Question

Two spherical point charges each carrying a charge of 40 μC are attached to the two ends of a spring of length 20 cm. If its spring constant is 120 Nm-1 , what is the length of the spring when the charges are in equilibrium?

Answer 1

Answer:

$x=3$

Explanation:

From the question we are told that:

Charge $Q=40 \mu C$

Length $L=20cm=0.20m$

Spring constant $k=120Nm^{-1}$

Generally the equation for Force between Charges is mathematically given by

 $F=k\frac{q_1 q_2}{r^2}$

 $F=9*10^9\frac{40*10^{-6}^2}{0.2^2^2}$

 $F=360N$

Therefore

 $F=kx$

 $x=\frac{F}{k}$

 $x=\frac{360}{120}$

 $x=3$