An unstrained horizontal spring has a length of 0.39 m and a spring constant of

Question

An unstrained horizontal spring has a length of 0.39 m and a spring constant of 270 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.023 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges. (a) Either both charges are positive or both charges are negative. (b) Number 1.025e-5 Units the tolerance is +/-2%

Explanation / Answer

Electric force is given by:

F = (1/4*pi*e0)*q1*q2/r^2

Spring force is given by:

F = kx

Given values are:

k = 270 N/m

x = 0.023 m

F = 270*0.023 = 6.21 N

Now this spring force should be equal to the electric force, so

1/(4*pi*e0) = 9*10^9

r = 0.39 m

q1 = q2 = q

q = sqrt [F*r^2*(4*pi*e0)]

q = sqrt (6.21*0.39/(9*10^9)) = 1.64*10^-5 C

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