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 0.083 N when their center to center separation is 51.00 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 0.046 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.)

Initial negative charge C.
Initial positive charge C.

Explanation / Answer

q1, q2 have opposite signs and their sum is positive.
When the spheres are connected the net charge is shared equally between them.
Assume q1 +, q2 -

Let q = (q1+q2)/2 (mean charge)
Let dq = |q1-q2|/2 (abs diff. of each charge from mean)
Then we can substitute q+dq for q1, q-dq for q2

|F1| = -k(q+dq)(q-dq)/r^2

|F2| = kq^2/r^2

q = sqrt(r^2|F2|/k) = sqrt(0.51^2 *0.046 / (9 *10^9)) = 1.15 *10^-6 C

Now,

|F1/F2| = -(q+dq)(q-dq)/q^2 = -(q^2-dq^2)/q^2 = dq^2/q^2-1

q^2(1+|F1/F2|) = dq^2

dq = sqrt(q^2(1+|F1/F2|)) = sqrt[ (1.15 *10^-6)^2 (1 + (0.083/0.046))] = 1.93 *0^-6 C

Initial + charge q1 = q+dq = 1.15 *10^-6 C + 1.93 *0^-6 C = 3.08 * 10^-6 C

Initial - charge q2 = q-dq = 1.15 *10^-6 C - 1.93 *0^-6 C = -7.8 *10^-7 C

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.