Two identical conducting spheres, fixed in place, attract each other with a forc

Question

Two identical conducting spheres, fixed in place, attract each other with a force of 0.088 N when their center to center separation is 47.00 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres have a net positive charge and repel each other with an electrostatic force of 0.042 N. What was the initial negative charge on one of the spheres, and what was the initial positive charge on the other?(Hint: Use charge conservation and solve for one of the initial charges. You will end up with a quadratic equation. The solutions give you the positive and negative charges.)

Explanation / Answer

F = kQq/r²

0.088 = (9 * 10^9)Qq/0.47²
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. . . . . . . . . . . . . . . (equation 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
0.042 = (9 * 10^9)(Q')²/0.47²
Q' = 1.015*10^(-6) N
Q + q = 2Q'
Q + q = 2.03 * 10^(-6). . . . . . . . . . . . . . . (equation 2)

substitute equation 1 to 2,

-2.16 * 10^(-12)/q + q = 2.03 * 10^(-6)
q = -7.71e-7 Coulomb
or
q = 2.8e-6 Coulomb

finally,
Q = -2.16 * 10^(-12)/(-7.71e-7 )  = 2.8*10^-6 Coulomb

or
Q = -2.16 * 10^(-12)/(2.8e-6)  =-7.71*10^-7 Coulomb

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