Respuesta :
Answer:
k = 1.23*10^{4} N/m
Explanation:
calculation of force when springs are 2.2 cm apart
[tex]F =\frac{q_1*q_2}{4\pi \epsilon r^2}[/tex]
we know that
[tex]\frac{1}{4\pi \epsilon} = 9*10^9 N-m^2/c^2[/tex]
[tex]F = \frac{2*10^6 *(4*10^6 *9*10^9 }{(2.2*10^{-2})^2}[/tex]
F = 148.76 N
spring force will balanced the coloumb force
kx = F
[tex]K = \frac{F}{x}[/tex]
[tex]= \frac{148.76}{1.2*10^{-2}}[/tex]
= 1.23*10^{4} N/m
The spring constant of mass less spring is ratio of force to the displacement of spring. The spring constant of the spring is [tex]1.23\times10^4 \rm N/m[/tex].
What is spring constant?
Spring constant is nothing but the stiffness of the spring. Spring constant is the ratio of force applied on the spring to the displacement of the spring.
The spring constant can be given as,
[tex]k=\dfrac{F}{x}[/tex]
Here, [tex]F[/tex] is the force and [tex]x[/tex] is the displacement of spring.
Given information-
The charge on the spring is 2.0 μC.
The charge of -4.0 μC is slowly brought near to spring.
The spring has stretched 1.2 cm when the charges are 2.2 cm apart.
Coulombs law states the the force of attraction or repulsion between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of distance between them.
It can be given as,
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
Here, [tex]k[/tex] is coulombs constant.
Put the values in the above formula as,
[tex]F=9\times10^9\times\dfrac{2\times10^6\times4\times10^6}{(2.2\times10^{-2})^2}\\F=148.76[/tex]
Spring constant is the ratio of force applied on the spring to the displacement of the spring. Thus,
[tex]k=\dfrac{148.76}{1.2\times10^{-2}}\\k=1.23\times10^{4}\rm N/m[/tex]
Thus, the spring constant of the spring is [tex]1.23\times10^4 \rm N/m[/tex].
Learn more about the spring constant here;
https://brainly.com/question/3510901
