# Question Video: Calculating the Electric Field between Parallel Plates

The wall of a cell is 8.81 nm thick and the potential difference across it is 74.0 mV. Modeling the cell wall as a pair of parallel plates, what is the magnitude of the electric field within the wall?

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### Video Transcript

The wall of a cell is 8.81 nanometers thick and the potential difference across it is 74.0 millivolts. Modeling the cell wall as a pair of parallel plates, what is the magnitude of the electric field within the wall?

We’re working then with a cell and we don’t know whether it’s a plant or an animal cell. But we know it’s a very very small object. Looking at this object, if we were to zoom in on the cell wall, we would see up close the thickness of this wall which we’re given as 8.81 nanometers. That’s a very small thickness, about 100 times smaller than the wavelength of visible light. Not only does the wall have this thickness, but from one side to the other, there’s a potential difference of 74.0 millivolts.

The idea here is that we model this cell wall as a pair of parallel plates. So one plate on one side and one plate on the other side. And we see they’re separated by a given distance. And there’s a potential difference between the plates. What we want to solve for is the magnitude of the electric field that would naturally arise between these plates, that is within the cell wall, given the fact that there is a potential difference across them and they’re separated by some distance.

The three variables we’re working with then are 𝐸, the electric field strength that we want to solve for, Δ𝑣, which is the potential difference across the cell wall or across our parallel plates, and 𝑑, which is the thickness of the cell wall. Our question is: given Δ𝑣 and given 𝑑, can we solve for the electric field strength 𝐸? Indeed, there’s a mathematical relationship that connects these three terms.

Since we’re modeling the cell wall as a pair of parallel plates, that means we can say that the electric field in between the plates is a constant. And that if we multiply it by the distance separating one plate from another, that’s equal to the potential difference across the plates. If we apply this relationship to our particular case and we rearrange the equation a bit, we see that the electric field strength is equal to Δ𝑣 divided by 𝑑, where we’ve seen that Δ𝑣 is 74.0 millivolts and 𝑑 is 8.81 nanometers.

Before we calculate this fraction, we wanna change the units of both of these values so that they accord with the SI standard units for voltage and length. First, the value on top, 74.0 millivolts, can be rewritten as 74.0 times 10 to the negative third volts. And then, when it comes to nanometers, our unit for distance, we know that one nanometer is equal to 10 to the negative ninth meters. So we rewrite our denominator as 8.81 times 10 to the negative ninth meters.

Now, expressed in units of volts and meters, we’re ready to calculate this fraction and solve for 𝐸. We find that, to three significant figures, it’s 8.40 times 10 to the sixth volts per meter. That’s the electric field strength between these parallel plates that is across the wall of this cell.