Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

Derive an expression for electric field inside a solid non-conducting sphere of charge

Q

, when the charge is distributed uniformly. Also calculate electric field at the centre of sphere and electric field on the surface of sphere.

Accepted Answer

Electric Field Intensity due to a Uniformly Charged Non-conducting Sphere:

When charge is given to non-conducting sphere, it uniformly spreads throughout its volume. If $q$ is the charge given and R is the radius of the sphere, then the volume charge density

$ρ = \frac{q}{\frac{4}{3} {πR}^{3}} \Rightarrow ρ = \frac{3 q}{4 {πR}^{3}} . . . . . . . . . . . . . . . . . (1)$

(a) Outside the sphere : In this case taking O as centre and r as radius, a spherical Gaussian surface is drawn. The point P will be situated at this surface. The direction of $\vec{E}$ will be outwards directed due to symmetry. The charge enclosed by the Gaussian surface

$Σ q = q$

Therefore, from Gauss's law

$ϕ_{E} = \frac{q}{ε_{\circ}} . . . . . . . . . . . . . (2)$

From the definition of the electric flux,

$ϕ_{E} = \oint \vec{E} . \vec{d S} = \oint E d S \cos 0^{\circ} = \oint E d S$

$ϕ_{E} = E \oint d S ϕ_{E} = E . 4 {πr}^{2} . . . . . . . . . . . . . . (3)$

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

$E \times 4 {πr}^{2} = \frac{q}{ε_{\circ}} \Rightarrow E = \frac{q}{4 {πr}^{2} ε_{\circ}} . . . . . . . . . . . . (4)$

From equation (1)

$ρ = \frac{3 q}{4 {πR}^{3}} q = \frac{4}{3} ρ {πR}^{3}$

Therefore, from equation (4)

$E = \frac{ρ R^{3}}{3 r^{2} ε_{\circ}}$

$\Rightarrow E = \frac{ρ}{3 ε_{\circ}} (\frac{R^{3}}{r^{2}}) . . . . . . . . . . . . (5)$

In vector form,

$\vec{E} = \frac{ρ}{3 ε_{\circ}} (\frac{R^{3}}{r^{2}}) \hat{r} . . . . . . . . . . . . . (6)$

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

(b) When point P is situated on the surface of the sphere : In this case we can consider the surface of the sphere as Gaussian surface. Therefore, the whole charge of the sphere will again be charge enclosed. Thus, by putting $r = R$ in equation (5), we get

$E = \frac{ρ}{3 ε_{\circ}} (\frac{R^{3}}{R^{2}}) \Rightarrow E = \frac{ρ R}{3 ε_{\circ}} . . . . . . . . . . (7)$

In vector form

$E = \frac{ρ R}{3 ε_{\circ}} \hat{r} . . . . . . . . . . (8)$

(c) When point P is inside the sphere ( $r < R$ ): Here also we draw a spherical Gaussian surface taking O as centre and r as radius. Point P will be situated at this surface. Now the charge enclosed by this Gaussian surface,

$Σ q = ρ \frac{4}{3} {πr}^{3}$

Substituting the value of $ρ$ from equation (1)

$Σ q = \frac{3 q}{4 {πR}^{3}} \times \frac{4 {πr}^{3}}{3} \Rightarrow Σ q = \frac{q r^{3}}{R^{3}}$

Therefore, from Gauss's law

$ϕ_{E} = \frac{Σ q}{ε_{\circ}} ϕ_{E} = \frac{q r^{3}}{ε_{\circ} R^{3}} . . . . . . . . . . . . (9)$

From the definition of the electric flux.

$ϕ_{E} = \oint \vec{E} . \vec{d S} = \oint E d S \cos 0^{\circ} = \oint E d S ϕ_{E} = E \oint d S = E \times 4 {πr}^{2} . . . . . . . . . . . . (10)$

Now, equating equations (9) & (10) we have,

$E \times 4 {πr}^{2} = \frac{{qr}^{3}}{ε_{\circ} R^{3}} ∴ E = \frac{1}{4 {πε}_{\circ}} \frac{qr}{R^{3}} . . . . . . . . . . . . (11)$

Again from equation (i), $q = \frac{4}{3} {πR}^{3} ρ$

Therefore, equation (11) becomes

$E = \frac{1}{4 {πε}_{\circ}} \frac{r}{R^{3}} \times \frac{4}{3} {πR}^{3} ρ \Rightarrow E = \frac{ρ}{3 ε_{\circ}} r . . . . . . . . . . . . . . . . . (12)$

In vector form,

$\vec{E} = \frac{ρ r}{3 ε_{\circ}} \hat{r} . . . . . . . . . . . . . . . (13)$

Derive an expression for electric field inside a solid non-conducting sphere of charge $Q$ , when the charge is distributed uniformly. Also calculate electric field at the centre of sphere and electric field on the surface of sphere.

Important Questions on Electric Charges and Fields

Learn and Prepare for any exam you want !

Derive an expression for electric field inside a solid non-conducting sphere of charge Q, when the charge is distributed uniformly. Also calculate electric field at the centre of sphere and electric field on the surface of sphere.

Important Questions on Electric Charges and Fields

Learn and Prepare for any exam you want !

Derive an expression for electric field inside a solid non-conducting sphere of charge $Q$ , when the charge is distributed uniformly. Also calculate electric field at the centre of sphere and electric field on the surface of sphere.