Potential is a function defined for conservative force(s). Now consider what happens when we connect two metallic spheres with a conducting wire. The whole system will act like a big metallic thing and hence the surface of the system will be equipotential. But saying that isn't enough. I must also specify which for which force the electric potential is defined. In this case, the potential at a point is defined for the net force due to each charged particle of the system at that point. Let me explain this another way. The equipotential surface due to a charged particle is any sphere that has its centre at the charge. Here obviously the potential is defined for the force due to the charged particle. Now consider a dipole, where the equipotential surface is the plane perpendicular to the dipole moment and passes through the centre of the line joining the two charges. Here the potential is defined for the net force due to both charges. But if we consider a charge in the dipole, the equipotential surface due to its force only is still any sphere with its centre at the charge and that perpendicular plane is not equipotential for this charge alone.

Let the name of the spheres be $1$ and $2$. They both have radius $r_1,$ $r_2$ and charge $q_1,$ $q_2$ respectively. Now it is written at many places$^*$ that since the surfaces are equipotential we can write that $V_1=V_2$, where $V$ is potential due to the spheres at a point on the surface of the spheres. Hence $\frac{Kq_1}{r_1}=\frac{Kq_2}{r_2}.$ But I think this is not correct.

I have explained in the first paragraph that the spheres have equipotential surface and the potential is defined for the net force. But in the second paragraph, it is assumed that potential at a point on a sphere is due to the sphere only, the effects of the wire and the other sphere are ignored. But on ignoring them we can't be sure whether the potential will be the same at the surface since, again, the surfaces are equipotential where potential is of the net force due to both spheres and the wire. So is the second paragraph wrong?

$*$1, 2, 3, 4, 5.


1 Answer 1


Okay let's make it clear:

When we say two bodies are at equal electric potential thus making the system electrostatic and we mean their individual potentials.

This is what you think - If potential is calculated with reference to the configuration of the whole system then why are we considering the individual potential here in this example.

See the electrostatic condition is a consequence of electric potentials being equal at all the points in the system. In your example Sphere 1, Sphere 2 and the wire altogether make the system.
When you connect the wire to both the spheres, altogether they make a conductor and a conductor will do anything to make electric field inside it equal to zero. Since electric field inside the system is zero so there will be no potential difference across the volume and potenial of each and every point is same.

The above explanation when said in revrse order will clear your doubt so here it is :

When the potential (individual) of each and every point in the system is same then there is no potential difference across the volume of the system which concludes there is no electric field which concludes the system is in electrostatic condition.

Therefore we use the individual potential in electrostatic condition as used in above question.

Hope this helps

  • $\begingroup$ "Since electric field inside the system is zero so there will be no potential difference across the volume and potenial of each and every point is same." It is wrong, all the points inside the conductors have same potential even though net field is zero there. But note that the potential of an inside point may be different from that of surface points. $\endgroup$
    – Osmium
    Dec 2, 2021 at 11:36
  • $\begingroup$ @Osmium In your case the potential at the surface is same as it is inside the sphere i.e. potential of a uniformly charged metallic sphere is kQ/R at its surface and also inside it. $\endgroup$
    – Spencer
    Dec 2, 2021 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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