Question Video: Calculating the Electric Field and Electric Potential between Parallel Plates

Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. What is the electric field strength between them, if the potential 8.00 cm from the zero-volt plate (and 2.00 cm from the other) is 450 V? What is the voltage between the plates?

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Video Transcript

Two parallel conducting plates are separated by 10.0 centimetres, and one of them is taking to be at zero volts. What is the electric field strength between them if the potential 8.00 centimetres from the zero-volt plate and 2.00 centimetres from the other is 450 volts? What is the voltage between the plates?

We have two questions about these two parallel conducting plates and we’ll start with the first one: what is the electric field strength at this particular location between the plates? Drawing these two plates in, we’re told that they are separated by a distance of 10.0 centimetres. And we’re told to consider a specific point in between these two plates; it’s eight centimetres from one plate, which we’ll say has zero volts, and two centimetres from the other plate.

We’ll take the plate on the left to be the one at zero volts and we’ll say the difference between the potential at that plate and the potential at the right plate is 𝑉 volts. We’ll solve for that voltage in part two. But for now, our focus is on the electric field strength in between these two plates.

Knowing that our plate on the right is at higher potential than our plate on the left, we know that when an electric field is set up between the plates, it will move parallel to them and it will go from right to left, higher potential to lower potential. And moreover, this field — even though it doesn’t quite look that way from our sketch — will be uniform in between the plates; it will have the same value everywhere.

It’s that field strength, which we can call 𝐸, that we want to solve for in this part. And to help us do it, we’re given some information about this point we’ve marked out with an 𝑥. First of all, we’re told that this point is 8.00 centimetres from the left plate, the one which is taken to be at zero volts. And second, we’re told that at this 𝑥 point, this location, the potential is 450 volts.

We can say then that we’re given the potential at this point; we’ll call it 𝑉 sub 𝑝. We’re also given the distance between this point and our zero potential point, the left plate. And we then want to solve for the electric field that exists between these parallel plates. Working with these three variables, there’s a mathematical relationship that ties them altogether.

When we have a uniform electric field in between two parallel conducting plates, then that field multiplied by the distance between the plates is equal to the potential across them. Working with our particular variables, we can say that 𝑉 sub 𝑝 is equal to 𝐸 times 𝑑. Well, remember it’s the electric field 𝐸 we want to solve for.

Dividing both sides by the distance 𝑑, we now have an equation that will let us do that. All that remains is to plug in 450 volts for 𝑉 sub 𝑝 and 8.00 centimetres for 𝑑 and then we can calculate 𝐸. When we’ve made these substitutions, before we calculate this fraction, there is one change we want to make and it has to do with the units of 𝑑.

Right now, 𝑑 is in units of centimetres, but the SI standard unit for length is metres. Recalling that 100 centimetres is equal to one metre, that lets us perform the conversion between these two length units. 8.00 centimetres is 0.08 metres. And when we calculate this fraction, we see that it’s 5.63 kilovolts per metre. This is a measure of just how much the potential changes for every metre of distance between our parallel plates.

Now, let’s move on to part two, where, as we mentioned before, we want to solve for the voltage between the plates, in other words, the voltage of the right most plate relative to the left plate which is at zero volts. There are a couple of ways we could go about doing this.

One way is to take our 𝑉 sub 𝑝 value, well 450 volts, which remember applied at a point 8.00 centimetres from our zero volt standard, and multiply that by a ratio of distances. On top, we have the total distance between the plates and on bottom we have the distance from the zero-voltage plate to this point, where 𝑉 sub 𝑝 is measured. This method is perfectly good and will give us the result we’re looking for.

A second way we can solve for this voltage is to reapply our equation 𝑉 is equal to 𝐸 times 𝑑. Recall that in part one, we solved for that electric field strength in between the plates. So now, knowing that electric field, we can multiply it by the distance between the plates to solve for 𝑉.

Either one of these two methods will work and both will lead us to the voltage between the plates. If we do choose the second method, one thing we want to keep in mind is that we want to rewrite our distance between the plates in units of metres from its given units of centimetres.

Both of these methods lead to the same result and that is 563 volts. This is the voltage across the total distance from one plate to the other.