Two identical conducting spheres, fixed in place, attract each other with an ele

Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.101 N when their center-to-center separation is 43.9 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 0.0413 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.) both units are C

Explanation / Answer

F = kQq/r²

0.101 = (9 * 10^9)Qq/0.439²

Qq = 2.16 * 10^(-12)

But since we are told that the charges attract one another, we know that q1 and q2 have
opposite signs and so their product must be negative.

Qq = -2.16 * 10^(-12)
Q = -2.16 * 10^(-12)/q. . . . . . . . . . . . . . . (1)

Then the two spheres are joined by a wire, If the new charge on each sphere is Q',

Q' + Q' = Q + q = 2Q'

F = kQ'q/r²
0.0413 = (9 * 10^9)(Q')²/0.439²
Q' = 9.4*10^(-7) N
Q + q = 2Q'
Q + q = 2 *9.4*10^(-7). . . . . . . . . . . . . . . (2)

substitute equation 1 to 2,

-2.16 * 10^(-12)/q + q = 2 *9.4* 10^(-7)

q2 - 1.88*10^-6*q - 2.16 * 10^(-12) =0

q = -8.05e-7 Coulomb or q = 2.69e-6 Coulomb

finally, Q = -2.16 * 10^(-12)/( -8.05e-7 ) =2.68*10^-6 C

or Q = -2.16 * 10^(-12)/(2.69e-6 ) =8.03*10^-7 C

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