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(a) What is the electric field 5.00 $\mathrm{m}$ from the center of the terminal of a Van de Graaff with a 3.00 $\mathrm{mC}$ charge, noting that the field is equivalent to that of a point charge at the center ofthe terminal? (b) At this distance, what force does the field exert on a 2.00$\mu \mathrm{C}$ charge on the van de Graaff's belt?

a)1.08$\cdot 10^{6} \frac{\mathrm{N}}{\mathrm{C}}$b)2.16 $\mathrm{N}$

Physics 102 Electricity and Magnetism

Chapter 18

Electric Charge and Electric Field

Cornell University

Simon Fraser University

Hope College

Lectures

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03:24

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(a) What magnitude point c…

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Point charges of $25.0 \m…

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(a) Find the total Coulomb…

06:58

Two $2.00-\mu \mathrm{C}$ …

03:12

An electron is projected h…

02:29

A uniform electric field o…

12:37

A uniformly charged ring o…

02:10

A Van de Graaff generator …

07:04

The electric field at the …

all right. So I feel like the best way to answer a physics question is to write down what you have and what you need, and see what formula you would use from there. So based on this question, we have an R which is the distance from the center of Terminal two point where electric field Isse found and the are IHS five meters. And when you continue under the question, you knows that a charge charges Q. And it is three Mela Coolum someone, something is Mellick alarms To use it in a big this ex occassion, you usually have to convert it to cools and to convert your basically take three and multiply it by 10 to the third. You know you have it in columns. So this is what tender then negative third and this is what you would use it. And then we are looking for the electric field, which is evil but a question market. So we're looking for and since there is Coolum, there will probably be a combs constant, which is K in the Coolers law. Constant is nine times 10 to the ninth and the units are nana meters squared over cool scored number. These right here are zeros, and that is a decimal point. All right, so this is now all the information that we have, we have the radius. We have the, um, charge electric fuel that we're looking for in the constant. So now we would end up using this formula, this formula Murray here. And we're basically to, as you would just put everything in So you would just take He equals e is the electric field. His, you know, showed as he capitalized e your put the constant, which is nine times 10 to the ninth. You'd multiply the constant by the charge in cool arms. Not Millicom school ums. So three times 10 to the negative three. Remember for your unison there. I'm just doing it to save sometimes in order to write it. And in the radius, which is five squared remembered a square, your radius. You do this, you just kept all into a calculator or you could do basic math. And so basically the top would equal 2.7 temps into the seventh of Felix writing all those zeros you can and the bottom will be 25 which would then equal 1.8 times, 10 to the six and the units for this of being Newton's over qualms. So this will be the answer to part a. Now we will move on to Part P just all right. So in part B, we will be looking at what has given and what's given. In part B is the charge que write down a queue again. And unless it is given in to my crew cool arms. So my car is basically like a funny looking, um, but you have to again converted to cools. You can wear something is in my girl to like Combs. You would do times 10 to be negative six, and now it is in cool Moms, and we also we're using columns law, and we will now calculate the electrostatic forces in terms of the electric field. That means we need the electric field, which we just calculated, and it was 1.8 times 10 to the sixth. Hayden's over. Cool. Okay, so rich where trying to find the electrostatic forces forces that's always an F question work, and there's a very simple equation. Do this. That simple equation listed a different color is for worse equals cool ums, times, electric field. So you're just multiply this and this together. So you just you two times 10 to the negative, six times 1.8 times 10 to the sixth and you would get 2.16 Newton's cause forces are always conflated in Nunes and this be your answer to part.

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