Question

Calculate electric field intensity due to uniformly charged non-conducting sphere:
(a) outside the sphere
(b) at the surface of sphere
(c)inside the sphere
(d) at the centre of the sphere
Plot at graph between electric Held intensity and distance.

Answer 1

Electric field intensity due to a uniformly charged non-conducting sphere

when charge is given to non-conducting sphere , it uniformly spreads through out its volume . if q is the charge given and R is the radius of the sphere, then the volume charge density.

$p=\frac{q}{\frac{4}{3}\pi R^3}$

or $p=\frac{3q}{4\pi R^3}............\left(1\right)$

(a) Outside the sphere : in this case taking O as center and r as radius a spherical Gaussian surface is drawn . the point P will be situated at this surface.

the direction of $\overrightarrow{E}$ will be outwards directed due to symmetry . the charge enclosed by the Gaussian surface.

$\Sigma q=q$

Therefore from Gauss's law

${\varphi }_E=\frac{q}{{ϵ}_0}$..........(2)

From the defination of the electric flux,

${\varphi }_E={\oint }_S\overrightarrow{E}\overrightarrow{d}s={\oint }_{S}Edscos{0}^0={\oint }_{S}Eds$

or ${\varphi }_E=E{\oint }_{S}ds$ ( as E will be same at each points of Gaussian surface )

or ${\varphi }_E=E4\pi r^2$...........(3)

Now equating equations (2) and (3) we get

$E\times 4\pi r^2=\frac{q}{{ϵ}_0}$

or $E=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}$..........(4)

From equation (1)

$p=\frac{3q}{4\pi R^2}\Rightarrow q=\frac{4}{3}\pi R^{3}p$

$\therefore$ From equation (4)

$E=\frac{1}{4\pi {ϵ}_0}frac4\pi R^{3}p3r^2$

or $E=\frac{p{R}^3}{3{ϵ}_0{r}^2}$

or $E=\frac{p}{3{ϵ}_0}\left(\frac{R^3}{r^2}\right)$...........(5)

in vector form

$\overrightarrow{E}=\frac{P}{3{ϵ}_0}\left(\frac{R^3}{r^2}\right)\hat{r}$

$\overrightarrow{E}=\frac{p}{3{ϵ}_0}\left(\frac{R^3}{r^2}\right)\hat{r}$...........(6)

where $\hat{r}$ = unit vector in direction $OP$

(b) When point is situated on the surface of the sphere ; in this case we can consider the surface of the surface as Gaussian surface . therefore the whole charge of the sphere will again be charge enclosed.thus by putting $r=R$ equation we get

$E=\frac{P}{3{ϵ}_0}\left(\frac{R^3}{R^2}\right)$

or $E=\frac{pR}{3{ϵ}_0}..........\left(7\right)$

In vector from

$\overrightarrow{E}=\frac{pR}{3{ϵ}_0}\hat{r}$...........(8)

(c) When point P is inside the sphere $(r<R)$

Here also we draw a spherical Gaussian surface taking $O$ as center and r as radius . point P will be situated at this surface. now the charge enclosed by this Gaussian surface.

$\Sigma q=p\frac{4}{3}\pi r^3$

Substituting the value of p form equation (1)

$\sigma q=\frac{3q}{4\pi R^3}\times \frac{4\pi r^3}{3}$

or $\Sigma q=\frac{q{r}^3}{R^3}$

Therefore from Gauss's law,

${\varphi }_E=\frac{\Sigma q}{{ϵ}_0}=\frac{q{r}^3}{{ϵ}_0{R}^3}$

or ${\varphi }_E=\frac{q{r}^3}{{ϵ}_0{R}^3}$..........(9)

From the defination of the electric flux,

${\varphi }_E={\oint }_S\overrightarrow{E}\overrightarrow{d}s={\oint }_{S}Edscos{0}^0$

(for spherical surface )

$=E{\oint }_{S}ds$

or ${\varphi }_E=E\times \pi r^2$........(10)

Now equating equation (9) and (10) we have

$E\times 4\pi r^2=\frac{q{r}^3}{{ϵ}_0{R}^3}$

$\therefore E=\frac{1}{4\pi {ϵ}_0}\frac{q{r}^3}{r^2{R}^3}$

or $E=\frac{1}{4\pi {ϵ}_0}\frac{qr}{R^3}$.........(11)

Again from the equation (1)

$q=\frac{4}{3}\pi R^{3}p$

$\therefore$ From equation (11)

$E=\frac{1}{4\pi {ϵ}_0}\frac{r}{R^3}\times \frac{4}{3}\pi R^{3}p$

or $E=\frac{p}{3{ϵ}_0}r$ .............(12)

In vector form ,

$\overrightarrow{E}=\frac{p}{3{ϵ}_0}r\hat{r}$...........(13)

(d) At the centre of the sphere : for central point

$r=0$

$\therefore$ From equation (12)

$E=0$.........(14)

The sketch showing the variation of $E$ with $r$ is given figure