Question

Two identical particles of charge $q$ each are connected by a massless spring of force constant $K$. They are placed over a smooth horizontal surface and the spring is unstretched. If the initial length of the spring is $r$ and the maximum extension of the spring is $r$ when the constrained is removed, then the value of $K$ is

$($Neglect gravitational effect$)$

• A. $\frac{q^2}{4\pi {ϵ}_0{r}^3}$
• B. $\frac{q^2}{2\pi {ϵ}_0{r}^3}$
• C. $\frac{2q^2}{\pi {ϵ}_0{r}^3}$
• D. $\frac{3q^2}{4\pi {ϵ}_0{r}^3}$

Answer 1

The correct option is A $\frac{q^2}{4\pi {ϵ}_0{r}^3}$

In both initial and final situation, the particles will be at rest.

Using conservation of mechanical energy,

$K_i+U_i=K_f+U_f$

$0+\frac{q^2}{4\pi {ϵ}_{0}r}=0+\frac{q^2}{4\pi {ϵ}_0\left(2r\right)}+\frac{1}{2}K{r}^2$

$\Rightarrow \frac{q^2}{4\pi {ϵ}_{0}r}(1-\frac{1}{2})=\frac{1}{2}K{r}^2$

$\Rightarrow K=\frac{q^2}{4\pi {ϵ}_0{r}^3}$