I don't understand why the working voltage of a capacitor increases with the plates' distance.

Suppose we have 2 capacitors in series where $C_1=12,0\mu F$ and $C_2=4,0\mu F$. The voltage over both capacitors is $100V$.
The charge is the same for capacitors in series and because of the relationship $$C={Q\over V}=>V={Q\over C}={Qd\over \epsilon_0 A}$$ the working voltages are $V_{1}=25V$ for $C_1$ and $V_2=75V$ for $C_2$.

Since the term $Q\over\epsilon_0$ is constant, what's defining the working voltage is the ratio $d\over A$. So by increasing the distance $d$ or decreasing the area $A$ you'd get a higher working voltage.
I can't picture the physical meaning of this relationship.


3 Answers 3


The working voltage of a capacitor depends on the dielectric strength of the insulator.

While electrical breakdown is actually a very complicated process with lots of non-linearities, you can simplify the design of a capacitor by saying "the electric field on the insulator must not exceed X".

Once you have said that, and you realize that the electric field strength scales with voltage divided by distance

$$E = \frac{V}{d}$$

it follows that if $E_{crit}$ is held constant, $V_{working}$ scales with $d_{plates}$.

Intuitively, if you think of a capacitor with a large spacing as being two capacitors with half the spacing, you should be able to reason that adding a "virtual plate" in the middle should not change the electric field.

enter image description here

If you consider that each of the "intermediate" capacitors on the right has half the thickness (thus double the capacitance), but two in series gives me the same net capacitance, it should be easy to see that

  1. the electric field is the same in both cases
  2. the capacitance is the same in both cases
  3. the voltage is half for the each of the two "half" capacitors
  4. therefore the working voltage scales with the separation
  • $\begingroup$ Thanks for your insight. Though I've asked the intuitive meaning of direct proportionality between $V$ and $d$ not a derivation. $\endgroup$ May 20, 2015 at 13:06
  • $\begingroup$ @yurihbss - I have tried to add some more "intuition" for you $\endgroup$
    – Floris
    May 20, 2015 at 15:31
  • 1
    $\begingroup$ that's the intuition I was talking about $\endgroup$ May 20, 2015 at 15:35

Since the charge Q have to be the same for both capacitors and you need more voltage to to push that charge in the capacitor with less capacitance then you must have more valtage difference in $C_2$ The mechanical analogy is a configuration with 2 springs in parallel that move the same distance from their equilibrium position need more force on the spring with greater elastic constant to accomplish the same displacement


Consider a plane of charge. For distances from the plane which are small compared to the size of the plane the electric field will be constant. In the capacitor, with two oppositely-charged plates and relatively small gap, the electric field is a constant, depending only on the overlapping plate area and the amount of charge, $E_{gap}=Q/(k\epsilon_o A)$, where $k$ is the dielectric constant of whatever is in the gap.

For a constant electric field, the voltage between two planes perpendicular to the field depends linearly on the distance between the planes, $V_{gap}=E_{gap}d$.

The capacitance tells us the ratio of stored charge to resulting voltage, and is strictly a mechanical/geometrical result. The capacitance is independent of the actual charge or voltage on/across the device. If you change the area or the gap size, you change the capacitance.

It's similar to having a room. Depending on the number of people (charges) in the room, the people density (voltage) will change. If you change the dimensions of the room, you change the volume of the room, and the people density will change even if you don't change the number of people. While the actual dimensional meanings aren't the same, the idea that changing the mechanical properties changes the capacitance is the important point. Voltage and charge vary in accordance with the actual capacitance, so that their ratio is a constant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.