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

In a solid uniformly charged sphere of total charge

Q

and radius

R

, if energy stored outside the sphere is

U_{0}

Joules then find out self energy of sphere in terms of

U_{0}

?

Accepted Answer

We know that energy density at a particular point in electrostatics is given by,

$\frac{1}{2} \in_{\circ} E^{2}$ , where energy density is defined as energy per unit volume.So Energy at any point is given by,

$\frac{1}{2} \in_{\circ} E^{2} . V$ , where $V$ is the corresponding volume.

Let $E_{out}$ be the energy stored outside charged sphere and $E_{in}$ be energy stored inside sphere.

Then self energy of sphere is equal to $E_{out} + E_{in}$ .

In order to calculate $E_{in}$ let us assume a spherical shell inside whose radius is $x$ and thickness is $d x$ so that its volume $d v = 4 {πx}^{2} dx$ .

Now electric field inside solid charged sphere at any point distant $x$ from centre with radius of sphere equal to $R$ is given by,

$\frac{K Q x}{R^{3}}$

Now energy stored in this shell $d E_{in} = \frac{1}{2} \frac{\in_{\circ} K^{2} Q^{2} x^{2}}{R^{6}} . 4 {πx}^{2} dx$

Total energy stored inside sphere is given by,

$E_{in} = \int d E_{in} = \int_{0}^{R} \frac{1}{2} \frac{\in_{\circ} K^{2} Q^{2} x^{2}}{R^{6}} . 4 {πx}^{2} dx = \frac{1}{2} \frac{\in_{\circ} K^{2} Q^{2}}{R^{6}} . 4 π \int_{0}^{R} x^{4} d x = \frac{K Q^{2}}{10 R} . . . (i)$

Similarly, in order to calculate energy stored outside sphere, we choose a spherical shell of radius $r$ and thickness $d r$ outside given sphere. Now Electric field outside sphere at a distance of $r$ from centre of sphere is given $\frac{K Q}{r^{2}}$

Therefore, using energy density concept, energy stored outside sphere is given by,

$E_{out} = \int d E_{out} = \int_{R}^{\infty} \frac{1}{2} \frac{\in_{\circ} K^{2} Q^{2}}{r^{4}} . 4 π r^{2} d r = \frac{1}{2} \in_{\circ} K^{2} Q^{2} . 4 π \int_{R}^{\infty} r^{- 2} d r = \frac{K Q^{2}}{2 R} . . . (ii)$

Using equation $(i), (ii)$ , self energy of sphere is given by,

$E_{in} + E_{out} = \frac{K Q^{2}}{10 R} + \frac{K Q^{2}}{2 R} = \frac{K Q^{2}}{2 R} \frac{6}{5} . . . (iii)$

Now according to the question energy stored outside is equal to $U_{\circ}$

Therefore $U_{\circ} = E_{out} = \frac{K Q^{2}}{2 R} . . . (iv)$

Using equation $(iv)$ in $(i i i)$ , we have self energy stored in spring is equal to $\frac{6}{5} U_{\circ} = 1.5 U_{\circ}$

This is the required answer.

In a solid uniformly charged sphere of total charge $Q$ and radius $R$ , if energy stored outside the sphere is $U_{0}$ Joules then find out self energy of sphere in terms of $U_{0}$ ?

Important Questions on Electrostatics

Learn and Prepare for any exam you want !

In a solid uniformly charged sphere of total charge Q and radius R, if energy stored outside the sphere is U0 Joules then find out self energy of sphere in terms of U0?

Important Questions on Electrostatics

Learn and Prepare for any exam you want !

In a solid uniformly charged sphere of total charge $Q$ and radius $R$ , if energy stored outside the sphere is $U_{0}$ Joules then find out self energy of sphere in terms of $U_{0}$ ?