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I know that electric field $E=-∇V.$ But since the only given parameter is the potential on the surface. Is it possible to calculate the potential everywhere (positions besides the surface)?

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    $\begingroup$ You should put the shape in the question. Don't ask the full question in the title. Is the cylindrical shape finite or infinitely long? The title mentions the electric field, but the question text asks about the potential. $\endgroup$ – Bill N Nov 4 '18 at 13:42
  • $\begingroup$ I think more information about the problem needs to be given. Where are you wanting to determine the field? Is this a physical cylinder with certain properties? $\endgroup$ – BioPhysicist Nov 4 '18 at 16:07
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I'm fairly sure you get elliptic integrals in the solution to this ... assuming of course you are not talking about an infinitely long cylinder. These are a class of somewhat tricky but extremely thoroughly studied integral. They certainly arise in similar problems in electrostatics & magnetostatics.

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There is not a unique solution without more given information. To make a simpler contrived analogy, it is like asking to find the function $f(x)$ and its derivative $f'(x)$ such that $f(3)=5$. Multiple functions have this property, so there is not a unique solution.

The issue is you are just specifying a function (potential) in some region of space. So as long as the field has that gradient on the cylinder, then it can do anything (consistent with electrostatics) anywhere else. You would need to introduce more constraints into the problem to determine a unique solution. (If there were other constraints you were assuming, you should put them in the question).

For example, if you are given the potential on the cylinder and are only interested in the field outside of the cylinder where the charge density $\rho=0$, then you are essentially solving the Laplace equation $\nabla^2V=0$, and that solution would be unique.

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  • $\begingroup$ The potential is not the derivative. The electric field components are derivatives of the potential. The field is the gradient of the potential. $\endgroup$ – Bill N Nov 4 '18 at 13:48
  • $\begingroup$ @Bill N Oops my bad! I got turned around. Will edit accordingly when I can. $\endgroup$ – BioPhysicist Nov 4 '18 at 13:59

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