I have learned that electric field lines goes from higher potential to lower potential. First of all, how does this happen?

And if the above is the case then suppose I have two uniformly positively charged conducting spheres. The charge on one is greater than on the other, and they are initially separated by infinite distance. Here, the electric potential of the sphere having more charge is greater than the potential of other having lesser charge, so according to the above statement, the electric field should go from higher to lower potential. However, since both spheres are positively charged, the electric field lines should not be continuous from one sphere to another; in fact, there will be a null plane between them. (Both spheres produce their own electric fields directed away from the spheres.) So how is this contradiction resolved? Please explain the core concept.

Also if in above case both the spheres are joined through a conducting wire, then charge flows till the potential are equal. How does this happen? If the field is not continuous, then how does charge flow from higher potential sphere to the lower potential sphere?

  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Aug 27, 2021 at 13:45

1 Answer 1


The question is not clear, but what I understood is that you want to know the reasons for the following :

$1.$ Why is it so that $\displaystyle \vec{E}$ points towards decreasing potential ?

$2.$ Why there is a flow of charges from the higher potential to lower when two positively charged conducting spheres are connected together ?

Reason for the first lies in the fact :

$\displaystyle W_{ext} = \int \vec{F_{ext}} \cdot \vec {\mathrm {d}r} \implies W_{ext} = - Q \int \vec{E} \cdot \vec {\mathrm {d}r} $

And we can also derive a covector field $\displaystyle \mathrm {d}\phi = \frac {\partial \phi}{\partial r} \ \mathrm {d}r $ to mark the potential of the Field.

Thus $ \displaystyle W_{ext} = \int\frac {\partial \phi}{\partial r} \cdot \mathrm {d}r \implies \vec{F_{ext}} = \frac {\partial \phi}{\partial r}$

And $\displaystyle -Q\vec{E}\cdot \mathrm {d}r = \mathrm {d}\phi $

This tells that $\mathrm {d}\phi $ points in opposite direction of the $\vec {E}$ , This in turn suggest that Electric field moves from high to low potential.

For the second part: $\vec {E}$ doesn't exist in isolated conductor, and Electric field is generated when an external conducting wire is connecting both the spheres. Here the whole system (both spheres and wire) act as 'conductor in isolation'. Thus to make the net potential generated across the wire zero a field is created.

Initially there were two separate conductors at fairly enough distance so as to avoid the electric influence, with an insulating air gap thus they had a null point between them.

  • $\begingroup$ I didn't understand the second part. An isolated charged conductor also produces electric field (only outside the conductor). If we draw field lines of two positively charged spherical conductor placed at some distance then the field lines from each conductor will terminate on the null point (plane) lying between them. BUT what different happens when we connect them with wire ?? How charge continuously flow from higher potential to lower potential even if it will experience a opposite force as it gets near the other sphere (+ve +ve repulsion) ?? $\endgroup$
    – Spencer
    Aug 28, 2021 at 4:45
  • $\begingroup$ This what I am telling : the most intrinsic property of a conductor is that it would do just ANYTHING (say it to be charge movement or something else) to make field inside the whole conducting system zero. There will no longer exist an opposing external electric field (Say it to be the property of the conductor surface that it is an equipotential surface.). The field line pattern changes as soon as you connect the wire causing the null point to vanish and transformation whole system as a new conductor. $\endgroup$ Aug 28, 2021 at 7:36
  • 1
    $\begingroup$ Thank you very much. Now I completely understood. $\endgroup$
    – Spencer
    Aug 28, 2021 at 11:35

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