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TrivialPants
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Homework Statement
A large electroscope is made with "leaves" that are 78-cm-long wires with tiny 24-g spheres at the ends. When charged, nearly all the charge resides on the ends of the spheres. (See diagram attached)
If the wires each make a 26° angle with the vertical, what total charge Q must have been applied to the electroscope? Ignore the mass of the wires?
Diagram of the Electroscope:
Homework Equations
What does it mean to ignore the mass of 24g? Does that mean that I can find the charge by finding the distance between the two points?
The Attempt at a Solution
I split the triangle into two right triangles. Then I used the trig properties to deduce that:
sin26° = opp/hyp = .4387, opp/78cm = .4387 = opp =.4387*(78cm) =34cm
cos26° = adj/hyp = .8988, adj/78cm = .8988 = adj =.8988*(78cm) =70cm
Now using this data, I would input the distance between the repelling positive charges into the Coulombs Law:
F = (k|Q1||Q2|)/(r12)^2 Where k = 8.988x10^9 N*m^2 / C^2 and r = 68 which is derived above
(34*2) = 68 cm or 68*10-2m
How do I find the value of Force? I will be needing it to complete the problem this way. Thank you!