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Potential Energy Of A System Of Charges

Potential Energy Of A System Of Charges :- The electrostatic potential energy of a system of charges is the work required to assemble the system by bringing each charge in from infinity—where the electric field is zero—to its final position, against the forces of electrostatic attraction or repulsion.

For a system of three or more charges, the total potential energy is the sum of the potential energy of every unique pair :

$\displaystyle {{U}_{Total}}=\sum\limits_{i<j}{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{i}}{{q}_{j}}}{{{r}_{ij}}}}$

Where :

and are the charges,
$_{j}$ is the distance between them,
The sum is over all distinct pairs of charges.

Example (Three Charges) :

Let charges be located at different positions. The total potential energy of the system is :-

$\displaystyle U=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{q}_{1}}{{q}_{2}}}{{{r}_{12}}}+\frac{{{q}_{1}}{{q}_{3}}}{{{r}_{13}}}+\frac{{{q}_{2}}{{q}_{3}}}{{{r}_{23}}} \right)$

Similarly potential energy of a system of four charges,

$\displaystyle U=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{q}_{1}}{{q}_{2}}}{{{r}_{12}}}+\frac{{{q}_{1}}{{q}_{3}}}{{{r}_{13}}}+\frac{{{q}_{1}}{{q}_{4}}}{{{r}_{14}}}+\frac{{{q}_{2}}{{q}_{3}}}{{{r}_{23}}}+\frac{{{q}_{2}}{{q}_{4}}}{{{r}_{24}}}+\frac{{{q}_{3}}{{q}_{4}}}{{{r}_{34}}} \right)$

Example 1.

Figure shows an arrangement of three point charges. The total potential energy of this arrangement is zero. Calculate the ratio q/Q.

Solution :

$\displaystyle U=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{\left( +q \right)\left( -Q \right)}{r}+\frac{\left( +q \right)\left( +q \right)}{2r}+\frac{\left( -Q \right)(+q)}{r} \right)=0$

$\displaystyle -\frac{qQ}{r}+\frac{{{q}^{2}}}{2r}-\frac{qQ}{r}=0$

$\displaystyle \Rightarrow \frac{q}{Q}=4$

Example 2.
Two point charges, each of mass m and charge q are released when they are at a distance r from each other. What is the speed of each charged particle when they are at a distance 2r ?

Solution :

According to conservation of momentum, both the charge particles will move with same speed.

Now applying law of conservation of energy :-

$\displaystyle {{\left( M.E. \right)}_{initial}}={{\left( M.E. \right)}_{final}}$

$\displaystyle 0+0+\frac{k{{q}^{2}}}{r}=2\times \left( \frac{1}{2}m{{v}^{2}} \right)+\frac{k{{q}^{2}}}{2r}$

$\displaystyle v=\sqrt{\frac{k{{q}^{2}}}{2mr}}$

Example 3.

Three equal charges q each are placed at the corners of an equilateral triangle of side a.

(i) Find out potential energy of charge system.
(ii) Calculate work required to decrease the side of triangle to a/2.
(iii) If the charges are released from the shown position and each of them has same mass m then find the speed of each particle when they lie on triangle of side 2a.

Solution :

(i)

$\displaystyle U=k\left( \frac{{{q}_{1}}{{q}_{2}}}{{{r}_{12}}}+\frac{{{q}_{1}}{{q}_{3}}}{{{r}_{13}}}+\frac{{{q}_{2}}{{q}_{3}}}{{{r}_{23}}} \right)=k\left( \frac{{{q}^{2}}}{a}+\frac{{{q}^{2}}}{a}+\frac{{{q}^{2}}}{a} \right)=\frac{3k{{q}^{2}}}{a}joules$

(ii) Work required to decrease the sides

$\displaystyle W={{U}_{f}}-{{U}_{i}}=\frac{3k{{q}^{2}}}{\left( {}^{a}\!\!\diagup\!\!{}_{2}\; \right)}-\frac{3k{{q}^{2}}}{a}=\frac{3k{{q}^{2}}}{a}joules$

(iii) Applying the law of conservation of energy

$\displaystyle {{\left( M.E. \right)}_{initial}}={{\left( M.E. \right)}_{final}}$

$\displaystyle 0+0+0+\frac{3k{{q}^{2}}}{a}=3\times \left( \frac{1}{2}m{{v}^{2}} \right)+\frac{3k{{q}^{2}}}{2a}$

$\displaystyle \frac{3}{2}m{{v}^{2}}=\frac{3k{{q}^{2}}}{2a}$

$\displaystyle v=\sqrt{\frac{k{{q}^{2}}}{ma}}$

Example 4.
Four identical point charges q each are placed at four corners of a square of side a. Find out potential energy of the charge system.

Solution :

$\displaystyle U=k\left( \frac{{{q}_{1}}{{q}_{2}}}{{{r}_{12}}}+\frac{{{q}_{1}}{{q}_{3}}}{{{r}_{13}}}+\frac{{{q}_{1}}{{q}_{4}}}{{{r}_{14}}}+\frac{{{q}_{2}}{{q}_{3}}}{{{r}_{23}}}+\frac{{{q}_{2}}{{q}_{4}}}{{{r}_{24}}}+\frac{{{q}_{3}}{{q}_{4}}}{{{r}_{34}}} \right)$

$\displaystyle \Rightarrow U=k\left( \frac{{{q}^{2}}}{a}+\frac{{{q}^{2}}}{\sqrt{2}a}+\frac{{{q}^{2}}}{a}+\frac{{{q}^{2}}}{a}+\frac{{{q}^{2}}}{\sqrt{2}a}+\frac{{{q}^{2}}}{a} \right)$

$\displaystyle \therefore U=\frac{2k{{q}^{2}}}{a}\left[ 2+\frac{1}{\sqrt{2}} \right]joules$

Example 5.
Six equal point charges q each are placed at six corners of a hexagon of side a. Find out potential energy of charge system.

Solution :

$\displaystyle U=k\left( \frac{{{q}_{1}}{{q}_{2}}}{{{r}_{12}}}+\frac{{{q}_{1}}{{q}_{3}}}{{{r}_{13}}}+\frac{{{q}_{1}}{{q}_{4}}}{{{r}_{14}}}+\frac{{{q}_{1}}{{q}_{5}}}{{{r}_{15}}}+\frac{{{q}_{1}}{{q}_{6}}}{{{r}_{16}}}+\frac{{{q}_{2}}{{q}_{3}}}{{{r}_{23}}}+\frac{{{q}_{2}}{{q}_{4}}}{{{r}_{24}}}+\frac{{{q}_{2}}{{q}_{5}}}{{{r}_{25}}}+\frac{{{q}_{2}}{{q}_{6}}}{{{r}_{26}}}+\frac{{{q}_{3}}{{q}_{4}}}{{{r}_{34}}}+\frac{{{q}_{3}}{{q}_{5}}}{{{r}_{35}}}+\frac{{{q}_{3}}{{q}_{6}}}{{{r}_{36}}}+\frac{{{q}_{4}}{{q}_{5}}}{{{r}_{45}}}+\frac{{{q}_{4}}{{q}_{6}}}{{{r}_{46}}}+\frac{{{q}_{5}}{{q}_{6}}}{{{r}_{56}}} \right)$

$\displaystyle \Rightarrow U=k\left( \frac{{{q}_{1}}{{q}_{2}}}{{{r}_{12}}}+\frac{{{q}_{1}}{{q}_{3}}}{{{r}_{13}}}+\frac{{{q}_{1}}{{q}_{4}}}{{{r}_{14}}}+\frac{{{q}_{1}}{{q}_{5}}}{{{r}_{15}}}+\frac{{{q}_{1}}{{q}_{6}}}{{{r}_{16}}} \right)\times 3$

$\displaystyle \Rightarrow U=k\left( \frac{{{q}^{2}}}{a}+\frac{{{q}^{2}}}{\sqrt{3}a}+\frac{{{q}^{2}}}{2a}+\frac{{{q}^{2}}}{\sqrt{3}a}+\frac{{{q}^{2}}}{a} \right)\times 3$

$\displaystyle \therefore U=\frac{3k{{q}^{2}}}{a}\left( 2+\frac{2}{\sqrt{3}}+\frac{1}{2} \right)joules$

Potential Energy Of A System Of Charges

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