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I mean potential of any thing requires a reference point to be defined. Potential at a distance r from a point charge q is given by;

V=electrostatic constant × q/r

;but I dont understand how can I calculate potential of an isolated conductor or of any isolated object using this definition.

I have been trying to understand self-capacitance which is the amount of charge to raise the potential of an isolated conductor by one volt. But this needs me to understand what is potential of an isolated conductor. Means, I have got to know how is it defined? Don't direct me to the google or wikipedia for I have tried my best to find about it on these sites already. Although wikipedia got very close to defining the potential of isolated conductor. It also tries to tell about the reference point but I still don't understand the story. I want a very simple illustration.

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If you have an isolated finite conductor (either charged or uncharged), its surface is an equipotential. The potential of this conductor you'll find by moving a small positive test charge $q$ along any path $\gamma$ from infinity to the surface of the conductor and determine the work $W$ done against the electrostatic force $\vec F$ in doing so. Then the electrostatic potential $\phi$ of the body is $$\phi=-W/q=-\frac{1}{q}\int_{\gamma}\vec F d\vec r=-\int_{\gamma}\vec E d\vec r$$ Depending on the net charge on the conductor and on induced charges and fields by nearby charges, the potential $\phi$ can be positive, negative or zero for example when there is no electric field.

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  • $\begingroup$ Where does the electric field come from? $\endgroup$ – user104909 Dec 7 '16 at 13:43
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    $\begingroup$ @Suyan Naeem - The electric field comes from any charges on the conductor and on any other charges, especially those near the conductor. $\endgroup$ – freecharly Dec 7 '16 at 14:52
  • $\begingroup$ I just wanted to add minus sign after $W$ in the equation but as long as 6 character edit is required, I had to add some extra words in last which I hope are meaningful. $\endgroup$ – user104909 Dec 8 '16 at 7:43
  • $\begingroup$ I also want to ask why the conducting surface has have to be equipotential? Is it necessary that it lies perpenduclar to the field at every point or is it a property of a conductor? $\endgroup$ – user104909 Dec 8 '16 at 7:49
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    $\begingroup$ @Sufyan Naeem - The surface of any conductor has to be equipotential. If it were not, there would be an electric field component parallel to the surface moving (und thus redistributing) the charges (electrons) until there is only a vertical electric field at the surface. In any electrostatic field, the electric field is always perpendicular to the equipotential surfaces in 3D space. $\endgroup$ – freecharly Dec 8 '16 at 21:20

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