# Electrostatic force electroscope

electroscope

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 spheres. If the wires each make a 3.0E1° angle with the vertical (see figure), what total charge Q must have been applied to the electroscope? Ignore the mass of the wires.

*here is what I did....and got stuck in the middle of doing this problem*

Well, first I found the distance between Q/2 and Q/2. Then I applied this equation http://i5.photobucket.com/albums/y178/besuu4/b836e8b8cc7c00400c5910d3eca76e37.png to find Q. The problem is I don't have F (force) value to solved it by using the equation I provided. Anyway...I am just confusing myself. Hints are welcome. Thanks.

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LeonhardEuler
Gold Member
You know that the charges are not moving. This means they must be in equillibrium. Set up a free body diagram.

Chi Meson
Homework Helper
You are told to ignore the mass of the wires. But don't ignore the mass of the charges.

Assume this occurs on Earth. (Weight a minute, is this a clue to the "missing force"?)

Can you say "trigonometry"?

I think I get your clue, Chi Meson.

I know this is oscesles (sp?) triangle so the other two angle must be equal which is 60 degrees. I then used Sin(30)=X/78=39 to find the length and mutliple by 2 to get the full length (happened to be 78cm).

Then I used F=ma to find the force. F=(.024kg)(9.8m/s^2)=.2352N.
after that, I used this equation http://i5.photobucket.com/albums/y178/besuu4/b836e8b8cc7c00400c5910d3eca76e37.png to solve for Q.

.2352N=1/4*pi*8.85e-12*(Q^2/78cm^2)
So the answer is Q=1.59e-7 C, correct?

so...right or not? Thanks.

Not quite! The forces don't have to be equal for equilibrium as they're not in opposite directions. Also, the charge on each ball isn't Q. Inquire further for more help.

Doc Al
Mentor
joanne,

Start by identifying all the forces acting on one of the spheres. (I count three forces acting.) Label the forces in a diagram, then apply the conditions for static equilibrium: The net force must be zero. (Hint: Write two equations. One for vertical components; one for horizontal components.)

Well not quite for vertical and horizontal... right? But yeah for two perpendicular directions. See if you can find out which :)
Edit: well I guess it doesn't matter, but one way you can ignore 1 of the 3 forces.