I am trying to connect my understanding of electrostatics in terms of the Coulomb force to a thinking in field energies and potentials.

For example, let us consider a parallel-plate capacitor whose plates are charged to $+Q$ and $-Q$, respectively. Intuitively, these plates should attract each other because of the Coulomb force between opposite charges. We can also understand this by looking at the energy stored in the capacitor, i.e. its electric field: $$ E = \frac{1}{2}CV^2=\frac{1}{2}\frac{Q^2}{C} $$ with energy $E$, voltage $V$, and capacitance $C$. For an ideal parallel-plate capacitor, the capacitance is $$ C = \varepsilon \frac{A}{d} $$ with the plate area $A$ and the distance $d$ separating the plates. With this we can deduce that the capacitance will decrease with an increasing separation $d$. Hence, the energy will also decrease if the voltage is held constant and will increase if the charge is held constant. So for the case of charged plates not connected to a voltage supply the latter case has to be applied and it follows that the plates attract each other to minimize the field energy. If I understand it correctly, you could also calculate the force quantitatively by taking the spatial derivative of the field energy.

Now for the question which I still don't quite understand: Let's assume that the two plates considered above have arbitrary charges $Q_1$ and $Q_2$. Intuitively, with the Coulomb force in mind, we should find a repulsive force for equally signed charges and an attractive one for opposite charges. Is there a way to make an argument in terms of field energies and/or voltages to prove that?

I think that the voltage between the two plates alone is not adequate to describe the problem. Consider equally charged plates $Q_1 = Q_2 = Q$ (like in an electroscope). The potiential difference/voltage between the two plates has to be zero, simply for symmetry reasons. (The work needed to move a test charge from infinity to each of the plates is the same. Hence, the difference is zero.) On the other hand, charges of equal sign should repel each other; so there has to be a repulsive force in spite of a zero voltage difference.

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    $\begingroup$ What's the question? Why is there a force if there is no potential difference? (I'm not sure what the question is.) If that's it, consider that everything you said also applies to small charged spheres. (Does that help?) $\endgroup$
    – garyp
    Commented Aug 5, 2016 at 10:22
  • $\begingroup$ @garyp Broadly speaking, my question is how charge, potential/voltage, energy and forces are connected for the case of two charged conductors. The argument for the equally charged plates shows that the potential difference between the plates does not determine the energy, am I right? This is quite different from the capacitor case, where the voltage determines the field energy. So I wonder what the connection is in general, maybe stated as: What is the field energy of two charged plates? For your last question: I don't see how considering small spheres might help. Could you explain? $\endgroup$ Commented Aug 5, 2016 at 11:03

2 Answers 2


When the plates have charges of the same sign on them, we can't apply the stored energy formulae $$U= \tfrac12 CV^2=\tfrac12\frac{\epsilon_0 A}X V^2 =\tfrac12\frac{Q^2}C=\tfrac12\frac X{\epsilon_0 A}Q^2.$$ This is because we don't have a uniform field in the gap between the plates. Try sketching the pattern of field lines! The force between the plates is still given by $$F_x=-\frac{\partial U}{\partial x}$$ But the partial derivative is not easy to calculate, as $U$ is no longer an easy function of the plate separation, $X$.


Alberto Nicolis derives a relationship for force on a surface with a given charge density in this pdf: http://phys.columbia.edu/~nicolis/Surface_Force.pdf

I'll echo the result here:

$\frac{d\vec{F}}{dA} = \sigma\vec{E}_{average}$


$\vec{E}_{average} = \frac{1}{2}(\vec{E}_{above}+\vec{E}_{below})$


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