I tried 3 online calculators for capacitance of parallel plates,all three show that halving the distance doubles the capacitance.I was expecting the capacitance to quadruple becose I read that electric field decreases with inverse square of distance.

Why does it double? Why it doesnt quadruple? Doubling would mean exponentional decrease with distance,not inverse square as I have read.


1 Answer 1


I read that electric field decreases with inverse square of distance.

That is for a point charge or a charged spherical conductor.

The capacitance of a parallel plate capacitor $C=\dfrac {\epsilon_0 A}{d}$ where $A$ is the area of the plates and $d$ their separation and it is also $C=\frac QV$ where $Q$ is the charge on the plates and $V$ is the potential difference across the plates.
The first formula predicts that halving the separation will double the capacitance.

The electric field $E$ between the parallel plates is $\frac Vd$ and also $E=\frac{Q}{A\epsilon_0}$.

If the separation of the plates is halved then with plate plates isolated the charge on the plates stays the same as does the electric field between the plates.
This means that the potential difference between the plates must halve and becomes $\frac V2$ so the new capacitance is $\frac{Q}{V/2}=2\frac QV$ which twice the original capacitance.

  • $\begingroup$ Can you please explain it without equations? So flat plate rolls off exponentialy with distance instead of square like spherical/point charge,but why? $\endgroup$ Sep 11, 2017 at 13:39
  • $\begingroup$ @wavscientist The electric field between the two condenser plates is constant. $\endgroup$
    – Farcher
    Sep 11, 2017 at 13:41
  • $\begingroup$ Thats true that its constant,but I fail to imagine how that makes the field fall off 2x slower than "normal". $\endgroup$ Sep 11, 2017 at 18:14
  • $\begingroup$ @wavscientist If the electric field is constant how can it fall off ......? $\endgroup$
    – Farcher
    Sep 11, 2017 at 18:15
  • $\begingroup$ Remember,I was asking about decrease/fall off with distance,by increasing the distance between plates,the electric field strenght and capacitance fall off,its probably bad choice as word,I am not native english speaker,I use it becose it tends to be said that acoustic pressure falls of with distance. $\endgroup$ Sep 12, 2017 at 5:25

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