If we have two charges of +1C a distance of 1m apart, then if we fix one and bring the other from infinity, the work done = +k. Now, if we fix the other one and bring this one from infinity, the work done = +k again. So to bring the two charges to a distance of 1m, the energy we must provide = 2k, and the electric potential energy of the system is 2k. Which step is wrong?
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$\begingroup$ Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$– Kyle KanosCommented May 10, 2017 at 10:12
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1 Answer
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EDIT: The scenario is not fully clear. But I am going to assume that these are the only two charges in the system.
Your doubling of the work done is wrong. You do not have to do twice this amount of work.
Remember, electric potential is defined as the work done to bring a positive test charge to a particular point in space from infinity (or whatever your zero potential reference point is).
The first charge you bring will require no work. There are no other particles in your system providing forces for it to overcome.
When you bring your second charge, you will have to do work because it experiences a repulsive force from the 1C charge of magnitude $$\frac{1}{4 \pi \varepsilon_0}$$
In this case, the electric potential energy of the system will then be the same and I will leave you to verify that with the relevant integrals or intuition.
answered May 9, 2017 at 21:40
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$\begingroup$ Thank you. But then what will be the energy of one charge ( in the same system)? Will it be half the total energy? If so, then why is the energy shared equally? $\endgroup$– JohnCommented May 9, 2017 at 21:52
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$\begingroup$ The electrostatic potential energy of each of the charges will be $\frac{1}{4 \pi \varepsilon_0}$ ('), not half of this. Consider it from the reference point of $q_1$, then it is receiving a repulsive force from $q_2$, giving it potential energy = ('). Likewise, the same is true from the reference point of $q_2$, giving it potential energy = ('). $\endgroup$ Commented May 9, 2017 at 21:57
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$\begingroup$ I thought I was somewhat clear on it, but now I'm really confused... the energy of the entire system is the energy of one charge? Generally, if we have two objects with kinetic energies E1 and E2, the energy of the 'system' is E1+E2, so this is something rather new for me.. $\endgroup$– JohnCommented May 9, 2017 at 22:21
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$\begingroup$ The energy of one charge is not defined. Potential energy is defined for pairs of charges. You are on to something: kinetic energy is defined only for single particles. Potential energy is defined only for pairs of objects. (Someone recently pointed out to me that this is not strictly true, but for an assembly of point charges, it is correct.) $\endgroup$– garypCommented May 10, 2017 at 1:22
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$\begingroup$ There are two charges after bringing the first one so (+k) one now for the third there would be a work done against the field of +1 and +k so there will be total 3 energies $\endgroup$ Commented May 10, 2017 at 18:00