Video: Determining the Potential Difference across Charged Parallel Plates

Two parallel plates, each with sides of length 10 cm are given equal and opposite charges of magnitude 5.0 Γ— 10⁻⁹ C. The plates are 1.5 mm apart. What is the potential difference between the plates?

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Video Transcript

Two parallel plates, each with sides of length 10 centimeters are given equal and opposite charges of magnitude 5.0 times 10 to the negative ninth coulombs. The plates are 1.5 millimeters apart. What is the potential difference between the plates?

We can call the length of the sides of the plates, 10 centimeters, capital 𝐿 and the charge magnitude on the plates, 5.0 times 10 to the negative ninth coulombs, 𝑄. The distance the plates are separated, 1.5 millimeters, we can name 𝑑. We want to solve for the potential difference between the two plates. We can call this 𝑉.

Let’s start our solution by drawing a sketch. We have two parallel plates, each with both sides of length 𝐿, separated by a distance 𝑑. We’re told the plates have equal and opposite charges. The top plate, we can say, has a charge of positive 𝑄, 5.0 times 10 to the negative ninth coulombs. And the bottom plate has a charge of negative 𝑄. This setup creates an electric field in between the two plates that we can call 𝐸. The electric field is connected to the potential difference 𝑉 from one plate to the other. Specifically, for oppositely-charged parallel plates like we have here, the constant electric field that exists in between the plates is equal to the potential difference from one plate to the other divided by the distance between them. So 𝑉 is equal to 𝐸 times 𝑑. We’re given 𝑑 but don’t know the electric field strength 𝐸 between the plates.

To find it, we can recall yet one more relationship for the electric field between oppositely-charged plates. That field 𝐸 is equal to 𝜎, the charge density on the plates, divided by πœ– naught, the permittivity of free space, a constant which we’ll take to be exactly equal to 8.85 times 10 to the negative 12th farads per meter. Since electric field equals 𝜎 over πœ– naught, we can substitute that in for 𝐸 in our equation for 𝑉. So 𝑉 equals 𝜎 over πœ– naught times 𝑑.

We can now recall what 𝜎 is. It’s a charge density of some amount of charge 𝑄 spread out over an area 𝐴. In our problem statement, we’re told what 𝑄 is and we’re also told the dimensions of the plates, 𝐿 times 𝐿. So 𝜎 equals 𝑄 over 𝐿 squared, where 𝑄 is 5.0 times 10 to the negative ninth coulombs and 𝐿 is 10 centimeters. We can now replace 𝜎 with this expression for 𝜎. When we do, and we simplify the equation, we find that 𝑉 equals 𝑄 times 𝑑 over 𝐿 squared πœ– naught.

We know each one of these four values and can now plug them in to the equation. When we do, we’re careful to plug in our distances with units of meters. When we calculate 𝑉, we find that, to two significant figures, it is 85 volts. That’s the difference in electrical potential from one plate to the other.

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