Asked Dec 28, 2019

A positive charge q1 = 2.70 μC on a frictionless horizontal surface is attached to a spring of force constant k as in Figure P15.12. When a charge of q2 = -8.60 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.

Figure P15.12

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92 Figure P15.12


Expert Answer

Step 1


Physics homework question answer, step 1, image 1
Step 2

Since the charges are of opposite nature then there will act an electrostatic force which will be of attractive nature and will try to stretch the spring while the inertia force will act in opposite direction to electrostatic force and will try to bring the spring in equilibrium position. In equilibrium position the net force acting will be zero which can be mathematically expressed as below,

Physics homework question answer, step 2, image 1
Step 3

By substituting the ...

Physics homework question answer, step 3, image 1

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