In the two plates of a parallel plate capacitor, the charges are equal and opposite i.e $q$ and $-q$. Can I say the potentials of these plates as $V_1$ and $-V_1$, so the difference in potentials will be $-V_1$ $-V_2$= -2$V_1$? What could be a formulae for $V_1$? What exactly it means by potential at a point on the plate? If there are 1000 charges on the plate and we want to know the potential of a point where already a charge is there, so is it like if I did some amount of work $w$ in bringing this single charge $q$ from infinity to the point against the repulsive forces of 999 charges? If a unit charge it taken there how much work is to be done i.e $w/q$? If so, why we don't have a formulae for the potential at a point on the surface of a plate?

  • $\begingroup$ In English, end each sentence with a period or a question mark. Capitalize "I". Also please use mathjax. $\endgroup$
    – DanielSank
    Commented Mar 18, 2017 at 16:35
  • $\begingroup$ -1. You are asking too many questions here. Please focus on one question. $\endgroup$ Commented Mar 18, 2017 at 17:57

1 Answer 1


Potential is relative and you can set the 0 anywhere, just like with gravitational potential. For instance, it's often convenient to say that gravitational potential at the Earth's surface is 0, and then calculate the potential anywhere else using that as reference. With the capacitors, it probably makes sense to say the potential at the surface of one of them is 0, and then you can calculate the potential at the surface of the other using q=CV

What's important is always the difference in potential. So actually, the V you calculate with q=CV is the potential difference between the two surfaces. You can pick whatever value you want for the potential at one of them, and add/subtract accordingly to get the potential at the other.


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