A spring with spring constant ks and rest length L has positive charges Q attach

Question

A spring with spring constant ks and rest length L has positive charges Q attached to either end, as shown. Returning to the situation described in Discussion Question 2 above, find the point(s) Find an equation that will determine the length D of the spring, once the charges have come to rest. Repeat part (a), this time assuming that the charges on either end are both negative. Repeat again, this time assuming that the charges on either end have opposite signs. on the x-axis where your point charge q would remain at rest.

Explanation / Answer

a)The force between the two point charges is F = k * (Q * Q/D2) ------------(1) k = (1/4o) = 9 * 109Nm2/C2 The force acting on the spring can also be written as F = ks * D or ks * D = k * (Q * Q/D2) or D3 = (k/ks) *(Q2/D2) or D = (k/ks)1/3 *(Q/D)2/3 b)When the charges on either end are both negative then wereplace Q by -Q therefore we get F1 = k * (-Q * -Q/D2) = k *(Q2/D2) Therefore,the distance D remains unchanged. c)When the charges have opposite signs then the force actingbetween them is F2 = k * (Q * -Q/D2) Therefore,the distance D is D = -(k/ks)1/3 * (Q/D)2/3 The negative value of distance indicates that the forcebetween the charges is attractive. Let p be a point at a distance d from the charge on the leftend of the spring.The force acting between charge Q on the rightend and the charge q is F = k * (Q * q/(L + d)2) and the force acting between the charge Q on the left end andthe charge q is F1 = k * (Q * q/(d)2) For the charge q to remain at rest we have F + F1 = 0 or k * (Q * q/(L + d)2) + k * (Q *q/(d)2) = 0 solving the above equation for d we get d = -(L/2) The charge q should be placed at distance (L/2) on theopposite side of charge Q on the spring at right end. The negative value of distance indicates that the forcebetween the charges is attractive. Let p be a point at a distance d from the charge on the leftend of the spring.The force acting between charge Q on the rightend and the charge q is F = k * (Q * q/(L + d)2) and the force acting between the charge Q on the left end andthe charge q is F1 = k * (Q * q/(d)2) For the charge q to remain at rest we have F + F1 = 0 or k * (Q * q/(L + d)2) + k * (Q *q/(d)2) = 0 solving the above equation for d we get d = -(L/2) The charge q should be placed at distance (L/2) on theopposite side of charge Q on the spring at right end.

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