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Total Electrostatic Energy Of A System

Total electrostatic energy of a system :- Total Electrostatic Energy of a System is the sum of the self-energy of individual charged conductors and the interaction energy between different charges or conductors.

Total Electrostatic Energy (U) = Sum of all self-energies (U_self) + Sum of all interaction energies (U_interaction)

Let us consider some examples :

(1). Two charged conducting spheres

Let us consider two spherical conductors with radii, and , carrying charges and respectively. Let the distance between their centers be

U = U_self + U_interaction

$\displaystyle U=\frac{kQ_{1}^{2}}{2{{R}_{1}}}+\frac{kQ_{2}^{2}}{2{{R}_{2}}}+\frac{k{{Q}_{1}}{{Q}_{2}}}{r}$

(2). Two charged non-conducting spheres

Let us consider two non-conducting spheres with radii, and , carrying charges and respectively. Let the distance between their centers be

U = U_self + U_interaction

$\displaystyle U=\frac{3kQ_{1}^{2}}{5{{R}_{1}}}+\frac{3kQ_{2}^{2}}{5{{R}_{2}}}+\frac{k{{Q}_{1}}{{Q}_{2}}}{r}$

(3). A system of two charged concentric shells

Consider a system of two concentric spherical shells with radii and , uniformly charged with charges and , respectively.

U = U_{self energy of inner shell} + U_{self energy of outer shell} + U_interaction

$\displaystyle U=\frac{kQ_{1}^{2}}{2{{R}_{1}}}+\frac{kQ_{2}^{2}}{2{{R}_{2}}}+\frac{k{{Q}_{1}}{{Q}_{2}}}{{{R}_{2}}}$

Example 1.

Three spherical shells are shown carrying charges and and of radii and respectively. If the middle shell expands from radius to , find the work done by the electric field during this process.

Solution :

We know that :

Work done by conservative forces = Decrease in electrostatic potential energy

Work done by the electric field (W) = ( Total initial electrostatic potential energy of the system, U_i ) – ( Total final electrostatic potential energy of the system, U_f)

Initial electrostatic potential energy of the system :

$\displaystyle {{U}_{i}}=S{{E}_{1}}+S{{E}_{2}}+S{{E}_{3}}+I{{E}_{12}}+I{{E}_{13}}+I{{E}_{23}}$

Here SE = Self Energy and IE = Interaction Energy

$\displaystyle {{U}_{i}}=\left( \frac{kq_{1}^{2}}{2a}+\frac{kq_{2}^{2}}{2b}+\frac{kq_{3}^{2}}{2c}+\frac{k{{q}_{1}}{{q}_{2}}}{b}+\frac{k{{q}_{1}}{{q}_{3}}}{c}+\frac{k{{q}_{2}}{{q}_{3}}}{c} \right)$

Final electrostatic potential energy of the system :

$\displaystyle {{U}_{f}}=\left( \frac{kq_{1}^{2}}{2a}+\frac{kq_{2}^{2}}{2b'}+\frac{kq_{3}^{2}}{2c}+\frac{k{{q}_{1}}{{q}_{2}}}{b'}+\frac{k{{q}_{1}}{{q}_{3}}}{c}+\frac{k{{q}_{2}}{{q}_{3}}}{c} \right)$

Hence

$\displaystyle W={{U}_{i}}-{{U}_{f}}=\left( \frac{kq_{2}^{2}}{2b}-\frac{kq_{2}^{2}}{2b'}+\frac{k{{q}_{1}}{{q}_{2}}}{b}-\frac{k{{q}_{1}}{{q}_{2}}}{b'} \right)$

Example 2.

Figure (a) shows a spherical shell of radius with charge uniformly distributed over it. A point charge is placed at the center of the shell. Now, as shown in Figure (b), the radius of the shell is increased from to R_₁. Find the work required to increase the radius of the shell from to R_₁.

Solution :

Work done by the electric field (W) = ( Total initial electrostatic potential energy of the system, U_i ) – ( Total final electrostatic potential energy of the system, U_f)

Initial electrostatic potential energy of the system :

$\displaystyle {{U}_{i}}=\frac{kq_{1}^{2}}{2R}+\frac{kq{{q}_{1}}}{R}$

Final electrostatic potential energy of the system :

$\displaystyle {{U}_{f}}=\frac{kq_{1}^{2}}{2{{R}_{1}}}+\frac{kq{{q}_{1}}}{{{R}_{1}}}$

Hence

$\displaystyle W={{U}_{i}}-{{U}_{f}}=k\left( \frac{q_{1}^{2}}{2R}+\frac{q{{q}_{1}}}{R}-\frac{q_{1}^{2}}{2{{R}_{1}}}-\frac{q{{q}_{1}}}{{{R}_{1}}} \right)$

Example 3.

A wire with linear charge density is placed along the axis of a cylindrical shell of inner radius , outer radius , and length $l$ . Find the electrostatic energy stored in the cylindrical shell.

Solution :

Electric field intensity at a distance from the wire :

$\displaystyle E=\frac{2k\lambda }{x}$

Here a ≤ x ≤ b.

Energy Density (u_E) :

$\displaystyle {{u}_{E}}=\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}=\frac{1}{2}{{\varepsilon }_{0}}{{\left( \frac{2k\lambda }{x} \right)}^{2}}=\frac{2{{\varepsilon }_{0}}{{k}^{2}}{{\lambda }^{2}}}{{{x}^{2}}}$

Volume of a cylindrical shell of radius , thickness , and length :

$\displaystyle dV\text{=}\left( 2\pi xl \right)dx$

Now, the total energy stored in the given cylinder between (a ≤ x ≤ b) is :

$\displaystyle U=\int{{{u}_{E}}}dV=\int\limits_{a}^{b}{\left( \frac{2{{\varepsilon }_{0}}{{k}^{2}}{{\lambda }^{2}}}{{{x}^{2}}}\times \left( 2\pi xl \right)dx \right)}$