Question

Find the electric field outside the shell at a distance $r(>>a)$ from the centre.

• A. $E=\frac{Q}{4\pi {ϵ}_0{a}^2}$
• B. $E=\frac{Q}{4\pi {ϵ}_0{r}^2}$
• C. $E=\frac{Q}{4\pi {ϵ}_0(r+a{)}^2}$
• D. $E=0$

Answer 1

The correct option is B $E=\frac{Q}{4\pi {ϵ}_0{r}^2}$

For a point P outside the charge distribution we have $a>R$.

We draw a spherical surface passing through the point P and concentric with the given charge distribution.Taking this as the Gaussian surface ,

The electric field distribution is radial by symmetry and is outward as Q is positive.Also, its magnitude at all points of the Gaussian surface must be equal.Let this magnitude be $E$. This is also the magnitude of field at P.

As the field $\overrightarrow{E}$ is normal to the surface element everywhere,

$\overrightarrow{E}.\overrightarrow{dS}=EdS$ for each element .

The flux of the electric field through this closed surface id

$\varphi =\int \overrightarrow{E}.\overrightarrow{dS}$

$=\int E.dS=E\int dS=E4\pi r^2$

This should be equal to the charge contained inside the Gaussian surface divided by ${ϵ}_o$. As the entire charge $Q$ is contained inside the Gaussian surface , we get

$E4\pi r^2=\frac{Q}{{ϵ}_o}$

$⟹E=\frac{Q}{4\pi {ϵ}_o{r}^2}$

Answer-(B)