Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.116 N when their center-to-center separation is 65.4 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 0.0273 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.)

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aristocles aristocles · Answer 1 · 2019-09-23T02:47:57+02:00

Answer:

Part a)

[tex]q_1 = -1.47 \times 10^{-6} C[/tex]

Part b)

[tex]q_2 = 3.75 \times 10^{-6} C[/tex]

Explanation:

Let the charge on two spheres is q1 and q2

now the force between two charges are

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

[tex]0.116 = \frac{(9\times 10^9)(q_1)(q_2)}{0.654^2}[/tex]

[tex]q_1 q_2 = 5.51 \times 10^{-12}[/tex]

now when we connect then with conducting wire then both sphere will equally divide the charge

so we will have

[tex]q = \frac{q_1-q_2}{2}[/tex]

now we have

[tex]0.0273 = \frac{(9\times 10^9)(\frac{q_1- q_2}{2})^2}{0.654^2}[/tex]

[tex]q_1 - q_2 = 2.28\times 10^{-6} C[/tex]

now we will have

Now we can solve above two equations

Part a)

negative charge on the sphere is

[tex]q_1 = -1.47 \times 10^{-6} C[/tex]

Part b)

positive charge on the sphere is

[tex]q_2 = 3.75 \times 10^{-6} C[/tex]