Question

In the figure shown $A\&B$ are two charged particles having charges $q$ and $-$ $q$ respectively. They are placed on a nonconducting fixed horizontal smooth plane. $B$ is fixed and $A$ is attached to a non conducting massless spring of spring constant $k.$ The other end of the spring is fixed. Mass of $A$ is $m,A$ and $B$ are in equilibrium when the distance between them is $r$. Choose the correct options

• A. time period of small oscillation of block A about means position is $=2\pi \sqrt{\frac{m}{k-\frac{q^2}{2\pi e_0{r}^3}}}$
• B. time period of small oscillation of block A about means position is $=2\pi \sqrt{\frac{m}{k+\frac{q^2}{2\pi {ϵ}_0{r}^3}}}$
• C. to perform SHM, K must be greater than $\left(\frac{q^2}{2\pi {\epsilon }_0{r}^3}\right)$
• D. to perform SHM, K must be greater than $\left(\frac{2q^2}{\pi {\epsilon }_0{r}^3}\right)$

Answer 1

Answer: (A), (C)

The correct options are
A time period of small oscillation of block A about means position is $=2\pi \sqrt{\frac{m}{k-\frac{q^2}{2\pi e_0{r}^3}}}$
C to perform SHM, K must be greater than $\left(\frac{q^2}{2\pi {\epsilon }_0{r}^3}\right)$

Let $x_0=$ extension in the spring when $A$ is in equilibrium. Then,
$$k{x}_0=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{r^2}---\left(1\right)$$
Now let A be shifted by a small distance $x$ towards $B$. Then the resultant force towards A is,

$F_{\text{res}}=k(x_0+x)-\frac{q^2}{4\pi {ϵ}_0(r-x{)}^2}=k(x_0+x)-\frac{q^2}{4\pi {ϵ}_0{r}^2}{(1-\frac{x}{r})}^{-2}$

$=k(x_0+x)-\frac{q^2}{4\pi {ϵ}_0{r}^2}(1+\frac{2x}{r});x<<r:$ By binomial expansion, we get
$$kx-\frac{q^2}{2\pi {ϵ}_0{r}^3}x;\text{using}\left(1\right)F_{res}=(k-\frac{q^2}{2\pi {ϵ}_0{r}^3})x$$
$F\propto x$
$\therefore$ It is a SHM with $T=2\pi \sqrt{\frac{m}{k-\frac{q^2}{2\pi {ϵ}_0{r}^3}}}$
For real $T$, $k>\frac{q^2}{2\pi {ϵ}_0{r}^3}\therefore k_{min}=\frac{q^2}{2\pi {ϵ}_0{r}^3}Ans$