I was just introduced to the concept of capacitance, and I am having a hard time understanding what my textbook is trying to convey…

It says (I am translating):

"A conductor with a high capacitance can bear a high charge while remaining at relatively low potential".

Where is the potential? Between what and what?

My second doubt is that it would seem to be a greater achievement to maintain a high charge with a high difference in potential than with a low difference in potential.

I am clearly not totally understanding the concept itself: could somebody just explain the gist of what I am missing?

• The potential refers to the difference in charge of the two plates, simply said. – QuIcKmAtHs Aug 24 '18 at 11:52
• The word potential should strictly speaking be potential difference. – Qmechanic Aug 24 '18 at 11:53

As said in the other answer, if we have only one conductor, the potential is between the conductor and ground (which in the case of a single conductor would be infinity distance from the conductor).

A capacitor is typically made out of two conductors with a dielectricum in the middle, and the potential is the potential between these conductors.

Where is the question abouit your second questiong from? Capactiance is $C=\epsilon\frac{A}{d}$ where $\epsilon$ is the permittivity of the dielectrum, d is the distance between the plates (if we assume a paralell plate capacitor) and A would be the area of the plates. Capacitance can also be defined as $C=\frac{q}{V}$. As you can see, charge and voltage are dependent on one another, so changing one causes a change in the other typically, unless you redesign your capacitor.

I hope this help. I am no capacitance-professional myself so if something here is wrong i'd appreciate some input.

• It's often the case that the concept of capacitance is introduced before before capacitor. From the quote in the question, I suspect that the context is the capacitance of an isolated conductor rather than, e.g., a parallel plate capacitor. For example, an isolated spherical conductor of radius $R$ has capacitance $C = 4\pi\epsilon_0R$ – Alfred Centauri Aug 24 '18 at 12:15
• That is correct, and you must be correct about the concepts as well. I'll need to rethink my answer or perhaps edit it into a comment instead, – DakkVader Aug 24 '18 at 12:16
• I see that I was writing my comment above, another answer has been posted which addresses the point I was making above. – Alfred Centauri Aug 24 '18 at 12:16
• Yeah i just saw it as well @AlfredCentauri . – DakkVader Aug 24 '18 at 12:17

Aside from the typical capacitor made of two conductors and explained in @DakkVader response, one often encounters a single conductor holding charge.

In that case, potential difference is measured between the conductor surface and infinity. Potential in infinity is usually stated as zero. So one often hears: "potential of the conductor" rather than the more accurate term: "potential difference to infinity (which happens to be zero there)".

As for your second doubt, the ideal is usually to hold large charges. Suppose you already hold $Q_0$ charge and you want to increase it. The higher the potential difference is, the more energy you need to apply to cause this increase in charge. Hence the challenge.

Also, from practical engineering point of view, the higher the potential difference is, more likely the capacitor will breakdown losing its charge. Hence a limiting factor and another challenge.

If you take the general definition of capacity $C = \frac{q}{V}$ (for simplicity let's take the reference level of the voltage at 0 V), you see the quantities C capacitance, q charge, V voltage. If you isolate the carge q you have $q = C\cdot V$ which shows you that with a material with high capacitance, even if you put low voltage, the product q (charge stored) could be relatively high. It is just another sentence that can describe the behaviour of a material with defined capacitance.

For insinct you can think at having low voltage and high charge stored as a benefit because having high voltages for storing the same charge means that you have to use more energy in your house to keep that high potential difference, bigger and more expensive batteries etc.

Physically, the voltage difference is $dV = dU/q$ with U being the potential energy but $-dU = W$ with W being the work, so more difference in tension means more work needed.

• Thanks for your answer! But wouldn't low voltage also limit the energy that can be extracted? Or do you mean that the high charge would compensate for that? – Pregunto Aug 24 '18 at 13:17
• In this system the energy is related to the voltage and certainly working with less voltage means less energy. You can also think that a certain voltage produce an electric field, and the electric field appears in the equation of the energy density w: $w = \frac{1}{2}\cdot ε0 \cdot E0^2$. – Costantino Aug 24 '18 at 18:49

My second doubt is that it would seem to be a greater achievement to maintain a high charge with a high difference in potential than with a low difference in potential.

Capacitance (in the context of an isolated conductor) can be thought of as the capacity for electric charge at a given potential $V$ (with respect to infinity).

That is, the greater the capacitance of the conductor, the greater the electric charge on the conductor for the given potential.

For example, let's say there are two different isolated conductors and each has potential $V$ (with respect to infinity) and there is electric charge $Q_1$ on the first conductor while there is electric charge $Q_2$ on the second conductor.

If, say, $Q_2$ is greater than $Q_1$ then we say the capacitance of the second conductor is greater than the capacitance of the first conductor since $Q_1 = C_1V$ and $Q_2 = C_2V$.