I had been working on a problem recently and i stumbled upon something I did not quite get. So the relevant part of the problem is like this:

Suppose we have two conducting spheres with radii a and b separated by a distance, r, much greater than either radius( refer to diagram below). A total charge Q is shared between the spheres. we show that when the electric potential energy of the system has a minimum value, the potential difference between the spheres is zero.<The total charge Q is equal to q1 + q2, where q1 represents the charge on the first sphere and q2 the charge on the second>

I started this part as follows:

--> Firstly because the spheres are so far apart, I assume an uniform charge distribution on either one. Next, I calculate the energy associated with a single conducting sphere by starting from Gauss's Law $\Phi_E=\iint_sE.dA=\frac{q}{\epsilon_0}$. The field due to a sphere carrying charge q is then $E=\frac{\kappa_e*q}{R^2}$ with R being the sphere's radius. From this i get the potential on the surface by using $\Delta V=-\int_cE.dr$ where the path of integration would be from R to $\infty$(we assume an infinite conductor with charge -q surrounding the sphere in question) from which i get $\Delta V=\frac{\kappa_e}{R}$; with the charge and the potential known i get the associated energy as $U_E=\frac{\kappa_eq^2}{2R}$

-->Applying all of the above to my system(from the problem), i get that the total energy corresponding to my system of two conducting spheres is: $$U_{E,total}=\frac{\kappa_e}{2}[\frac{\kappa_eq_2^2}{2b} +\frac{\kappa_eq_1^2}{2a}]=\frac{\kappa_e}{2}[\frac{\kappa_e*(Q-q_1)^2}{2b} +\frac{\kappa_eq_1^2}{2a}]$$

-->Now differentiating with respect to $q_1$ and setting the derivative to 0, i get $q_1=\frac{Qa}{a+b}$ and using the given information Q=$q_1+q_2$ i get that $q_2=\frac{Qb}{a+b}$ using these in the expression for the potential of a single sphere i get $V_1 ,V_2$ the difference of which really is 0. Now i must ask:

  1. what are the physical interpretations of what i did just now? that is, why is the potential difference 0 when the energy is minimized?
  2. Is there a more fundamental approach to determining this relation? 3)can any one give me an intuition behind this? why does it work like it does? why does a minimum electric potential energy mean no potential difference?.

2 Answers 2


Initially, we know minus gradient of potential yields the force on a point where the point lives in vector field which satisfies the physical necessities. Considering your question, minimizing potential energy means the getting the minimized gradient magnitude for specific point(s). This method is basically a way to find equilibrium for the forces. So, which allows to reduced or minimal forces that can effect each of the charges and as we know equilibrium can be reached if there is no unbalanced force. The main reason that gradient of potential yields force is, in the small neighbourhood of a point if potential is changing and in total some points have higher values of potential then, from classical thermodynamics law energy flows higher potential to lower so, if potential difference is not approaching 0 in the infinitesimally small neighborhood then unbalanced forces will breaks the equilibrium.


Sorry for the loooong answer. Please read it completely before voting.

If you have a non-minimum potential energy configuration, it means that something is holding the particles or charges that disables them of attaining minimum-energy configuration.

Also, minimum energy configuration is the stable equillibrium of a system. If you wish to find a minimum energy configuration for a system of particles or charges, just let them free. They will finally achieve an equillibrium which is stable and that is the minimum energy configuration.

Now getting on to your question, We have to find such a distribution that minimizes the potential energy. So, let us set the charges free and see what happens.

Give all the charge to the first conductor. We see that the charges are not yet completely free under our given constraints. However, we can free them if we connect both the conductors with a connecting wire. In that case, the charges will be completely free in the constraint mentioned (that the charges must not leave the conductors).

After all the charge transfer, remove the wire. Now, we are sure that the configuration is the minimum energy configuration and also the potential difference is zero.

Conclusion: The charge distribution shown by a conductor is the minimum possible energy configuration possible in the given conditions.

  • 1
    $\begingroup$ I see. Thanks for your input, i will just study it once more and i think i will have a better understanding. $\endgroup$
    – F.N.
    Sep 22, 2020 at 4:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.