2. Electric Potential and Capacitance
medium

Derive the formula for the electric potential energy of system of three charges.

Option A
Option B
Option C
Option D

Solution

Let charge $q_{1}, q_{2}$ and $q_{3}$ are bring from infinity distance to at the point $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ located at distance $r_{1}, r_{2}$ and $r_{3}$ respectively.

All three charges are brought as shown in figure.

To bring $q_{1}$ first from infinity to $\mathrm{P}_{1}$, the work done $\mathrm{W}_{1}=0$ $…..1$

because there is no external force to bring $q_{1}$ to $\mathrm{P}_{1}$ Electric potential at $\mathrm{P}_{2}$ due to charge $q_{1}$,

$\mathrm{V}_{1}=\frac{k q_{1}}{r_{12}}$

… $(2)$

Now work done to bring charge $q_{2}$ at point $\mathrm{P}_{2}$,

$\mathrm{W}_{2}=\mathrm{V}_{1} \times q_{2}$

$\therefore \mathrm{W}_{2}=\frac{k q_{1} q_{2}}{r_{12}}$$…..3$

Electric potential at $\mathrm{P}_{3}$ due to charge $q_{1}+q_{2}$, $\mathrm{V}_{2}=\frac{k q_{1}}{r_{13}}+\frac{k q_{2}}{r_{23}}$

$\therefore$ Work done to bring charge $q_{3}$ to $\mathrm{P}_{3}$

$\mathrm{W}_{3}=$ potential at $\mathrm{P}_{3}$ due to $q_{1}+q_{2} \times q_{3}$ charge

$=k\left[\frac{q_{1}}{r_{13}}+\frac{q_{2}}{r_{23}}\right] \times q_{3}$

$\quad=k\left[\frac{q_{1} q_{3}}{r_{13}}+\frac{q_{2} q_{3}}{r_{23}}\right]$$…..4$

$\therefore$ Total potential energy of charges $q_{1}+q_{2}+q_{3}$ $\mathrm{U}=\mathrm{W}_{1}+\mathrm{W}_{2}+\mathrm{W}_{3}$ $\mathrm{U}=\mathrm{W}_{1}+\mathrm{W}_{2}+\mathrm{W}_{3}$

 

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.