# Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.119 N...?

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.119 N when their center-to-center separation is 71.9 cm.

The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0325 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other? (Assume the negative charge has smaller magnitude.)

### 2 Answers

- NCSLv 73 months agoFavorite Answer
electrostatic force F = kQq / d²

where k = 8.99e9 N·m²/C²

first condition:

-0.119 N = 8.99e9N·m²/C² * Q * q / (0.719m)²

Q*q = -6.84e-12 C²

second condition:

0.0325 N = 8.99e9N·m²/C² * Q'² / (0.719m)²

Q'² = 1.87e-12 C²

Q' = 1.367e-6 C

and so the total charge present is Q_tot = 2.734e-6 C = Q + q

and so

q = 2.734e-6C - Q

return to the first condition

Q*(2.734e-6C - Q) = -6.84e-12 C²

which has roots at

Q = -1.58e-6 C = -1.58 µC ◄ initial negative

and Q = 4.32e-6 C = 4.32 µC ◄ initial positive

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- JOHNLv 73 months ago
Answer

EDIT: Line 3 - the value for Q1Q2 should have a - sign for the reason given by NCS in Comments.

Line 6 the first - sign should be changed to a + and changes mutatis mutandi made in the calculation. The answers will come out the same as NCS's.

Yes, thanks NCS. I missed that. Too late, though, for an edit. BA should go to you.