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Could anyone tell me what and where is the origin and vector dr that prof. Walter Lewin used to calculate work here?

With origin I mean the coordinate origin to define vector r and dr.

Why vector dr is same for both Coulomb force and Walter Lewin force?

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  • $\begingroup$ In the general case, a curve is taken to be parametrised by $\vec r(t)$ for some $t\in [a,b]$. For example, for a unit circle, we have $\vec r = (\cos t, \sin t)$ and $t \in [0, 2\pi]$. Then, $d\vec r = \frac{d\vec r}{dt}dt = (-\sin t, \cos t)dt$. $\endgroup$
    – JamalS
    Dec 30, 2016 at 12:50
  • $\begingroup$ How do you define dr in the case above? $\endgroup$
    – emnha
    Dec 30, 2016 at 15:44

1 Answer 1

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I would suggest that the position of charge $q_1$ is a reasonable place to have the origin.

$d\vec r$ is an incremental change of position with the path taken determined by the limits of the integration.

So in the second integral the force acting on charge $q_2$ is in the same direction as the path taken from $R$ to infinity so the dot product of force and incremental change in position (work done) is positive.

For the first integral the dot product will be negative but once the integration from infinity to $R$ is performed the work done by Walter Lewin comes out to be positive.

It might have been clearer if Walter Lewin had evaluated the work done by the electric field in going from infinity to $R$ which would have come out as a negative quantity and then used the idea that the change in potential energy is minus the work done by the electric field.
The limits of integration (infinity to $R$) would then have been the same way around in both integrals.

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  • $\begingroup$ Thanks. So the direction of vector dr depends one the integration limit direction? For example, if the integration from infinity to R then vector dr will have direction from infinity to R. Similarly if the integration limit from R to infinity then dr will also have the direction from R to infinity? $\endgroup$
    – emnha
    Dec 30, 2016 at 12:35
  • $\begingroup$ "For the first integral the dot product will be negative": in this case for the direction of integration from infinity to R, vector dr will have the same direction with Walter Lewin force. However, the magnitude of dr is negative. So the product of two vectors is negative. Is that right? $\endgroup$
    – emnha
    Jan 2, 2017 at 20:16
  • $\begingroup$ For a moment forget about the integration. Taking the positive direction as to the right then dr is to the right and Walter Lewin force is to the left so the dot product will be negative. Having sorted out the dot product now do the integration. $\endgroup$
    – Farcher
    Jan 2, 2017 at 20:35
  • $\begingroup$ So the direction of vector dr can be chosen as to the right no matter of what the integration limits are? $\endgroup$
    – emnha
    Jan 2, 2017 at 21:14
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    $\begingroup$ @anhnha It might help if you look at the answer to similar question about gravitational potential? physics.stackexchange.com/a/302728/104696 $\endgroup$
    – Farcher
    Jan 5, 2017 at 6:00

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