If there exists a charged conductor, the surface has a potential. This potential at a point on the surface is created by the charge distribution of all the other points on the surface. This means that all the electron except for the point where the potential is calculated contribute to the potential. But if that is so, when atoms are so close to each other, even if there is barely any charge right beside the point, the potential will be turn out to be extremely high. Now, where am I going wrong? If I am wrong, then what potential is it when we are talking about equipotential surfaces (no external electric field)?? Thanks..
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$\begingroup$ "even if there is barely any charge right beside the point, the potential will be turn out to be extremely high." But you have to also consider that the adjacent atoms contain very little amount of charge; which will not help make the potential go very high. $\endgroup$– MockingbirdCommented Dec 23, 2016 at 16:09
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$\begingroup$ An uncharged conductor also has a potential. $\endgroup$– my2ctsCommented Sep 5, 2019 at 5:59
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3 Answers
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The (equi)potential at the surface of a conductor (relative to 0 at infinity) is not only a function of the net charges on the surface, but depends also on the charges in the 'vicinity' of the conductor. However it's always an equipotential surface (in electrostatics). Let's therefore focus on the potential created by the surface charges and let's assume we add electrons to the conductor. By keeping adding them, they will (almost instantaneously) redistribute themselves such that the electric field inside the volume of this conductor is zero. However, these electrons will try to keep away from each other as much as possible, so they won't be residing on neighboring atoms until you add about $10^{16}$ electrons to a surface with an area of about $1 \mathrm{cm}^2$. Then, indeed you would create a huge potential at the surface, in the order of: $$\sum_i \frac{q_i}{4 \pi \epsilon_0 r_i}\approx\frac{-1.6 \times 10^{-19}\cdot 10^{16}}{1.1 \times 10^{-10} \cdot \frac{1}{2} \times 10^{-2}} V\approx -3 \times 10^{9} \, V, $$ where $q_i$ here stands for the charge of one electron, $\epsilon_0$ the permittivity of vacuum and $r_i$ the distance of this charge to the (arbitrary) point on the surface where you want to know the potential. Note that in this approximation I used a value of $\frac{1}{2} \times 10^{-2}$ m as an 'average' distance to the charges.
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The potential of a point is not a function of only the charges in vicinity of the point.
to solve the ambiguities, let's look at the definition:
$$V(\mathbf{r})=-\int_{\mathbb{infinity}}^{\mathbf{r}}\mathbf{E}\ \cdot d\mathbf{l}$$
The alternative for the equation above is:
The electric potential of a point is the work that needs to be done on an infinitesimal positive test charge to move it slowly from infinity to that specific point, divided by the magnitude of the test charge.
assuming one electron on the surface of the conductor, if you take it from infinity to its position, slowly (Not for it to gain velocity and therefor kinetic energy), you will have to do a not-very-large work. and its reasonable.
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$\begingroup$ So how do I calculate it. I have tried doing that but since there is a charge density at the point we want to calculate the potential at, it turns out to be infinity. Now, i dont know how to calculate the sum of all other potentials of points except the point of calculation in an integral. $\endgroup$ Commented Dec 24, 2016 at 3:39
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It all depends on scale. Averaged over a few atomic distances the potential is constant. At atomic scale and below it obviously is not.
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$\begingroup$ The negative voter should explain himself. $\endgroup$– my2ctsCommented Sep 7, 2019 at 14:35