# Distribution of charge when 2 metallic spheres are connected

It is given in my textbook that when 2 charged conducting sphere of different radius are connected by finite wire the redistribution of charges takes place such that the potential just outside of both spheres become equal.

But why potential is the necessary condition?

Like if net electric field just outside one sphere is 0 then even if there is some potential difference charge will not flow ,so why doesn't electeic field is necessary condition ??

And what should be the relation between the charges of two sphere after equillibrium is established when they are just touched?

• Why should electric field be 0 outside a charged sphere?
– V.G
Mar 29, 2021 at 5:09
• @Light Yagami I mean "if "net field is 0.there is no need for potential to be 0 because charge will not flow after that. Mar 29, 2021 at 5:23
• The "if" is also not possible because there is definitely an electric field outside the sphere.
– V.G
Mar 29, 2021 at 6:18
• @Light Yagami but net field due to both sphere can be 0. Mar 29, 2021 at 6:28

When you touch two spheres, you can consider them as one system, in other words, one big conductor. Now, if a conductor has different potentials on either side, then current (charges) flows through it from higher potential to lower potential. This happens as a system always wants to be in the lowest energy state, hence it will try and minimize potential energy by redistribution of charges. If we talk about net electric field, then it can be zero at only one point between the spheres, it is non zero everywhere else. Assuming distance between the spheres as d, we get: $$\frac{kq{1}q{2}}{r^{2}}=\frac{kq{1}q{2}}{\left(d-r\right)^{2}}$$ where r is the distance from one sphere, upon solving we find only 1 value of r for which the net field is zero. Since net field is zero at only one place, the charges can still move under the influence of non zero electric field elsewhere.
Consider the equation $$E=\frac{dV}{dr}$$ written in different format: $$V=\int_{ }^{ }E\cdot dr$$ now, V is the integral of E•dr, hence it is the area under the curve of electric field vs distance. In such a situation E can be negative and positive, depending on direction, which means that the net area under the curve can be zero. This doesn't imply that E is 0 everywhere, it just means the positive area cancels with negative area
When we connect two spherical conductors with a conducting wire, essentially the potential on surface of both of the conductors become the same. Basically they are equipotential surfaces. Now given $$E = dV/dl$$, and in this case, $$dV =0$$, $$E=0$$. Now you are telling that electric field is zero at only one point between the two conductors, but given what I just mentioned, shouldn't it be $$0$$ everywhere between the surface of the two conductors?