Question

Determine the electric field everywhere outside the sphere at a distance $r(>>a)$ from the centre.

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

Answer 1

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

Given a non-conducting solid sphere of radius '$a$' and charge '$Q$' is uniformly distributed over it. We have to find the electric field $\overrightarrow{E}$ at a distance $r(r>>a)$ from center.

Let P be a point outside the sphere at a distance $r$ from center of sphere.

Draw a gaussian surface in form of sphere of radius $r$ with O as a center.

According to Gauss' theorem,

${\oint }_S\overrightarrow{E}.\overrightarrow{dS}=\frac{Q}{{\epsilon }_0}$

or ${\oint }_{S}ES\cos \theta =\frac{Q}{{\epsilon }_0}$

Since $\overrightarrow{E}$ and $\overrightarrow{dS}$ are along same direction, so $\theta =0^{\circ }.$

Since $E$ is same at all points on the surface of gaussian surface,

$E{\oint }_{S}dS=\frac{Q}{{\epsilon }_0}$

Now, ${\oint }_{S}dS=$ Surface area of spherical gaussian sphere.

$=4\pi r^2$

$\Rightarrow E\times 4\pi r^2=\frac{Q}{{\epsilon }_0}$

$\Rightarrow E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}$