Suppose a uniform electric field $E$ exists between to oppositely charged metal plates (one is positively charged and one is negatively charged). If the plates move apart, and the charges on each plate stay the same, why does the potential difference increase?

From my understanding (which I know is somehow flawed), the electric potential of each plate seems to vary inversely with distance $r$. However, as distance increases, the potential difference between the plates increases. I think the potential on both plates would decrease, but this does not determine anything about the difference between the potentials.

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    $\begingroup$ What would happen if you let go of the two plates? What does that mean for the case when you move the plates farther apart? $\endgroup$
    – CuriousOne
    May 10, 2015 at 23:37
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    $\begingroup$ Wouldn't the plates attract, because they're oppositely charged? So, perhaps as the plates move further apart, the distance they travel when released is greater, kind of like gravitational potential of a tennis ball held at waist-level versus the gravitational potential of a tennis ball held at shoulder-level? @CuriousOne $\endgroup$
    – user80932
    May 10, 2015 at 23:47
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    $\begingroup$ Very good! The two plates do attract, which means that one has to exert a force to keep them apart. Even better, you have identified that movement against a (conservative) force (like gravity) is the hallmark of a potential. When we move something against such a force we have to expend energy, which then becomes potential energy of the system. So when you are moving the plates of a capacitor apart, you are giving it potential energy. The remaining hard part is for you to identify the mechanical definition of potential with the electrical one (charges moving in electric fields). Does that help? $\endgroup$
    – CuriousOne
    May 10, 2015 at 23:54
  • $\begingroup$ Are you thinking of an external electric field, or the electric field due to the charge on the capacitor? In the second case, see physics.stackexchange.com/q/183137/44126 $\endgroup$
    – rob
    May 11, 2015 at 1:39

2 Answers 2


we know, $$E=V/d$$ (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html ) Where $V$,$E$ and $d$ are potential difference ,electric field and distance between the two plates respectively.

Electric field between the plates only depends on charges of the plates and since charges must be conserved so when the plates are moved apart charges(the amount) on the plates donot change. So $E$ is constant. ( http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html )

Now,look at the equation, you increased the distance $d$ so in order to make the LHS constant(which is $E$ ) $V$ must also increase.


Potential depends upon the dielectric medium. In your case air acts as a medium. Capacitance is directly proportional to Permittivity of free space, multiplied by the Area of the plate, and inversely proportional to the distance between the plates. The capacitance decreases with increase in distance.