While studying the unit of electrostatics, I came across the formula $$C=\dfrac QV,$$ Where C refers to the capacitance of the conductor, $Q$ refers to the charge present on the conductor and $V$ refers to the potential of the conductor. I don't understand what is meant by the potential of a conductor. Isn't electric potential defined for a point in space? How can an extended object have a potential? Could someone please explain what exactly is meant by the potential of a conductor?

P.S. Is it assumed that the surface of the conductor is an equipotential surface, hence the common potential on all points on the surface of the conductor being referred to as the potential of the conductor?

Thanks a lot in advance!!!

  • $\begingroup$ so: a sequence of adjacent points defining a surface have the same potential. $\endgroup$ Aug 2, 2020 at 13:09

1 Answer 1


First, a very important point where you're wrong

Isn't electric potential defined for a point in space?

No, it's defined as the line integral of the electric field between two points $$V_{A\rightarrow B} = -\int_A^B \vec{E}.\text{d}\vec{l}$$ The thing is that usually you omit to say which is the second point. For example, the usual expression of the electric potential for a point charge $$V(\vec{r})=\frac{1}{4\pi \epsilon_0} \frac{Q}{r}\hat{r}$$ express the difference of potential between $\vec{r}$ (which represents $B$ in the above equation) and infinity.

Going to your question, as pointed in the comments, you can imagine a extended object as a bunch of points. In the case of a conductor, you know that all points have the same potential, so you only have to know the potential of one of those.

Finally, for circuits you usually say "potential of an element" when you're really talking about "difference of potential introduced by the element". For example, if your circuit have a battery of $12\,V$, it means that if you take point $A$ of the integral at one side of the battery, and $B$ at the other, the difference of potential between $A$ and $B$ is $12\,V$. Your expression can be interpreted in the same way.


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