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. $F=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{qQ}{R^2}>F>0forr<R$
• B. $F=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{qQ}{r^2}forr>R$
• C. $F=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{qQ}{r^2}forallr$
• D. $F=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{qQ}{R^2}forr<R$

Answer 1

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 $r<R$.

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 $\left(B\right)$.