Question

A massless spring is attached to a support at one end and has a 2.0 mC charge glued to the other end. A -4.0 mC charge is slowly brought near. The spring has stretched 1.2 cm when the charges are 2.6 cm apart. What is the spring constant of the spring?

Short answer

The spring constant is$k=8.9\times {10}^3\mathrm{N}/\mathrm{m}$

Step-by-step solution

Given information

The charge on the springis 2.0 mC.

The charge of -4.0 mC is slowly brought near to spring.

The springhas stretched 1.2cm when the charges are 2.6cm apart.

Explanation

Coulombslaw states the the force of attraction or repulsion between two charged particles is directly proportionalto the product of their chargesand inversely proportional to the square of distancebetween them.

It can be given as,

$F=k\frac{q_1{q}_2}{r^2}$

Here, $k$is coulombs constant.

Put the values in the above formula as,

localid="1651166694018" $\begin{array}{r}F=(9\times {10}^{9}kg\cdot m^3\cdot s^{-2}\cdot C^{-2})\times \frac{(2\times {10}^{-3}C)\times (4\times {10}^{-3}C)}{{\left(2.6\times {10}^{-2}m\right)}^2}\\ F=1065N\\ \end{array}$

Springconstant is the ratio of forceapplied on the spring to the displacement of the spring. Thus,

localid="1651166716398" $\begin{array}{r}k=\frac{1065N}{(1.2\times {10}^{-2}m)}\\ k=8.9\times {10}^3\mathrm{N}/\mathrm{m}\end{array}$

Thus, the spring constantof the spring is localid="1650558278808" $k=8.9k\mathrm{N}/\mathrm{m}$.