# Video: Calculating the Maximum Operating Voltage for a Capacitor

Two parallel conducting plates are separated by a 0.384 cm thick gap of air. If the greatest electric field magnitude in which the air can act as an insulator is 3.00 × 10⁶ V/m, what is the maximum potential difference that can be applied across the plates that allows them to function as a capacitor?

04:11

### Video Transcript

Two parallel conducting plates are separated by a 0.384-centimetre thick gap of air. If the greatest electric field magnitude in which the air can act as an insulator is 3.00 times 10 to the sixth volts per metre, what is the maximum potential difference that can be applied across the plates that allows them to function as a capacitor?

If we sketch in these parallel conducting plates, we can get an idea for what’s taking place here. The idea is that there is air between these two plates which are separated from one another by a distance we’ll call 𝑑. The question asks about the maximum potential difference that can be applied across the plates that still allows them to act like a capacitor.

And how does a capacitor act? Well, we recall that a capacitor is an element typically of a circuit, where charge of opposite types builds up on opposite sides of the capacitor. So if we had conventional current flowing in from the left, that would mean positive charges would build up on our left capacitor plate and negative charges would build up on the right capacitor plate.

Now, normally, this would be all well and good. But what if the charge continues to pile up more and more and more as this current continues to flow? At some point, with enough charge built up on these capacitor plates, the attraction between the positive charges and the negative charges is so strong that it overcomes the fact that these two plates aren’t even connected by a conductor.

At that point, the charges will literally jump this gap even though there’s an insulator air in between. But up to that point, the air does act as an effective insulator, keeping the positive charges away from the negative charges and vice versa. So long as these opposite charges stay on their opposite sides of the capacitor, they’ll set up between them an electric field which is constant between these plates.

We’ll label the electric field between these two plates capital 𝐸. And we know that this field strength will increase as the amount of charge on the capacitor plates grows as well. We’ve said though that at some point after enough charge builds up, it will jump this air gap and the air will no longer be effective as an insulator.

Just before that point — that is, just as the capacitor has the maximum amount of charge for which the air can serve as an insulator — we’re told the electric field has a certain maximum magnitude and we’ll call that 𝐸 sub max; that’s the greatest electric field we’ll ever find between these capacitor plates. Anything greater than that and the plates would stop working as a capacitor.

Knowing this, we want to solve for the maximum potential difference that can exist across these two plates. To figure out what this is, let’s start by recalling a relationship that connects electric field with potential difference and distance between two parallel conducting plates.

In this situation, we can say the potential difference across our plates is equal to the electric field between them multiplied by the distance separating the plates. That tells us then that 𝑉 sub max — this maximum potential difference we want to solve for — is equal to the product of 𝐸 sub max and the distance between the plates.

We’re well set up to do this because we’ve been given 𝐸 sub max as well as 𝑑, the distance separating the plates; it’s 0.384 centimetres. The one thing we want to do before plugging these numbers in and calculating 𝑉 sub max is to convert the units of 𝑑 from centimetres to metres. That way, its units will agree with those of 𝐸 sub max and the units for the whole expression will be consistent.

We know that to convert centimetres to metres involves moving the decimal point two spaces to the left, which tells us that 𝑑 is equal in metres to 0.00384. We’re now ready to plug these values in to our expression for 𝑉 sub max. Before we write out our result, notice that when we multiply these numbers, the units of metres will cancel out and we’re left just with the units of volts, which is good because that’s the units of potential difference.

To three significant figures, our answer comes out to 11.5 thousand volts or 11.5 kilovolts. That’s the maximum potential difference that can exist across these capacitor plates with the plates still functioning as a capacitor.