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

`x=3`

Step-by-step 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(q_1 q_2)/(r^2)`

`F=9*10^9(40*10^(-6)^2)/(0.2^2^2)`

`F=360N`

Therefore

`F=kx`

`x=(F)/(k)`

`x=(360)/(120)`

`x=3`