Two objects are identical and small enough that their sizes can be ignored relat
ID: 2230680 • Letter: T
Question
Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.255 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.61 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object, the answer to part (a) being the one with the greater (and positive) value?Explanation / Answer
We can consider the two initial charges as q1 and q2. Since the force is attractive, and the sign of each charge is opposite, I will consider F1 (the original force) to be -1.61N (you would get a negative sign if you multiplied a positive charge by a negative charge in the equation F = kq1q2/r^2). So,
F1 = -1.61N = kq1q2/d^2
where d = 0.255m.
Now, after the charges are brought into contact, the charge equally distributes between the two of them, leaving each of them with the same charge Q. This means
q1 + q2 = 2Q
Now, when the two redistributed charges are taken back to their original separation of d, the force between them is once again 1.61N, but repulsive. So, now I will consider F2 (the new force) to be positive:
F2 = 1.61N = kQ^2/d^2
So, Q^2 = F2d^2/k = (1.61 N)(0.255 m)^2/(8.99*10^9 Nm^2/C^2) = 1.16*10^-11 C^2
and Q = 3.41*10^-6 C = 3.41 C
Now, we have two equations and two unknowns (q1 and q2), so this system is solvable:
Q^2 = -q1q2 (from the fact that F2 = -F1)
2Q = q1 + q2 --> q1 = 2Q - q2
---> Q^2 = -(2Q - q2)q2 = -2Qq2 + q2^2
---> q2^2 - 2Qq2 - Q^2 = 0
Using the quadratic equation, one will see that q2 can equal either -1.41C or 8.23C. This means that q1 could equal either
q1 = 2Q - q2 = 2(3.41 C) - (-1.41 C) = 8.23 C or
q1 = 2Q - q2 = 2(3.41 C) - (8.23 C) = -1.41 C
Notice how these two pairs of solutions are the same, just with opposite assignments of which is particle 1 and which is particle 2. So, the answer is that one particle was initially -1.41C and the other was initially 8.23C.
Hope this helps.