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?

  • 1
    $\begingroup$ "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. $\endgroup$ – Alfred Centauri Dec 11 '16 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 Dec 11 '16 at 13:40
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    $\begingroup$ Will they be equal and opposite? $\endgroup$ – stack exchange Dec 11 '16 at 13:43
  • $\begingroup$ Ideally, assuming no change in KE, yes. $\endgroup$ – Alfred Centauri Dec 11 '16 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. $\endgroup$ – stack exchange Dec 11 '16 at 15:41

The work done by the external force is in the direction of the force applied, hence work done by F(ext) is +ve & at the same time work done by the electric field due to the +ve charge (q) is -ve since work is done against the direction of the force (as you say). BUT F(ext)=-F(electric field) Therefore W(ext)=-W(electric field) The work done by force field is negative. MAY be you are confused with potential energy Potential energy = W(EXT) = -VE work done by Electric field = -(-W(electric field)) = W(electric field) HOPE you got it. Please VOTE :)

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

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