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I am having problems understanding the relationship between the concepts of Work and Energy in a electrostatic system.

As I know, the definition of Potential Energy is the ability to do work. In a gravitational field, that would be $$\mathrm{PE} = W(\mathrm{after}) = -W(\mathrm{before})$$ where $W(\mathrm{before})$ is the Work done to bring the object from infinity to a certain position in space, and $W(\mathrm{after})$ the contrary, from position to infinity.

Now, in the case of an electrostatic system, it seems that the Potential Energy is the Work required to bring the charges together (this is what most websites say, example). That would be $$\mathrm{PE} = W(\mathrm{before})$$ Does this mean that $\mathrm{PE} = -W(\mathrm{after})$? That would be for me minus the ability to do Work.

Am I completely missing the idea of $W(\mathrm{after})$ in a electrostatic system?

AccidentalFourierTransform
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  • @AnubhavGoel Sorry, i was not too clear. By W(before) i mean the work done to bring the object from infinity to a certain position, and by W(after) the contrary, from a position to infinity. That's why i wrote P= W(aft) = -W(bfr) –  Apr 10 '16 at 10:03
  • @AnubhavGoel minus the Work done to bring the object from infinity to a certain position in space? –  Apr 10 '16 at 10:09
  • @AnubhavGoel yes that is what i wrote. Look, we know that gravity PE is always negative. That is because the PE at infinity is zero, and the work to bring the object from infinity to a position (W(before)) is positive (since direction of movement and direction of attractive gravity force are the same). So the Work to bring it the other way, towards infinity (W(after)) is negative, which is equal to Potential Energy, the ability to do work (which is after all again, W(after)). –  Apr 10 '16 at 10:17
  • Who did the work Field or us? – Anubhav Goel Apr 10 '16 at 10:33
  • Potential energy is equal to the work done by the external agent for moving a body to its given position in space from infinity. – Anubhav Goel Apr 10 '16 at 10:37

2 Answers2

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Remember that every time we talk about work being done, we must know which force that is doing it as well as how the signs are defined.

where $W(begin)$ is the Work done to bring the object from infinity to a certain position in space, and $W(after)$ the contrary, from position to infinity."

What force is doing this work?

That would be some external force pushing them together, which they are resisting (if their signs are equal, which I assume here) giving:

$$PE=W_a=W$$

The electric force between them is repelling and is doing the same amount of negative work! But since it is the system itself that is doing this work, then we put a minus in front:

$$PE=-W_b=-(-W)=W$$

As a rule of thumb, energy into the system is positive. That's why work by external forces is positive, since that could typically bring energy in from the outside. Energy out of the system is negative. That's why work done by the system is given a minus sign, since that could typically mean work spent from the system on something external. But that is very often defined either this or the opposite way in different textbooks.

So it depends

  • first of all on which force is doing the work, and
  • second if all on how the sign on work is defined (is work done on the system positive or negative).

Often this is merely a matter of words, since if the energy content rose then we know without more math that work was done on the system, which in the case of charges means that they are closer together. Then the direction is known and doesn't have to be determined from signs.

Steeven
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  • then why is work done by external agent negative when we raise an object by a height H –  Feb 24 '18 at 02:03
  • @harambe It is not? You apply positive work to the object-earth system to raise the potential energy. – Steeven Feb 24 '18 at 02:16
  • https://physics.stackexchange.com/questions/271031/why-is-the-potential-energy-equal-to-the-negative-integral-of-a-force.The answer by zhutchens1 –  Feb 24 '18 at 02:19
  • I am getting confused at this part...I solved my textbook questions using your analysis and I got right answers but I am getting confused about the whole work done and potential energy scenario?? –  Feb 24 '18 at 02:25
  • When we lift an object by a height H with reference with the ground then Work done by gravity is mgh which is equal to the external force .So which one of them will have negetive sign and why –  Feb 24 '18 at 02:27
  • @harambe Hmm, I do not fully agree with that answer in the other question, unless I misunderstood it. We always talk about two forces in regards to potential energy: one which is doing the external work such as your lifting force - this is the external force - and one which is the "go back to normal" force such as gravity (or the spring force in a spring, or the electric repulsion between charges etc) - this is the conservative force.... – Steeven Feb 24 '18 at 12:15
  • ... When the potential energy is increased, work done by the conservative force is negative and work by the external force positive. Increasing the potential energy is the same as "charging" the system with energy (such as compressing a spring and thus storing energy in it), so we need energy from the outside. Therefore work by the external forces is positive. When the potential energy is decreased, work done by the conservative force is positive and that by any external force would be negative... – Steeven Feb 24 '18 at 12:16
  • ... That also makes sense since you new "spend" the stored potential energy meaning that the energy is given to the outside. The outside takes energy away from this system. Taking energy away corresponds to negative work. Maybe the issue is that these two simultaneous conservative and external forces are not always clearly separated. – Steeven Feb 24 '18 at 12:16
  • So in short the work done by external force is positive when it increases potential energy and is negative when potential energy is decreased because the work done by external force is spending the enrgy of the system –  Feb 25 '18 at 23:50
  • Also work done by system is negative when potential energy is increased because the system generally tries to lower the energy of the system by spending the energy the system intially so when energy is decreased the work done by the system is positive because the enrgy is given to outside which is we want the system doe everytime so the work done by it should also be positive as this is what we want and when energy is decreased the system does negative work because it is trying to decrease the energy which is contrary to what we want –  Feb 25 '18 at 23:57
  • @harambe Something like that, yes (your sentence I pretty looong there but I think I get your meaning) – Steeven Feb 26 '18 at 06:46
  • But If I take the system to be a book and Earth and raise the book to some height then work done by system will be negative but if I take the system to be book only then also the work done will be negative.In the first case,the system was acting internally but in the second case an external force was acting on the object but both of them have same sign.....How is that possible – Abhinav Apr 28 '18 at 10:41
  • @Abhinav Why should that not be possible? – Steeven Apr 28 '18 at 13:55
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There are two equivalent ways of defining the potential at a point.

  • The potential at a point is the work done by an external force in bring unit positive charge from the zero of potential (often taken as infinity) to the point.
  • The potential at a point is minus the work done by the electric field in bring unit positive charge from the zero of potential (often taken as infinity) to the point.
Farcher
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