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Point charges of 5.00$\mu \mathrm{C}$ and $-3.00 \mu \mathrm{C}$ are placed 0.250 $\mathrm{m}$ apart. (a) Where can a third charge be placed so that the net force on it is zero? (b) What if both charges are positive?

a)$\left.0.859 \mathrm{m}, \text { closer to }-3 \mu \mathrm{C} \text { (on a line connecting } q_{1} \text { and } q_{2}\right)$b)between charges, 0.109 $\mathrm{m}$ from $-3 \mu \mathrm{C}$

Physics 102 Electricity and Magnetism

Chapter 18

Electric Charge and Electric Field

University of Michigan - Ann Arbor

Hope College

McMaster University

Lectures

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(III) $\mathrm{A}+4.75 \mu…

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Point charges of $25.0 \m…

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Three point charges are pl…

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Two $2.00-\mu \mathrm{C}$ …

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(a) Two point charges tota…

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Two point charges are plac…

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Identical charges $q = +$5…

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05:23

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A charge of $-$3.00 nC is …

19:15

A charge $q_{1}=+5.00 \mat…

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$\bullet$$\bullet$ Three p…

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A point charge $q_{1}=+5.0…

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Two tiny objects with equa…

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04:35

$\bullet$ $\bullet$ Three …

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Three negative point charg…

09:38

Two point charges $q_{\mat…

To put part a were given a 5 miccolum charge and were given a negative 3 micocoulcharge, so we know by intuition that our third charge is going to have to be placed somewhere here. The distance between them is 0.25 meters and let this distance be equal to x, because, no matter what magnitude or sine of charge we place in between that's not going to work right. So we place it further away from the 1 with the charge which has more magnitude. Hence this side and not this side and now just solving it. The magnitude of the forces has to be equal because that's how they will cancel out right and force is equal to k. Q 1 q 2 over r square at r is the distance between them, and so the force between the 5 microcolum and our unknown charge will be k, cube plus x, squared this has to be equal to k c, the force 3 micro colum charge, and since We just need the magnitude we don't actually take the negative sign of the 3 microcurie charge at and solving this we get 5 x. Squared is 3 just take the square root on both sides. He implies axis. This is the answer to the first part, and now part tells us that both the charges are positive in magnitude, so we have 5 nitro column charge here and 3 micro colum charge here again. The distance between them is .25 meters, and our now third charge can be placed somewhere in between them. Not, as is the position so just take it to be x as if a 4 charge was placed here right and now. Similarly, we need the magnitude of the force is due to each of the 5 and the 3 micropholm charges to be equal. So k, q, unknown 5 microcoat over x square has to be equal to k, cube 3 micocoulem over 0.25 minus x, squared as that distance between r q and r 3 micropholim charge solving this equation. We small x is equal to 0.12 m.

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