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What is the force on the charge located at $x=8.00 \mathrm{cm}$ in Figure 18.51$($ a) given that $q=1.00 \mu C ?$

12.8 $\mathrm{N}$

Physics 102 Electricity and Magnetism

Chapter 18

Electric Charge and Electric Field

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

McMaster University

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Let's first run the distribution of charge. So there's a charge negative to Q at eight, and that's cute at 10 and rescue at at three. Now we're supposed to find the fools on this judge. Negative to Cube. That is the full charge. Located at X equal to eight centimeters. Now, each charge value of cube is one microt cooler. As you know, Force is a vector quantity on DDE. This judge is going to pull this judge in this direction, and this judge here is going to apply the force in left inside. So there are both in opposite direction. Let's do Let's number this charges. Let's just let this charge be one. Let this judge be to discharge with three year. So let's first find force on one D recharge too. Since this force is directed to the positive X direction, it is positive. So the charge is given by one divided by full pie force. Sorry, forces given my one divided by four by absolutely not Q one Que tu divided by distance, squared distant between one and two. Now see that the distance here is two centimeters Okay when divided by four perhaps alone not is nine times 10 power nine. No charges are One charge is one micro cologne. So since the forces attractive, let's give it a negative sign first so that negative sign will be canceled when we substitute the values. So Q one is negative to micro cooler and cue to Is one my group. Hello. So just in Parliament and the 26 divided by do you want to square, that is point zero to square. Let's do this calculation. So nine times two is 18. It's divided by 0.2 square. So just taking the numbers into account, we get the value. That's fool. Chief, 5000 times 10 Power Negative. Three. Since the charges are very close to force. Seems very large. But let let's write the force as Force One to US 45 Newton. So this becomes 45 Newton because it's 45,000 times 10 power. Negative. Three. I got negative. Three from the factor. 10 power negative. Six year in power. Negative. Six year and 10. 5000 to 9 here. So this is 10 power Negative. 12 on the cistern. Power nine. So that makes it 10 Power, not 10 power positive. Negative three on dhe. This nine times two is 18 divided by 180.4 which is the denominator gets you 45,000. Now let's find out the force on charge one. Do you to charge three? See that how the forces here. Positive? Because it was attractive force. So we had a negative sign before on DDE. One of the charges. Negative. So it becomes positive. No forced 13 on the other hand, is given by one over four pi Upsilon. Not as before. Q. One Q. Three. Again, this force is also attractive and the distance squared. This time it's a distrust. The one thing square. No, we have to make sure that the answer is negative to make sure that we know the force here is in the direction off. Negative X. So if you if you substitute negative sign here because the force is attractive and one of the charges being negative, the value here will be positive. But in the end, we will deal with negative sign to denote that the force is an ex negative. X direction So again. Nine times 10 Power nine. It's a one on one divided by four pi Upsilon. Not you Want you to. The product gives you two times 10 Power Negative 12 divided by the distance, which is here, five centimeters. So that's 0.0 by square. So this gives you the value. 72 Newton's. No, we have to give this negative side. So Force 13 Since it's a vector, it's negative. 72 Newton Now The net force on charge, located at eight centimeters, is given by the Force, but 45 new tune minus 72 new tents. Oh, so this should be 7.2 mutants, not 72. Since the charges that large distance it will be, it will be small force in magnitude. So the first force was 45. So it's right it as 45 minus the second force is 7.2. This is equal to 37.8 Newton. So the net force is acting in the positive X direction

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