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?
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:
[tex]x=3[/tex]
Explanation:
From the question we are told that:
Charge [tex]Q=40 \mu C[/tex]
Length [tex]L=20cm=0.20m[/tex]
Spring constant [tex]k=120Nm^{-1}[/tex]
Generally the equation for Force between Charges is mathematically given by
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
[tex]F=9*10^9\frac{40*10^{-6}^2}{0.2^2^2}[/tex]
[tex]F=360N[/tex]
Therefore
[tex]F=kx[/tex]
[tex]x=\frac{F}{k}[/tex]
[tex]x=\frac{360}{120}[/tex]
[tex]x=3[/tex]