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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 1.0 \times 10^{-4}\,\,(kg), are suspended at a common point by very thin nonconducting threads of a length 0.2 \,\,(m). A charge Q 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 10 \,\, (deg). Assuming a gravitational force of 9.80 \,\, (N/kg) and a negligible mass for the threads, find Q.
My Work)
We first will deal with:
\vec F_{12} = \frac{\hat R_{12} k q_1 q_2}{R^2_{12}}
\sum \vec F_i = m \vec a
Since equilibrium is reached, \vec a = \vec 0. Thus,
\sum F_i = \vec T_2 + \vec F_G + \vec F_{12} = \vec 0
We now find the forces.
\vec F_G = -\hat y (9.8 \times 10^{-4})
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:
\vec S_2 = \hat x(0.2 \sin 5^{\circ}) - \hat y(0.2 \cos 5^{\circ})
The vector quantities for coloumbs law are as follows:
\vec R_{12} = \hat x (2(0.2\sin 5^{\circ}))
R = 0.4 \sin 5^{\circ}
\hat R = \hat x
Thus, since the spheres have an equal charge
\vec F_{12}=\frac{\hat x k Q^2}{0.16 \sin^2 5^{\circ}}
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 \vec T_2 = -\vec S_2?
I'm not really sure what to do. Is what I'm doing even correct?
thanks in advance
Q) Two very small conducting spheres, each of a mass 1.0 \times 10^{-4}\,\,(kg), are suspended at a common point by very thin nonconducting threads of a length 0.2 \,\,(m). A charge Q 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 10 \,\, (deg). Assuming a gravitational force of 9.80 \,\, (N/kg) and a negligible mass for the threads, find Q.
My Work)
We first will deal with:
\vec F_{12} = \frac{\hat R_{12} k q_1 q_2}{R^2_{12}}
\sum \vec F_i = m \vec a
Since equilibrium is reached, \vec a = \vec 0. Thus,
\sum F_i = \vec T_2 + \vec F_G + \vec F_{12} = \vec 0
We now find the forces.
\vec F_G = -\hat y (9.8 \times 10^{-4})
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:
\vec S_2 = \hat x(0.2 \sin 5^{\circ}) - \hat y(0.2 \cos 5^{\circ})
The vector quantities for coloumbs law are as follows:
\vec R_{12} = \hat x (2(0.2\sin 5^{\circ}))
R = 0.4 \sin 5^{\circ}
\hat R = \hat x
Thus, since the spheres have an equal charge
\vec F_{12}=\frac{\hat x k Q^2}{0.16 \sin^2 5^{\circ}}
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 \vec T_2 = -\vec S_2?
I'm not really sure what to do. Is what I'm doing even correct?
thanks in advance
