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The plates of a parallel-plate capacitor are 2.50 $\mathrm{mm}$ apart, and each carries a charge of magnitude 80.0 $\mathrm{nC}$ . The plates are in vacuum. The electric field between the plates has a magnitude of $4.00 \times 10^{3} \mathrm{Vm}$ (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the capacitance?

(a) $V_{a b}=10.0 \mathrm{kV}$(b) $A=22.6 \mathrm{cm}^{2}$(c) $C=8.00 \mathrm{pF}$

Physics 101 Mechanics

Physics 102 Electricity and Magnetism

Chapter 18

Electric Potential and Capacitanc

Kinetic Energy

Potential Energy

Energy Conservation

Electric Charge and Electric Field

Gauss's Law

Electric Potential

Capacitance and Dielectrics

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

University of Winnipeg

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is this question. We have another parallel plate capacitor hands a capacitor has a distance. Key equals 2.5 millimeter apart is in each of some carrier charge off 81 0 Nada Coolum Since a selective idea point Sarah inadequate. So the place you're in vacuum and we know that electric field between the two plates has a magnitude of equals four point 00 time's tend his assert, two votes per meter and these are the other numbers mean? Oh, who wanted now a. What is a potential difference between the plates? So, in order to find the potential difference, we won't. You know, potential difference can be created by it, actually feel across the distance. So we had V across the surface equals a stress off the electric field between the two points times the distance between them. So we have e equals 100 times 10 to the as six six votes per meter and times 2.5 times 10 to the negative three meter. And this gives us a total one point. Though there were times 10 to the four votes. Okay, so there is that and part being asked us what is the area of each plate. So how do we do that? In order to do that, we need to utilize the equation. So it is a book right before equation into 0.16. So it is about Look, it tells us in Chapter 17 that is, is a previous chapter. Willard circuses slow. We know that's the electric field equals the charge density, divided by absolute. Not in such harsh density is a total charge. Demanded my area. It's not. So if we rewrite this equation, we can have this relationship off the area equals Q. Over the terms, I'm still not so we already know the charge, and we know the electric field. This's just a constant said I elected constant. So from that we can calculate the surface area of this parallel plate. So a cruise really point to what extent is elective time cooler and divided by a full point? Sales zero times 10 to the six votes per meter Times 8.8 Fact for exist pre Find it, sir, after the heads of front cover of this book put on the square over 10 times meter scraped. This gives us 2.26 times 10 to the negative three meters square, which is 22.6 centimeter square. So the last question asks us, What is a capacitance off this capacitor? And so capacitance could be fine. Bye capacities, because the deposition is charged divided by voltage difference. And we already calculated partition office here. So over to you to do is to have charge is 80 Donald crude, um, divided by it's a voltage. So this is eight times 10 cheesy Detective 12 Sound, which is eight parent.

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