Asking this because I could not understand the correct answer for a question.

Question goes like this: A capacitor is charged with direct current, disconnected; plates are separated further afterwards.

I) Work done by electrical forces is equal to 0

My thought process was: Energy of a capacitor is $\frac{qV}{2}$ or $\frac{1}{2}CV^2$;
C = $\frac{q}{V}$ = $\varepsilon \frac{A}{d}$; because q is constant and d is increasing, V should be too.

So the potential energy of the system increases by $\frac{qV_2}{2}$ - $\frac{qV_1}{2}$
Is all of this work done by me? Is the system doing negative work?

First statement is accepted as false by the textbook.


1 Answer 1


Let's start with the work-energy theorem. Calling $W_\text{ext}$ the work done by external forces (i.e. you) and $W_e$ the work by electrostatic forces, $$W_\text{ext} + W_\text{e} = \Delta E_k$$ where $\Delta E_k$ is the change in kinetic energy. If the plates are moved slowly so $E_k \approx 0$ (or at least stays constant), $$W_\text{ext} + W_\text{e} = 0.$$ Now, since Coulomb forces are conservative, we define electrostatic potential energy $E_p$ as $$\vec F_e=-\vec\nabla E_p,$$ or alternatively, $$W_e = -\Delta E_p$$ so $$W_\text{ext} = \Delta E_p.$$

You have correctly calculated $\Delta E_p$, which is the same thing as the work done by you. The work done by Coulomb forces is the same thing with the opposite sign.

As an exercise, please draw the forces (external and electrical) on one of the plates, and as the plate is moved away from the other, calculate the work done by each force as force times displacement.

  • $\begingroup$ Could you explain the third line? The one after this: "Now, since Coulomb forces are conservative, we define electrostatic potential energy as" $\endgroup$
    – Quante
    May 6, 2023 at 18:17
  • $\begingroup$ @Quante This is the "differential form" of $W_e = -\Delta E_p$, they are one and the same. Electrostatic potential is defined such that electric field is its negative gradient, i.e. $\vec E = -\vec \nabla \phi$. If you multiply this equation by a charge, you get that equation. It relates the force on a charge to the gradient of electrostatic potential energy, i.e. how much the potential energy of the system changes as you move the charge by a tiny amount. $\endgroup$
    – Puk
    May 6, 2023 at 18:38

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