91Ó°ÊÓ

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.

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" $\begin{array}{r}F=(9×{10}^{9}kgâ‹\dots m^3â‹\dots s^{-2}â‹\dots C^{-2})×\frac{(2×{10}^{-3}C)×(4×{10}^{-3}C)}{{\left(2.6×{10}^{−2}m\right)}^2}\\ F=1065N\\ \end{array}$

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

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

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