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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?

Qmechanic
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emnha
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  • 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, $d\vec r = \frac{d\vec r}{dt}dt = (-\sin t, \cos t)dt$. – JamalS Dec 30 '16 at 12:50
  • How do you define dr in the case above? – emnha Dec 30 '16 at 15:44

1 Answers1

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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.

Farcher
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  • 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? – emnha Dec 30 '16 at 12:35
  • "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? – emnha Jan 02 '17 at 20:16
  • 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. – Farcher Jan 02 '17 at 20:35
  • So the direction of vector dr can be chosen as to the right no matter of what the integration limits are? – emnha Jan 02 '17 at 21:14
  • Can I understand vector dr as follows: assume the coordinate origin is at q1. At the time t (second), the charge q2 is at vector r1. Then later at the time t + delta(t), the charge q2 is at vector r2. Then vector dr = r2-r1 when delta (t) approaches zero. If this is correct then vector dr should have direction to the q1 which is opposite to what you assumed above. – emnha Jan 02 '17 at 21:24
  • To me $d\vec r$ means the displacement when going from position $\vec r$ to position $\vec r + d\vec r$. – Farcher Jan 02 '17 at 21:56
  • So is there any reason for you to choose direction of vector dr to the right of q1? – emnha Jan 02 '17 at 22:00
  • " Taking the positive direction as to the right then dr is to the right ...": why dr should be to the right? – emnha Jan 02 '17 at 22:22
  • Simply because Walter Lewin is moving charge q2 and the distances are measured from q1. – Farcher Jan 03 '17 at 05:28
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    @anhnha It might help if you look at the answer to similar question about gravitational potential? http://physics.stackexchange.com/a/302728/104696 – Farcher Jan 05 '17 at 06:00
  • Thanks. That is a similar question. If I understand correctly, whether dr being positive or negative depends on the limit of integration. If the integration from R to infinity then dr is positive, otherwise dr is negative. Another question is that if we assume that two charges q1 and q2 are different in polarity. So they will attract each other. q2 is pulled to R by q1 and we don't need to use any force to move q2 from infinity to R. So should the potential at R is 0 then? – emnha Jan 05 '17 at 19:30
  • @anhnha If the field does work on you, i.e. You do negative work then the potential is negative. – Farcher Jan 05 '17 at 20:19
  • That is a bit strange to me. If the field does work on me, then should it be "I do nothing" and my work is zero? – emnha Jan 05 '17 at 21:22
  • What it means is that force and the displacement are in opposite directions. – Farcher Jan 05 '17 at 22:20