I have read that potential energy of a dipole in an electric field is taken to be zero at 90 degree as they say that it is a reference.
But there must be some reason behind it?
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2$\begingroup$ Potential energy is relative, so you have to pick a reference. Usually, you either pick the top, or the bottom. What kind of "reason" would convince you? $\endgroup$– FlorisCommented Sep 28, 2015 at 16:17
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$\begingroup$ I asked from my physics sir the same problem and he said that while bringing a dipole from infinity to 90 degree in electric field the net potential created is zero but I was not able to understand it? Is he right? $\endgroup$– Lakshay GuptaCommented Sep 28, 2015 at 16:30
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4 Answers
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Imagine you have a single charge (not a dipole). We say it has zero potential energy at infinity. If we bring it close to another charge, it will end up with some non-zero PE:
$$V_+ = \frac{Q}{4\pi \epsilon_0 r}$$
Now if we have a second charge with opposite polarity, and we move it to the same distance $r$, it would have potential energy
$$V_- = -\frac{Q}{4\pi \epsilon_0 r}$$
The sum of these two is zero.
Now a dipole is really just the limit of two opposing charges brought infinitely close together while maintaining a constant product $p = Qd$ where $Q$ is the charge, and $d$ is the distance between them. The dipole vector points along the line joining the two charges.
Putting it all together, the distance $r$ for the two charges is the same if the dipole is at right angles; and so there is no net work done when you move a dipole from infinity to a distance $r$, and keep the dipole at right angles to the electric field. See this diagram:
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$\begingroup$ Thanks for answering! I don't understood the last part .. How r will be the same distance if the dipole is at right angle? $\endgroup$ Commented Sep 28, 2015 at 17:26
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$\begingroup$ @LakshayGupta - I wonder if the diagram I just added makes it clearer to you. $\endgroup$– FlorisCommented Sep 28, 2015 at 18:10
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$\begingroup$ @Floris There will be no work done How do you account for the work done when the two charges constituting dipole are brought close? $\endgroup$ Commented Sep 16, 2021 at 0:26
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$\begingroup$ @LalitTolani the charges are assumed to be constituted into a dipole “at some time”, and the energy released during that operation is ignored (we have to set a zero somewhere, and we set it for the dipole at infinity). From there, moving perpendicular to an electric field doesn’t require any work so the potential energy is zero everywhere. There is a big world of possible electrostatic problem out there - this is just one narrow one. $\endgroup$– FlorisCommented Sep 16, 2021 at 0:31
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The potential energy is zero since for a dipole the potential energy is zero at its equatorial plane, and potential energy is nothing but potential $\times$ charge.
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$\begingroup$ I guess your caps lock was on ..... $\endgroup$– ShubhamCommented Sep 28, 2015 at 16:45
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Another way to think about this problem is to consider the direction of the electric field along any point along the perpendicular bisector of the line between the dipole. It is, in fact, purely in the direction of the dipole (i.e. a test charge would repel from the positive end and go to the negative end) and it is 0 in the direction of this perpendicular bisector. This means that no work would be done on a test charge that moves along the path of the perpendicular bisector (since the field is always perpendicular to this test charge)
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When the dipole (its moment) is perpendicular to the electric field, they are on the same equipotential line. So the net work done to bring them from infinity to their places is $qV+(-q)V = 0$ (without considering their electrostatic interaction with each other). So, it is convenient to choose this orientation as zero potential energy. Also, the potential energy formula gives a nicer formula with this setup. But again you can choose any orientation as your reference point.
