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Two charged spheres repel (attached to strings)

  • Thread starter FrogPad
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It's been awhile since I've had physics I, so this problem is giving me a headache.

Q) Two very small conducting spheres, each of a mass [itex] 1.0 \times 10^{-4}\,\,(kg) [/itex], are suspended at a common point by very thin nonconducting threads of a length [itex] 0.2 \,\,(m) [/itex]. A charge [itex] Q [/itex] is placed on each sphere. The electric force of repulsion separates the spheres, and an equilibrium is reached when the suspending threads make an angle of [itex] 10 \,\, (deg) [/itex]. Assuming a gravitational force of [itex] 9.80 \,\, (N/kg) [/itex] and a negligible mass for the threads, find [itex] Q [/itex].

My Work)
We first will deal with:
[tex] \vec F_{12} = \frac{\hat R_{12} k q_1 q_2}{R^2_{12}} [/tex]
[tex] \sum \vec F_i = m \vec a [/tex]
Since equilibrium is reached, [itex] \vec a = \vec 0 [/itex]. Thus,

[tex] \sum F_i = \vec T_2 + \vec F_G + \vec F_{12} = \vec 0 [/tex]

We now find the forces.

[tex] \vec F_G = -\hat y (9.8 \times 10^{-4}) [/tex]

Setting up the coordinate system we assume the orgin as at the point of interesection of the two threads. Thus, a vector that points to sphere-two is:
[tex] \vec S_2 = \hat x(0.2 \sin 5^{\circ}) - \hat y(0.2 \cos 5^{\circ}) [/tex]

The vector quantities for coloumbs law are as follows:
[tex] \vec R_{12} = \hat x (2(0.2\sin 5^{\circ})) [/tex]
[tex] R = 0.4 \sin 5^{\circ} [/tex]
[tex] \hat R = \hat x [/tex]

Thus, since the spheres have an equal charge
[tex] \vec F_{12}=\frac{\hat x k Q^2}{0.16 \sin^2 5^{\circ}} [/tex]

Now I know the tension has to exert a force that holds the sphere in place, so gravity and the electric repulsion keep it from moving away. So do I just say that [itex] \vec T_2 = -\vec S_2 [/itex]?

I'm not really sure what to do. Is what I'm doing even correct?

thanks in advance :smile:
 

Answers and Replies

  • #2
809
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I figured it out
 

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