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October 23, 2014

October 23, 2014

Posted by **Alex** on Saturday, May 3, 2008 at 10:37pm.

Note: when one object with charge Q is brought in contact with a neutral object 1/2 the charge is transferred to the neutral object.

I don't understand how to do this.

Ans: 3.9x10^-6C

- Physics -
**drwls**, Sunday, May 4, 2008 at 5:38amAfter the spheres touch, each one acquires a charge of -q/2, and the Coulomb repulsion force pushes them away from each other. If each one hangs inclined A = 12 degrees, and T is the tension in the wire,

T cos A = M g

T sin A = k (Q/2)^2/(2L sinA)^2

k is the Coulomb constant, 8.99 x 10^9 N•m^2/C^2

2L sin A is the separation of the spheres

T can be eliminated by dividind one equation by the other

tan A = k (Q/2)^2/(2L sinA)^2/(Mg)

M g tan A = k Q^2/(16 L sin A)^2

This should let you solve for Q

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