# Relationship between work and energy in an electrostatic system

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?

• @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)
– user113932
Apr 10, 2016 at 10:03
• @AnubhavGoel minus the Work done to bring the object from infinity to a certain position in space?
– user113932
Apr 10, 2016 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)).
– user113932
Apr 10, 2016 at 10:17
• Who did the work Field or us? Apr 10, 2016 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. Apr 10, 2016 at 10:37

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.

• then why is work done by external agent negative when we raise an object by a height H
– user184271
Feb 24, 2018 at 2:03
• @harambe It is not? You apply positive work to the object-earth system to raise the potential energy. Feb 24, 2018 at 2:16