# For a parallel plate capacitor, why does decreasing the distance increase the charge?

When a capacitor is attached to a battery with a constant voltage,

where C = capacitance; Q = charge; V = voltage; E = electric field; d = distance; Є = permittivity; A = area,

there's the equation:

C = (A * Є) / d

in which distance is inversely proportional to capacitance.

However, there is also:

C = Q / V

V = E * d

C = Q / (E * d)

In C = Q / (E * d), capacitance is inversely proportional to distance, but if voltage is held constant by a battery, then wouldn't decreasing distance also cause the electric field to increase by the same proportion? If distance decreases but electric field increases by the same proportion, wouldn't there be no change to the capacitance? And therefore no change to charge, since the E * d = V is constant?

My interpretation is that decreasing distance would make it easier to hold charge separated by that distance, which would increase capacitance. But increasing electric field would make it harder to hold the separated charge, which would decrease capacitance by the same proportion.

What am I misunderstanding?

• Welcome syndromeofme, you may want to check out this MathJax guide to see how you can type math expressions in a more readable way, all the best.
– Amit
Jun 11, 2023 at 21:04

Why did you write,

But increasing electric field would make it harder to hold the separated charge . . . .

as a larger electric field means that the surface charge density $$\sigma = \epsilon \,E$$ is larger, ie more charge per unit area is being stored.

$$V= E\, d =\text{constant}$$, $$C= \dfrac{\epsilon \,A}{d}$$ and $$C= \dfrac QV = \dfrac{Q}{E\,d}$$

Whichever way you look at the situation if the separation $$d$$ decreases the electric field between the plates $$E$$ increases in proportion and the capacitance $$C$$ of the capacitor increases.

So looking at your equation $$C=\dfrac{Q}{E\,d}$$ then if $$d$$ decreases the following happens $$C\uparrow=\dfrac{\Rightarrow Q\uparrow}{(E\uparrow\,d\downarrow)}$$ which must mean that the charge stored increases ie $$Q\uparrow$$.

• Thanks for the response! I wrote "increasing electric field would make it harder to hold the separated charge" because if electric field is increased, then capacitance decreases, which means less charge would be stored per voltage difference. However, since voltage is held constant by the battery, then that means the increase in electric field is the same proportion as the decrease in distance. As you noted, electric field increases by the same proportion as distance decreases, so why do you have an increase symbol next to capacitance in the last line, as the denominator is a constant value? Jun 11, 2023 at 23:35
• Sorry to have caused confusion and as a consequence I have added a couple of symbols to my last equation. What I wanted to show is that as the capacitance increased this lead to $\Rightarrow$ and increase in the charge stored $Q\uparrow$ which I did in words before by writing which must mean that the charge stored increases ie $Q\uparrow$. Jun 12, 2023 at 7:23

You are saying V.E would be constant and their is no need of change in Charge on a capacitor. This is not possible. E will change only if Q is changed as E = Q/Aϵ. Please upvote if you got the mistake.

You can understand how charges, voltages and Electric field change very clearly with the online simulation of capacitor here.