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$\bullet$ A particle has a charge of $-3.00 \mathrm{nC}$ . (a) Find the magnitudeand direction of the electric field due to this particle at a point0.250 m directly above it. (b) At what distance from the parti-cle does its electric field have a magnitude of 12.0 $\mathrm{N} / \mathrm{C}$ ?

a) 431.4 $\mathrm{N} / \mathrm{C}$b) 1.499 $\mathrm{m}$

Physics 102 Electricity and Magnetism

Chapter 17

Electric Charge and Electric Field

Gauss's Law

Electric Potential

Cornell University

University of Sheffield

University of Winnipeg

McMaster University

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and this problem, we're going to use this formula for the electric field now e points towards so points towards negative and away away from positive. Another way you can think about this is e points in the direction that positive charge would move and since positive charges repel one another e points towards the minus in away from the positive and since the source that were given as a minus are, you will be pointing towards the source. And since the sources below said here and we're looking at this point here, our electric field plain downward towards it. So we have the direction sort out. But let's actually calculate the magnitude and we're just going to apply this here. And so are you in for K 8.99 times 10 to the ninth for Q. I get 3.0 times 10 to the minus sign and for our 8.25 and then I get squared doing this out. I get 432 and the units are nunes for cool. Um and so this is the magnitude and the directions downward. Because of this reasoning here in part B, we're going to take this formula and we're going to solve for R. So when we sell for our we get square root. Okay? Times ask the guy for you over the electric field and we happen to know everything here and so we can solve this. I'm just gonna plug in the values here. It's okay. Is 8.99 times 10 to the nine. Absolutely, thank you. Is three times 10 to the 99 actually, and the electric field, we're told is 12. And so this gives in our value of 1.5 meters, and that completes the role.

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