Q. 284.3( 6 Votes )

Two identically charged particles are fastened to the two ends of a spring of spring constant 100 N m–1 and natural length 10 cm. The system resets on a smooth horizontal table. If the charge on each particle is 2.0 × 10–8 C, find the extension in the length of the spring. Assume that the extension is small as compared to the natural length. Justify this assumption after you solve the problem.

Answer :

Natural Length of the spring: l = 10cm = 0.1m
Spring Constant : K = 100 N m–1
Charge on each particle: q = 2.0 × 10–8 C
Separation between two charges = l
Formula used:
Let the extension be ‘x’ m.
We use Coulomb’s Law:
Where Fe is the electrostatic force, k is a constant .
k =
= 9× 109 Nm2C-2 and r is the distance between two charges.
Since the electrostatic force is repulsive in nature, the spring will exert a restoring spring force Fr.
F= -Kx

Here, K is the spring constant and x is the extension.
Negative sign is because the restoring spring force us opposite to the applied force. The system would be in equilibrium when the Electrostatic force of repulsion between the two charges is equal to the spring force

Fe + Fr = 0

Yes, the assumption is justified. When two similar charges are present at two ends, they will exert repulsive force on each other. The spring will extend due to its elastic nature.
The repulsive force will have an opposite force called as restoring force in the spring. This force is directly proportional to the extension of the spring and depends on the elasticity if the material. If the extension is large compared to the natural length, then the restoring force would be proportional to the high powers of the extension.

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