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Finding charges in a pendulum in equilibrium 
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#1
Nov1212, 03:37 AM

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1. The problem statement, all variables and given/known data
Two small metallic spheres, each of mass 0.20 g, are suspended as pendulums by light strings from a common point. The spheres are given the same electric charge, and it is found that the two come to equilibrium when each string is at an angle of 5.0° with the vertical. If each string is 30.0 cm long, what is the magnitude of the charge on each sphere? 2. Relevant equations [tex]F = k \ \frac{q_1 q_2}{r^2} \ \mbox{ or } \ \frac{1}{4 \pi \epsilon_0} \ \frac{q_1 q_2}{r^2}[/tex] [tex]\begin{equation*} \begin{split} \frac{1}{4 \pi \epsilon_0} = k & = & & 8.988 \times 10^9 \ N \cdot m^2 / C^2 \\ \\ \end{split} \end{equation*}[/tex] 3. The attempt at a solution Finding the horizontal distance between the two spheres to give r: 1. An isosceles triangle is formed with the two strings, so the base angles are the same. 2. Therefore, the base angles are ((1802×5)/2) = 85°. 3. Using the sine rule, r = 0.3×sin(2×5)/sin(85). Finding the force of gravity that is acting horizontally on the two spheres to bring them to the vertical, which gives us F: 1. Treating the 30 cm strings as radii of a circle with centre from the common point, and the spheres as points on the circle, a straight line tangent to a sphere intersects the vertical. 2. The angle from one of the spheres to the other sphere and then to where the tangent intersects the vertical is thus 5° (radii are perpendicular to a tangent, adjacent angles). 3. Therefore, right angle trigonometry tells us that the force of gravity acting horizontally on a sphere is mg/tan(5) = 0.196/tan(5). q_1 = q_2 (given) So:[tex]F = k \ \frac{q^2}{r^2}[/tex] Therefore,[tex]q = \sqrt{\frac{Fr^2}{k}}[/tex] And when I sub my values in I get q = 8.3 x 10^7 C, which is apparently incorrect, the correct answer being 7.2 x 10^9 C. Help? 


#2
Nov1212, 04:02 AM

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Best to keep symbols as long as possible though. Helps with troubleshooting. I think you have overcomplicated this  what is wrong with drawing a freebody diagram for each mass and summing the forces to zero? (Put yaxis along the tension.) 


#3
Nov1212, 04:19 AM

P: 265

It would be easier to find r by using a single rightangle triangle.
Then r = 2Lsinθ, (where L is the string length). In all problems like this, it's always best to start with a freebody diagram, ( you only need to do this for 1 sphere because of symmetry ). I would put the yaxis along the direction of the weight. This gives 2 equations allowing you to eliminate the string tension and leaving the charge as the only unknown. As Simon Bridge says, you are making this more complicated than you have to. 


#4
Nov1212, 04:35 AM

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Finding charges in a pendulum in equilibrium



#5
Nov1212, 04:52 AM

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[tex]\begin{equation*} \begin{split} k \ \frac{q^2}{r^2}cos5°  mgsin5° & = & & 0 \\ ∴ kq^2cos5° & = & & r^2mgsin5° \\ ∴ q & = & & \sqrt{\frac{r^2mgsin5°}{kcos5°}} \\ & = & & \sqrt{\frac{\frac{0.3sin10°}{sin85°}^2×0.196×sin5°}{8.988\times 10^9×cos5°}} \\ & = & & 7.2\times 10^{8} \ C \\ \end{split} \end{equation*}[/tex] But... the answer is supposedly 7.2 x 10^9 C. I can't seem to find the missing 10^1. Am I overlooking it, or is the answer provided simply incorrect? Thanks so much for your help! 


#6
Nov1212, 05:13 AM

P: 265

Remember that the mass given is 0.2 grams, so mg=0.00196, not 0.196



#7
Nov1212, 05:29 AM

P: 4

*headdesk*
Thanks. 


#8
Nov1212, 07:27 PM

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Nice to see people using LaTeX though  kudos.
Just a tip: to get the trig to format correctly you should put a "\" in front of it ... because ##kq^2\cos(5^\circ)## reads better than ##kq^2cos(5^\circ)## ... technically units should be in normal, rather than italic, text ... which you can do, in mathmode, with the \text{} so 15m/s will be ##15\text{ms}^{1}## rather than ##15ms^{1}## ... but it is still much nicer than trying to format in plain text innit? On your math  the other thing to try was to realize that the sum of the force vectors being zero means that the tension (T), gravity (mg), and coulomb force (F) vectors, form a rightangle triangle with the tension on the hypotenuse, and the halfapex angle (##\alpha =5^\circ##) is between mg and T. In which case, ##F=mg\tan(\alpha)## right off ... from there you can see where you originally made your mistake. The power in the fbd method is that it does not rely on you drawing the right triangle right at the start. 


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