F$We know that, $F=qE$ $F=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{qQ}{{r}^{2}}$Hence the correct option is $\left(B\right)$.

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