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 -0.5610 N when separated by 50 cm, center-to-center. 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 0.3024 N. What were the initial charges on the spheres Since one is negative and you cannot tell which is positive or negative, there are two solutions. Take the absolute value of the charges and enter the smaller value here. Enter the larger value here.

Explanation / Answer

Initial condition,

F1= kq1q2/r1^1

Plugging values,

-0.5610 = 9*10^9*q1q2/(0.50^2)    => q1q2= -1.56*10^-11 C^2   --------------(1)

Final condition,

Then the two spheres are joined by a wire. The charge is now free to re–distribute itself between the two spheres and since they are identical the total excess charge (that is, 11+q2) will be evenly divided between the two spheres. If the new charge on each sphere is Q, then

Q+Q = 2Q = q1 + q2

F2= kQ^2/r2^1

0.3024 = 9*10^9*Q^2/(0.50^2)    => Q= 3*10^-6 C

We don’t know what the sign of Q is, so we can only say,

Q = +/- 3*10^-6 C

q1+q2= 2Q = +/- 6*10^-6 C

q2 = (+/- 6*10^-6 - q1) C --------(2)

Choosing + sign in (2),

q2 = ( 6*10^-6 - q1)

q1q2= -1.56*10^-11

q1(6*10^-6 - q1) = -1.56*10^-11    => q1 = -1.96*10^-6 C   or 7.96*10^-6 C

Choosing - sign in (2),

q2 = (- 6*10^-6 - q1)

q1q2= -1.56*10^-11

q1(-6*10^-6 - q1) = -1.56*10^-11    => q1 = -7.96*10^-6 C or 1.96*10^-6 C

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