# Question Video: Calculating the Resistivity of a Substance

Two square conducting plates of side length 50.0 cm are placed parallel to each other 9.0 mm apart. What magnitude electric field is generated between the plates after 10¹³ electrons are transferred from one plate to the other?

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### Video Transcript

Two square conducting plates of side length 50.0 centimeters are placed parallel to each other 9.0 millimeters apart. What magnitude electric field is generated between the plates after 10 to the 13th electrons are transferred from one plate to the other?

We have then these two square parallel conducting plates. And from one plate to the other, there’s a mass exodus of electrons, 10 to the 13th of them. After that transfer takes place, there’s a net charge on each one of the plates which now have a net surface charge. And these surface charges set up an electric field in between the plates we can call capital 𝐸. We want to solve for the magnitude of that field based on this number of electrons having migrated.

The magnitude of the electric field formed between two parallel plates is equal to the surface charge density on either plate divided by 𝜖 naught, the permittivity of free space. This constant has a value of 8.85 times 10 to the negative 12th farads per meter. Based on this relationship for electric field, we can say that the magnitude of the field we want to solve for is equal to the positive charge density divided by 𝜖 naught.

That positive charge density is equal to a total charge divided by the area over which the charge is spread. That area is the area of each plate, its length on either side squared. And the total charge is equal to the number of electrons, 10 to the 13th, multiplied by the magnitude of the charge on an individual electron.

When we look up that value for the charge on an individual electron and also recognize that 𝐿, the length of our sides, is 50.0 centimeters, we find that the magnitude of the field we’re looking for is equal to the number of electrons that transferred times the charge of an individual electron, the magnitude of that charge, all divided by the area of each plate converted to units of meters, 0.50 meters quantity squared. All that is divided by 𝜖 naught. And when we calculate up this fraction, we find that, to two significant figures, it equals 7.2 times 10 to the fifth newtons per coulomb. That’s the magnitude of the electric field created between these plates.