# Electric Charge Applied on Two Wires of Electroscope?

1. Sep 4, 2015

### TrivialPants

1. The problem statement, all variables and given/known data
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:

2. Relevant 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?

3. 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

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!

2. Sep 4, 2015

### Staff: Mentor

You are to ignore the mass of the wire, not the spheres. You'll need their mass to solve the problem.

Analyze the forces acting on each sphere. Hint: Three forces act.

3. Sep 4, 2015

### TrivialPants

Is the method I am using thus far correct to find the value of the Q? Since there is no mention of an electric field in this problem I assume I will be using the coulombs law equation I stated above: $$k\bullet\mid Q_1\mid\bullet\mid Q_2\mid\\r_{1,2}^2$$

4. Sep 5, 2015

### Staff: Mentor

Sure, you need to use Coulomb's law to calculate the total charge. (Note that Q_1 = Q_2 = Q/2.)

But you first need to do the force analysis to find the electrostatic force.