### Video Transcript

How far apart are two conducting plates that have an electric field strength of 7.25 times 10 to the third volts per meter between them if the potential difference across them is 15.0 kilovolts?

If we draw in these two conducting plates, we’re told that there’s an electric field between them. And we can sketch the field as moving from left to right between the plates. We can call that field capital 𝐸. And we know the strength of that field. It’s given to us. We also know that between these two conducting plates there is a potential difference. We’ll call it capital 𝑉. And just like the electric field 𝐸, the potential difference 𝑉 is a known quantity.

Knowing all this, we want to solve for the distance between these two conducting plates. There is a mathematical relationship between the distance in the two quantities we have here: electric field and potential difference. Let’s look at the units of these two terms to see if we can discern that relationship. First, as we look at electric field, we see that that’s given in units of volts per meter. And then looking onto potential difference, we see that that’s given essentially in units of volts.

So our one term has units of volts and our other term has units of volts per meter. Notice what happens if we divide these terms one by the other. If we do the division this way, then the units of volts cancel out and the units of meters moves up into the numerator. This suggests to us that we can take the potential difference 𝑉 and divide it by the electric field 𝐸 and then that will result in some distance which we can call 𝑑.

We’re making this argument purely on the basis of our units rearrangement, But in fact, what we’ve landed on is a correct mathematical relationship. It’s actually true, in general, that the potential difference across two parallel conducting plates is equal to the constant electric field between them multiplied by their separation distance. We will then use this relationship to solve for the separation between our two conducting plates.

If we plugin for our two values, first for the potential 𝑉 which we write as 15.0 times 10 to the third volts and then for the electric field 𝐸, we see that some cancellation goes on, namely of the factor 10 to the third. in both the numerator and denominator. Likewise, we see as before that the units of volts cancel out and meters moves up to the numerator. Our fraction then simplifies to 15.0 divided by 7.25 meters. To three significant figures, that’s 2.07 meters. That’s the distance that must be separating these two conducting plates.