Two large, parallel conducting plates carrying op­posite charges of equal magnitude are separated by 2.20 cm. (b) What is the potential difference between the two plates?

Video transcript

Welcome back, everyone. We are making observations about two large sheets of metal. You're told that the separation between them is 3.2 centimeters or 0.32 m with a, the surface charge density of 26. micro columns per meter squared. And we are tasked with finding what is the potential difference between these two sheets. Let's look at our answer choices here. We have a 9.47 times 10 to the second volts B 9.47 times 10 to the first volts C 9.47 times 10 to the third volts per meter or D 9.47 times 10 to the fourth volts per meter. All right. Well, we know that the magnitude of an electric field is equal to our potential difference divided by our distance. We also know that it is equal to a surface charge density divided by epsilon knot. So we can cut out the middle formula here and set these two formulas equal to one another. I want to isolate this V term here. And so I'm going to multiply both sides by D and you'll see that the D term cancels out on the left hand side. And what we are left with is that our potential difference is equal to Sigma D all over epsilon knot. Great. Well, we know all these values. So let's go ahead and plug them in. We have that. Our potential difference is equal to 26. times 10 to the negative six. And we're doing that because we're given micro columns, but we need OMs times our distance between the two plates of 20.32 all divided by epsilon knot, which is 8.85 times 10 to the negative 12th. When we plug this into our calculator, we get a potential difference of 9.47 times 10 to the fourth volts per meter which corresponds to our final answer choice of D. Thank you all so much for watching. I hope this video helped. We will see you all in the next one.