Question

The electric field inside a sphere which carries a charge density proportional to the distance from the origin $\rho$ = $\propto$ r ($\propto$ is a

constant) is:

Class-12th / Physics
Chapter- Electrostatics. ​

Answer 1

AnswEr :

The charge density of the sphere is directly proportional to the distance.

$\sf \rho \propto \: r$

From Guass' Law,

$\sf \: \phi = \dfrac{q}{ \epsilon_{o}} - - - - - - - - - (1)$

Electric Flux is also expressed as :

$\displaystyle \sf \phi = \oint \vec{E}. \vec{ds} \\ \\ \longrightarrow \: \displaystyle \ \sf \phi = \oint \: E.ds \cos(0) \\ \\ \longrightarrow \: \sf \: \phi = E( {4\pi r}^{2} ) - - - - - - - (2)$

Equating equations (1) and (2),

$\sf \: E = \dfrac{q}{4\pi \epsilon_o \: {r}^{2} }$

Charge Density is defined as the Charge per unit volume of the sphere

$\sf \: \rho = \dfrac{q}{V} \\ \\ \implies \: \sf \: q = \rho V \\ \\ \implies \: \sf \: q = \rho \times \dfrac{4}{3} \pi {r}^{3}$

Now,

$\sf \: E = \dfrac{ \rho \: \times \dfrac{4}{3} \pi {r}^{3} }{4\pi \epsilon_o \: {r}^{2} } \\ \\ \longrightarrow \: \sf \: E = \frac{ \alpha \times r \times r\times \cancel{4\pi {r}^{2}} }{ \cancel{4\pi {r}^{2} }\times 3 \epsilon_o} \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf E = \dfrac{ \alpha r^2}{3 \epsilon_o}} }$

Answer 2

GiveN

  Charge density inside the sphere at a distance r is given by $\alpha$ r.  

To find 

Electric field at any point inside the sphere.

ConcepT

First we will find total charge inside a sphere of radius r and then we will use Gauss' law.

•  Gauss' law :- The law states that $\frac{Q}{E_0} = \int E.dS$  

SolutioN

Finding total charge inside a spherical shell of radius r.

Assume a spherical shell of thickness dx at a distance x from the centre of the sphere. Charge density at a distance x =  $\alpha$ x  .

Total volume of this cavity = 4πx² dx

Here, Since the thickness dx is infinitesimally small. It is safe to assume that the charge density which is dependent on distance from the centre, is constant here

 Hence, total charge inside this cavity = Charge density × Volume  

={ $\alpha$ x } × 4πx²dx  = 4 $\alpha$ πx³ dx  

• Now, integrating this from 0 $\rightarrow$ r , we get total charge inside the sphere of radius r.

$\int \limits_{0}^{r} (4\alpha\:\pi){x}^{3} dx \\\\ = [ 4\pi \alpha (\frac{r^{4} }{4} - 0)\\\\= 4\pi \alpha \frac{r^{4} }{4} \\\\= \pi \alpha r^{4}$  

Now, applying gauss' law , we get total electric field at any point inside the sphere of radius r.

$\frac{Q}{E_0} = \int E.ds\\\\=> \frac{\pi \alpha r^4} {E_0} = E.4\pi r^{2} \\\\=> E = \frac{\alpha r^2 }{4E_0}$.

$\boxed{ \boxed{ \orange{Hence,\:E_in\:= \frac{\alpha r^2 }{4E_0} }}}$