Coulomb's law says electrostatic field intensity varies as $1/r²$ i.e ha singularity at location of charge. now if a charge distribution is accumulation of many point charges. then for a charged metallic sphere why do we get finite potential at surface? it should hav been infinity. since it is whwre charges are located. please answer as simply possible.
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$\begingroup$ As simple as possible: in the case of a finite point charge, the potential is infinite at the charge location because located on a single point, whereas for the sphere it is located on a surface, infinitely many points, hence it doesn't need to become infinite in order to generate a finite total charge. $\endgroup$– user130529Commented Feb 5, 2017 at 20:15
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2 Answers
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At a given point on the sphere, the rest of the surface is a positive distance away so contributes a finite potential. In theory you can compute the potential at a point by integrating over all these contributions, although that's not the simplest method. It shouldn't surprise you too much to learn the answer is finite, even though one point's contribution is infinite. (As a simple analogy, $x^{-1/2}$ diverges at $0$ but has finite $\int_0^1 dx$ integral.)
Suppose the sphere has radius $a$ and surface charge density $\sigma$. A concentric sphere of radius $r$ encloses charge $0$ if $r<a$ and $4\pi\sigma a^2$ if $r\ge a$. We need to find the electric field $E$, then use $E=-\dfrac{dV}{dr}$. By Gauss's law, the electric field strength at $r\ge a$ satisfies $4\pi r^2E=\dfrac{4\pi}{\epsilon_0}\sigma a^2$ so $E=\dfrac{\sigma}{\epsilon_0}\left(\dfrac{a}{r}\right)^2$, whereas for $r<a$ there is no electric field. We thus have a constant potential inside the sphere; this constant can be chosen arbitrarily. The most popular convention sets the potential to $0$ at $r=\infty$, so outside the sphere $V=\dfrac{\sigma a^2}{\epsilon_0r}$. Note this implies a surface potential of $\dfrac{\sigma a}{\epsilon_0}$.
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$\begingroup$ correct, but hardly as simple as possible an explanation ;) $\endgroup$– FloydCommented Feb 6, 2017 at 1:20
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$\begingroup$ yes J.G i agree with what u showed here......thats what i found in books. but i am failing to understand why is my logic clashing with mathametics? i mean where exactly is my logic going wrong? thanks for ur previous answer. $\endgroup$– watsonCommented Feb 6, 2017 at 3:04
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In short, the coulomb law which you correctly cite, applies only to point charges. If you go close to the electrons, this approximation is no longer valid and you have to integrate the charge distribution in space using Gauss' law.
The charge of the metal sphere can be imagined as being made up of individual electrons. These electrons are a finite distance apart and they are not point charges but 'smeared' over a very small but measurable volume. If the surface is defined as going through the center of all the outermost electrons, the electric field will still never be infinite, because the electrons are not point charges but best imagined as charge distributions, little clouds.
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$\begingroup$ As the saying goes, 'make things as simple as possible, but not simpler'. This is the latter. $\endgroup$ Commented Feb 6, 2017 at 1:57
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$\begingroup$ Or as the saying goes "Making the simple complicated is commonplace; making the complicated simple, awesomely simple, that's creativity." $\endgroup$– FloydCommented Feb 6, 2017 at 2:49
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$\begingroup$ thanks Floyd of ur answer.but what u said I think is practical problem. theoretically at r=a (surface) or v.near the value of field must have been explode.{it would be great if u could debug my logical errors.} $\endgroup$– watsonCommented Feb 6, 2017 at 3:10
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$\begingroup$ See my update answer, the logic error is assuming point charges when thinking about the field at the surface. $\endgroup$– FloydCommented Feb 6, 2017 at 3:37