Consider a point charge located above an infinite, grounded, conducting plane. The potential everywhere above the plane is the same as that of the image charge (located below it). Why is it wrong to say the total energy of the original charge is just its charge times the potential at that location? I'm aware there's a factor of one half, I just don't understand what's the problem with this specific argument, since the there isn't such a factor for the potential itself
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$\begingroup$ "The potential everywhere above the plane is the same as that of the image charge (located below it)." No. It is the same as the potential due to the image charge plus the potential due to the real charge. $\endgroup$– hftCommented Mar 8, 2023 at 23:50
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$\begingroup$ Yes, that's true. But we aren't supposed to consider the self energy of the real charge right? $\endgroup$– user35013Commented Mar 8, 2023 at 23:54
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2 Answers
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There is only one charge. You include the second charge to make the problem similar by including the image charge in such a way which satisfies the boundary condition for the problem of a point charge and an infinite plane. So, when we consider the energy of the plane and the charged particle, we are actually considering only half of the energy of the system with one real and one image charge. That's why the actual energy is going to be half of it.
Please let me know if you have any doubts.
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$\begingroup$ If I somehow set the charges on the plane first - generating a potential $V$ on the position the real charge is going to be - the charge does not aquire a potential energy of $qV$? How do I justify the factor $1/2$ mathematically (other than a physical consideration)? $\endgroup$ Commented Mar 9, 2023 at 0:19
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$\begingroup$ You can also consider the work done by an electric field for a charge configuration. $$\int_\limits{\mathbb{R^{3}}} \frac{\epsilon_{0}}{2} E^{2} dV.$$ If we consider the electric field produced by the infinite plane, it is only due to the side which has the charge. And the other-side of the plane would not have any charges(because if there was any charge then they would exert force on each other and then they would get as far away as possible to minimise the interaction between them.) So you can do the integral and find that it is 1/2 of the value with the image charges. $\endgroup$ Commented Mar 9, 2023 at 0:50
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I found out the issue.
The charge does indeed aquire a negative potential energy $qV$ due to the induced charges. But the induced charges interact with one another, producing a positive potential energy that cuts the total value by exactly half.