Question

A massless spring is attached to a support at one end and has a 2.0 μC charge glued to the other end. A -4.0 μC charge is slowly brought near. The spring has stretched 1.2 cm when the charges are 2.2 cm apart. What is the spring constant of the spring? Express your answer with the appropriate units.

Answer 1

Answer:

k = 1.23*10^{4} N/m

Explanation:

calculation of force when springs are 2.2 cm apart

$F =\frac{q_1*q_2}{4\pi \epsilon r^2}$

we know that

$\frac{1}{4\pi \epsilon} = 9*10^9 N-m^2/c^2$

$F = \frac{2*10^6 *(4*10^6 *9*10^9 }{(2.2*10^{-2})^2}$

F = 148.76 N

spring force will balanced the coloumb force

kx = F

$K = \frac{F}{x}$

   $= \frac{148.76}{1.2*10^{-2}}$

   = 1.23*10^{4} N/m