1. Jun 22, 2014

### fgp40

Hi everybody, I'm new at the forum. I've started learning physics on my school last year.

I've a question.

How can we measure the voltage stated on the definition of capacitance? Just V=Q/C ?

Or how to measure the capacitance? What's about voltage?

I didn't know that Q=C.V. What is that? The definition of capacitance, voltage or electric charge?

Generally I didn't understand that. Can you help me please?

I'm sorry for my bad English.

2. Jun 22, 2014

### Staff: Mentor

Welcome to the PF.

The voltage V of a capacitor is directly proportional to the charge stored on it (Q), and inversely proportional to the capacitance C. Given some capacitance value, if you double the charge on the capacitor, you double the voltage of the capacitor.

You measure the voltage of the capacitor with standard measuring devices, like oscilloscopes and digital voltmeters (DVMs).

3. Jun 22, 2014

### Staff: Mentor

This is the definition of capacitance. If a capacitor has a potential difference V between its two "plates" when they carry charges +Q and -Q, then its capacitance is C = Q/V.

4. Jun 22, 2014

### fgp40

Thank you for answer. And if the capacitor is a sphere or just a plate? How to know their capacitance? With experiments?

5. Jun 23, 2014

### Delta²

Given the design of the capacitor and the dielectric, one can calculate the capacitance using Gauss Law with some extra assumptions. Its better to think that capacitance depends on the design of the capacitor (e.g size and shape of plates and distance between them and dielectric constant of the dielectric material) and just view the relationship C=Q/V as an additional relationship that connects those 3 physical quantities of charge in plates, voltage between plates and capacitance.

Last edited: Jun 23, 2014
6. Jun 23, 2014

### Philip Wood

Perhaps the place to start is understanding what the capacitance of a single conductor means, in other words how you should interpret Q/V, the defining equation. Q is simply the charge on the conductor and V is the potential of the conductor with respect to infinity. In the case of a sphere, and no doubt for certain other shapes, this potential has a well defined value; for a sphere of radius a it is $V = \frac{Q}{4 \pi \epsilon_0 a}$. So the capacitance of the sphere is $C = \frac{Q}{V} = 4 \pi \epsilon_0 a$.

This interpretation may seem very unlike that for a two plate capacitor. But in fact, it's not so different… The electric field lines from our single, 'isolated conductor' will, in practice, end on negative charges induced on very distant objects. The total negative charge induced is, as Faraday showed, equal in magnitude to the positive charge on the 'single, isolated conductor'. So we can, if we wish, regard such a conductor, plus the distant surroundings, as a 'pair of plates' with equal and opposite charges, and a p.d. between them. Whether this helps is a matter of taste.

Last edited: Jun 23, 2014
7. Jun 23, 2014

### fgp40

OK. Thanks I understood.