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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.)

User Ronn
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1 Answer

3 votes

Answer:

Part a)


q_1 = -1.47 * 10^(-6) C

Part b)


q_2 = 3.75 * 10^(-6) C

Step-by-step explanation:

Let the charge on two spheres is q1 and q2

now the force between two charges are


F = (kq_1q_2)/(r^2)


0.116 = ((9* 10^9)(q_1)(q_2))/(0.654^2)


q_1 q_2 = 5.51 * 10^(-12)

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

so we will have


q = (q_1-q_2)/(2)

now we have


0.0273 = ((9* 10^9)((q_1- q_2)/(2))^2)/(0.654^2)


q_1 - q_2 = 2.28* 10^(-6) C

now we will have

Now we can solve above two equations

Part a)

negative charge on the sphere is


q_1 = -1.47 * 10^(-6) C

Part b)

positive charge on the sphere is


q_2 = 3.75 * 10^(-6) C

User Chris Bosco
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