Imagine the two terminal of a parallel-plate capacitor are connected to the two terminal of a battery with electric potential difference $V$. If the capacitance of the capacitor is $C$, and the area of each plate is $A$, then how can I express the charge stored on each plate ($\pm Q$) in terms of only $C$, $V$ and $A$?

My Attempt:

We know that the voltage across the capacitor is $V=\frac{Qd}{\epsilon A}$, where $d$ is the distance between the plates (since there is only one capacitor connected to the battery of electric potential $V$), and the capacitance is $C=\frac{\epsilon A}{d}$. We can rearrange this to get $d=\frac{\epsilon A}{C}$; then plugging it to the first equation we get: $V=\frac{Q}{\epsilon A}\cdot \frac{\epsilon A}{C}= \frac{Q}{C}$, so thatthe stored charge is $Q=CV$. Unfortunately, this is the elementary formula that relates only $C$ and $V$ to $Q$, but not the area $A$ of the plates.

I need a formula that relates $C$, $V$, and $A$ to $Q$.


3 Answers 3


The answer is $$Q(C,V,A)=CV,$$ which is why you're having trouble.

This is like the equation for a horizontal line: $$y=f(x)=2$$

Has no $x$ in the equation.


The individual quantities in $\frac d {\epsilon A}$ can freely vary as long as the overall quantity stays the same. For instance, if you double both the distance and the area, then $\frac d {\epsilon A}$ remains constant, and thus $Q$, $V$, and $C$ can remain constant. This shows that you can have two different capacitors with the same $Q,V,C$ but different $A$.

the stored charge is Q=CV. Unfortunately, this is the elementary formula that relates only C and V to Q, but not the area A of the plates.

But it does relate $Q$ to $C,V,A$. You can rewrite it $Q^1=C^1V^1A^0$ if you want. If you were doing a log-log linear regression on $\log Q = \beta_1\log C+\beta_2\log V+\beta_3 \log A$, you would find that $\beta_3=0$. That's the value of $\beta$. No further rearranging of terms is going to get a different $\beta$. That is the one correct answer. The effect size of $A$ on $Q$, after controlling for $C$ and $V$, is zero.

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    $\begingroup$ What would be the energy lost by the battery in the process of charging the capacitor? $\endgroup$ Commented Aug 30, 2021 at 6:31
  • $\begingroup$ @NazmulHasanShipon I think "work done" is better phrase than "energy lost". It's $\frac {Q^2}{2C}$. Is this a rhetorical question? $\endgroup$ Commented Aug 30, 2021 at 17:01

You have the formula in the question: $Q=CV$ (which holds for any capacitor, not just one with a parallel plate). There is no way to include any additional dependence on the plates' area $A$, because once you know $C$, and $V$, $Q$ is already completely determined. So, when written as a function of $C$, $V$, and $A$, the dependence of $Q(C,V,A)=CV$ on $A$ is trivial.

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    $\begingroup$ How should I approach the question? Will I point out $C=\frac{Q}{V}$ as the elementary formula then show that $Q=CV$ or It would be better to approach it in the way I did in My Attempts ? $\endgroup$ Commented Aug 29, 2021 at 16:42

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