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A positive charge q1 = 2.90 μc on a frictionless horizontal surface is attached to a spring of force constant k as in the figure shown below. when a charge of q2 =-9.100 μC is placed 9.50 cm away from the positive charge, the spring stretches by 5.00 mm, reducing the distance between charges to d = 9.00 cm. Find the value of k. The response you submitted has the wrong sign. N/m 1 42

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Answer #1

Force, F = kx = k qq/r^2

k (0.005) = (9 x 10^9)(2.9 x 9.1 x 10^-12)/0.09^2

k = 5864.44 N/m

Comment in case any doubt please rate my answer ....

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