# Equation of capacitance of a capacitor

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 $$Q$$.

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.

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

• 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 ? Aug 29, 2021 at 16:42