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Zayin
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(solved) Finding charges in a pendulum in equilibrium
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?
[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]
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?
Homework Statement
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?
Homework 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]
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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