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Potential Energy Of A System Of Two Point Charges

Potential Energy Of A System Of Two Point Charges :- The potential energy of a system of two point charges can be defined as the work done in bringing them from infinity to the given separation.

Below figure shows two +ve charges q₁ and q₂ separated by a distance r. The electrostatic interaction energy of this system can be given as work done in bringing q₂ without acceleration from infinity to a point P against the electrostatic force of q₁.

Work done by external agent during small displacement dx against the electric field E of q₁ ,

$\displaystyle dW=(\overrightarrow{{{F}_{ext}}}).(\overrightarrow{dx})=(-\overrightarrow{{{F}_{E}}}).(\overrightarrow{dx})=(-{{q}_{2}}\overrightarrow{E}).(\overrightarrow{dx})$

$\displaystyle dW=-{{q}_{2}}(\overrightarrow{E}).(\overrightarrow{dx})=-{{q}_{2}}\left( E\text{ }dx\text{ }cos{{180}^{o}} \right)$

$\displaystyle dW=-{{q}_{2}}\left[ E\text{ }dx\text{ }\left( -1 \right) \right]={{q}_{2}}E\text{ }dx={{q}_{2}}\times \frac{k{{q}_{1}}}{{{x}^{2}}}\times dx$

$\displaystyle dW=k{{q}_{1}}{{q}_{2}}{{x}^{-2}}dx$

Total work done to move q₂ from infinity to point P :

$\displaystyle W=\int_{\infty }^{r}{k{{q}_{1}}{{q}_{2}}{{x}^{-2}}\left( -dx \right)}$

As we move from infinity to point P, we are actually moving in a direction in which x decreases, so we have to replace (dx) by (- dx) and then we get total work.

$\displaystyle W=-k{{q}_{1}}{{q}_{2}}\int_{\infty }^{r}{{{x}^{-2}}dx}=-k{{q}_{1}}{{q}_{2}}\left[ \frac{{{x}^{-1}}}{-1} \right]_{\infty }^{r}$

$\displaystyle W=k{{q}_{1}}{{q}_{2}}\left[ \frac{1}{x} \right]_{\infty }^{r}=k{{q}_{1}}{{q}_{2}}\left[ \frac{1}{r}-\frac{1}{\infty } \right]=k{{q}_{1}}{{q}_{2}}\left[ \frac{1}{r}-0 \right]$