# Video: Calculating the Electric Field between Parallel Plates

Two parallel conducting plates are separated by 1.00 cm and have a potential difference between them of 2.36 × 10⁴ V. What is the magnitude of the electric field between the plates?

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### Video Transcript

Two parallel conducting plates are separated by 1.00 centimetre and have a potential difference between them of 2.36 times 10 to the fourth volts. What is the magnitude of the electric field between the plates?

If we sketch in these parallel conducting plates, we’re told they’re separated by a given distance. We can call that 𝑑. And that if we measure the potential of one plate relative to the other plate, there is a difference between them that we can call Δ𝑉. And both this distance and the change in potential are given to us.

From a physical perspective, if we have two conducting parallel plates separated by some distance with a potential difference between them, that means that an electric field will be created in between these plates. Now, let’s say it’s the plate on the left that has the higher potential — that is the higher potential relative to the plate on the right. That means that the electric field formed between these plates will go from left to right and will be uniform in between them.

It’s the magnitude of that field that we’ve called capital 𝐸 that we want to solve for in this exercise. We can see then that we’re working with three different variables: we have the change in potential Δ𝑉, we have the distance between the plates, and we have the electric field magnitude between them that we wanna solve for.

It turns out there’s an equation that ties together all three of these variables for a parallel conducting plate setup, like we have here. The electric potential between parallel plates is equal to the strength of the field between them multiplied by their separation distance.

In terms of our variables, we can write that Δ𝑉 is equal to 𝐸 times 𝑑. But it’s not Δ𝑉 we want to solve for; it’s the electric field 𝐸. So let’s rearrange by dividing both sides by the distance 𝑑. At this point, we have a symbolic relationship that gives us what we wanted to solve for, the electric field 𝐸. Now, all we need to do is plug in and calculate to get the numerical answer.

But before we do, notice one thing about our distance 𝑑: it’s given in units of centimetres, but we would prefer first to convert that to units of metres. That way we can get an electric field in SI standard units of volts per metre.

So let’s recall the conversion from centimetres to metres. There are 100 centimetres in one metre. And that tells us that 1.00 centimetres is equivalent to 0.01 metres. We’re now ready to insert our values of 𝑑 and Δ𝑉. See that we’ve done that here and we’re ready to calculate this fraction, finding a result of 2.36 times 10 to the sixth volts per metre. That’s the magnitude of the electric field between the two plates.