Question

Deduce the expression for the P.E. of a system of two charges ${{q}_{1}}$and ${{q}_{2}}$ located $\overset{\to }{\mathop{{{r}_{1}}}}\,$and$\overset{\to }{\mathop{{{r}_{2}}}}\,$ respectively, in an external electric field.

Answer 1

The work done in moving a charge equals the product of the charge and the potential difference between the initial and final positions of the charge. The potential energy of a system of 2 charges equals the sum of work done in moving the charges from infinity to the located positions and the work done in moving the second charge in the electric field due to the first charge.
Formula used:
$W=q{{V}_{r}}$

Complete answer:
From the given information, we have the data as follows.
The work done to bring the charge ${{q}_{1}}$ from infinity to the located position $\overset{\to }{\mathop{{{r}_{1}}}}\,$ is, ${{W}_{1}}={{q}_{1}}V(\overset{\to }{\mathop{{{r}_{1}}}}\,)$
The work done to bring the charge ${{q}_{2}}$ from infinity to the located position $\overset{\to }{\mathop{{{r}_{2}}}}\,$ is, ${{W}_{2}}={{q}_{2}}V(\overset{\to }{\mathop{{{r}_{2}}}}\,)$
As, firstly, we move the charge ${{q}_{1}}$ from infinity to the located position $\overset{\to }{\mathop{{{r}_{1}}}}\,$, thus, while bringing the charge ${{q}_{2}}$there will electric field due to the charge ${{q}_{1}}$present that influences the movement of the charge ${{q}_{2}}$.
The work done to move the charge ${{q}_{2}}$under the influence of the electric field of charge ${{q}_{1}}$is, ${{W}_{{{q}_{2}}\to {{q}_{1}}}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{12}}}$
The total work done equals the potential energy of the system of charges. This potential energy of a system of 2 charges equals the sum of work done in moving the charges from infinity to the located positions and the work done in moving the second charge in the electric field due to the first charge. So, we have,

& PE={{W}_{1}}+{{W}_{2}}+{{W}_{{{q}_{2}}\to {{q}_{1}}}} \\\ & \therefore PE={{q}_{1}}V(\overset{\to }{\mathop{{{r}_{1}}}}\,)+{{q}_{2}}V(\overset{\to }{\mathop{{{r}_{2}}}}\,)+\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{12}}} \\\ \end{aligned}$$ $$\therefore $$ The expression for the P.E. of a system of two charges $${{q}_{1}}$$and $${{q}_{2}}$$located $$\overset{\to }{\mathop{{{r}_{1}}}}\,$$and$$\overset{\to }{\mathop{{{r}_{2}}}}\,$$ respectively, in an external electric field is $${{q}_{1}}V(\overset{\to }{\mathop{{{r}_{1}}}}\,)+{{q}_{2}}V(\overset{\to }{\mathop{{{r}_{2}}}}\,)+\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{12}}}$$. **Note:** There will be some amount of work done in moving the second charge in the electric field due to the first charge. The potential energy will be measured as the work done in moving the charges from the located positions to the final positions.