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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 external 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 by external force in bringing them from infinity to a separation r from each other without acceleration. Let first the charge q₁ is brought from infinity to the current position. But since initially there is no electric filed, so no work is done in bringing q₁ to its current position. Now suppose that q₂ is brought from infinity to the point P. In this case we have to do some external work to bring q₂ because there is an electric field produced by q₁ and q₂moves against this electric this field. Hence the electrostatic interaction energy of this system can be given as work done in bringing q₂ without acceleration from infinity to the point P against the electrostatic field of q₁.

Small amount of 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 by integration.

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

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

$\displaystyle W=\frac{k{{q}_{1}}{{q}_{2}}}{r}$

This work done is the interaction energy/potential energy of a system of two point charges (U),

$\displaystyle U=W=\frac{k{{q}_{1}}{{q}_{2}}}{r}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{r}$

If the two charges here are of opposite sign, the potential energy of the system will be negative :

$\displaystyle U=-k\frac{{{q}_{1}}{{q}_{2}}}{r}$

Example 1.

Two charged particles each having equal charges 2 × 10^–5 C are brought from infinity to within a separation of 10 cm. Calculate the increase in potential energy during the process and the work required for this purpose.

Solution :

Increase in potential energy = ΔU

$\displaystyle \Delta U={{U}_{f}}-{{U}_{i}}={{U}_{f}}-0={{U}_{f}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{r}=\frac{9\times {{10}^{9}}\times {{\left( 2\times {{10}^{-5}} \right)}^{2}}}{10\times {{10}^{-2}}}$

$\displaystyle \Delta U=36joules$

Work required = change in potential energy of the system (ΔU) = 36 J

Potential Energy Of A System Of Two Point Charges

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