Question

You wrapped the string around the top of the spinner which has the radius $r = 12 mm$. You brought another charged rod (not shown in the picture) distance $d = 1 cm$ from the very tip of your test rod and hung a small weight $m = 468.3 g$ on the end of the string. The entire system is in balance, that is neither the weight nor the rod are moving. The length of the rod is $L = 25 cm$, gravitational constant $g = 9.8 m/s^2$ and vacuum permeability $\epsilon_0 = 8.85 \times 10^{-12} F/m$. Making the same assumptions about the interaction between the rods as you did in the lab (treating the tips of the rod as point charges) what is the amount of charge, $Q$, on each rod? Q = $\times 10^{-8} C$

          You wrapped the string around the top of the spinner which has the radius $r = 12 mm$. You brought another charged rod (not shown in the picture) distance $d = 1 cm$ from the very tip of your test rod and hung a small weight $m = 468.3 g$ on the end of the string. The entire system is in balance, that is neither the weight nor the rod are moving.
The length of the rod is $L = 25 cm$, gravitational constant $g = 9.8 m/s^2$ and vacuum permeability $\epsilon_0 = 8.85 \times 10^{-12} F/m$. Making the same assumptions about the interaction between the rods as you did in the lab (treating the tips of the rod as point charges) what is the amount of charge, $Q$, on each rod?
Q =  $\times 10^{-8} C$
        
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You wrapped the string around the top of the spinner which has the radius r = 12 mm. You brought another charged rod (not shown in the picture) distance d = 1 cm from the very tip of your test rod and hung a small weight m = 468.3 g on the end of the string. The entire system is in balance, that is neither the weight nor the rod are moving.
The length of the rod is L = 25 cm, gravitational constant g = 9.8 m/s^2 and vacuum permeability ϵ0 = 8.85 × 10^-12 F/m. Making the same assumptions about the interaction between the rods as you did in the lab (treating the tips of the rod as point charges) what is the amount of charge, Q, on each rod?
Q =  × 10^-8 C

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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please help me answer this and provide details!! You wrapped the string around the top of the spinner which has the radius r=12mm. You brought another charged rod (not shown in the picture) distance d=1cm from the very tip of your test rod and hung a small weight m=468.3g on the end of the string. The entire system is in balance, that is neither the weight nor the rod are moving. did in the lab (treating the tips of the rod as point charges) what is the amount of charge , Q, on each rod? Q X10-8C Submit Answer
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Transcript

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00:01 So this question deals with some issues of electrostatics.
00:08 The first one wants to know what is the direction of the net electric force on a positive test charge placed at three different locations and they label them a, b, and c.
00:21 Well, let's take a look here.
00:22 What i've done, my version of the diagram, is i've drawn two little points over here.
00:28 And let's not consider the force from the entire semicircle here or a quarter circle.
00:37 Let's just consider the force from the charge at this location right here and the charge at this location right here.
00:46 And what we'll do is we'll draw a straight line because according to kulam's law, that force is going to go in a straight line.
00:55 And the force on a test charge put right over here at position a caused by just a little bit of the charge, whatever charge happens to be at that location i put on a dot at, the direction of that force is going to be along this line, and because it's a positive test charge and positive charge up there, the force is going to go this way.
01:21 Now similarly, down here we have a exactly symmetric situation, but we've located this point down here is exactly at the same relative position on its arc as the positive one was on its art.
01:38 But this charge is negative.
01:40 And so the force caused by the little bit of charge down here on the test charge is going to be going off this way.
01:55 Now, if it is correctly, i should have drawn the magnitude of the two charges forces should be easy.
02:01 Because they're both the same distance away from the test charge.
02:05 Each of these two charges should have equal magnitudes of forces, but they're going these different directions.
02:12 Because they're equal in magnitude, the horizontal component of these two forces is the same.
02:21 And so they end up canceling each other out, if you will.
02:24 The perpendicular component, the vertical component of these forces, though, they're going in the same direction.
02:32 So they kind of add up.
02:35 And so the net force at point a is straight down.
02:41 But you can do the same thing at the other two points as well.
02:44 Let's just erase some of these forces here.
02:51 Let's go over here and let's say what is the net force at point b here? well, again, if you just consider the net force caused by this charge right here and the corresponding charge on the bottom quarter circle.
03:24 The top one is going to produce a force going this way, and like that.
03:32 And then the bottom one is going to have a force going this way.
03:40 So it goes like that.
03:42 And again, because those two forces are equal in magnitude, perhaps didn't draw that very well, but this should be equal in magnitude.
03:48 Again, the net force in those two charges is going to be.
03:51 Going to be down like this in that direction.
03:55 But that could be said for any pair of charges like that.
04:00 You could take this little bit of charge up here and figure out its net force compared to this little charge down here.
04:09 So every charge in the positive quarter circle is going to have a corresponding charge in the negative quarter circle that ends up giving you a net force in the perpendicular direction, sorry, in the vertical direction...
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