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For finding potential at a point due to a +ve charge $(q)$, we find work done to move a unit +ve charge $(q_o)$ from infinity to that point in the presence of +ve charge $(q)$

Since both charges being +ve, the force would be repulsive and hence while bringing unit +ve charge $(q_o)$ from infinity to that position, the path would be against force field. Thus the work done by force field should be negative.

But the following calculation/derivation in my book shows that work done is positive:

$$W$$ $$=\int_\infty^r F.dr$$ $$=-\int_\infty^r F dr\\$$ (since path is against force field) $$=-\int_\infty^r \frac{1}{r^{2}}dr\\$$ $$=-\left( {-\frac{1}{r}} \Big |_{\infty}^{r}\right)$$ $$={\frac{1}{r}} \Big |_{\infty}^{r}$$ $$=\frac{1}{r}-\frac{1}{\infty}$$ $$=\frac{1}{r}$$

Why is this contradiction? Where am I (or my book) wrong?

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    $\begingroup$ "Thus work done should be negative" - the work done by the field is negative but the work done by the force bringing the charge in is positive. $\endgroup$ Commented Dec 11, 2016 at 13:37
  • $\begingroup$ The work done by the electrostatic force will be negative whereas work done by the external agent to move the charge is positive $\endgroup$
    – cobra121
    Commented Dec 11, 2016 at 13:40
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    $\begingroup$ Will they be equal and opposite? $\endgroup$ Commented Dec 11, 2016 at 13:43
  • $\begingroup$ Ideally, assuming no change in KE, yes. $\endgroup$ Commented Dec 11, 2016 at 13:45
  • $\begingroup$ In the derivation, $F$ represents the electrostatic force and not the external force. That is why the path is against the field. So in the derivation, $W$ should represent work done by electrostatic force and not work done by external force. This is turning out to be positive while in reality, it should be negative. $\endgroup$ Commented Dec 11, 2016 at 15:41

1 Answer 1

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The work done by the external force is in the direction of the force applied, hence work done by F(ext) is positive & at the same time work done by the electric field due to the positive charge (q) is negative since work is done against the direction of the force (as you say).

But, $$ F_\text{ext}=-F_\text{electric field}$$

Therefore
$$W_\text{ext}=-W_\text{electric field}$$

The work done by force field is negative. Maybe you are confused with potential energy, \begin{align} PE&=W_\text{EXT}\\ &= ∫ F_\text{ext}.dr\\ &=“-“∫ F_\text{electric field}.dr\\ &= -W_\text{electric field} \end{align}

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  • $\begingroup$ I'm new to the app, so had a problem typing $\endgroup$ Commented Jun 18, 2019 at 18:00

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