(adsbygoogle = window.adsbygoogle || []).push({}); (solved) Finding charges in a pendulum in equilibrium

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 ((180-2×5)/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?

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# Homework Help: Finding charges in a pendulum in equilibrium

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