Question

The given plot shows the variation of $U$, the potential energy of interaction between two particles w.r.t. the distance of separation, $r$.


$1.$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{B}$ and $\text{D}$ are equilibrium points.
$2.$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{C}$ is a point of stable equilibrium.
$3.$ The force of interaction between the two particles is attractive between points <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{C}$ and $\text{D}$ and repulsive between <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{D}$ and $\text{E}$.
$4.$ The force of interaction between particles is repulsive between points <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{E}$ and $\text{F}$.

Which of the above statements are correct?

• A. $1$ and $2$
• B. $1$ and $4$
• C. $2$ and $4$
• D. $2$ and $3$

Answer 1

The correct option is C $2$ and $4$

In stable equilibrium, PE is minimum (local minima),

At $C$, potential energy is minimum. So it is a stable equilibrium position.

Further, the relation between force and the potential energy is given by,

$F=-\frac{dU}{dr}=-\left[\text{slope of (U-r) graph}\right]$.

Negative force means attraction and positive force means repulsion.

Clearly, from the graph, the slope of $(U-r)$ graph is negative between $E$ to $F$. So, the force will be positive.

Hence, the force of interaction between particles is repulsive between points <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{E}$ and $\text{F}$.

Hence, option $\left(c\right)$ is the correct answer.

Key concepts- In a conservative force field, the force and potential energy at a position are related as: $\overrightarrow{F}=-\frac{dU}{dr}\hat{r}$ [ along position vector r] In this expression at the state of equilibrium of body we use $\overrightarrow{F}=0\Rightarrow \frac{dU}{dr}=0$ Stable euilibrium (local minima): When a particleis displaced slightly from a equilibrium position and a force acting on it brings it back to its initial position, it is said to be stable equilibrium position. Unstable euilibrium (local maxima): When a particleis displaced slightly from the equilibrium position and a force acting on it pushes it further it is said to be unstable equilibrium position. Necessary condition : For local minima : Just after the equilibrium point slope of $\frac{dU}{dr}$ is $=+ve$ and $\frac{d}{dr}\text{(slope}(U-r))=\frac{d^{2}U}{d{r}^2}=+ve$ $F=-\frac{d}{dr}\frac{dU}{dr}=\frac{d^{2}U}{d{r}^2}=-ve$ For local maxima : Just after the equilibrium point slope of $\frac{dU}{dr}$ is $=-ve$ and $\frac{d}{dr}\text{(slope}(U-r))=\frac{d^{2}U}{d{r}^2}=-ve$ $F=-\frac{d}{dr}\frac{dU}{dr}=\frac{d^{2}U}{d{r}^2}=+ve$ Hence, $\text{C, E}$ and $\text{F}$ are equilibrium points. $\text{C,}$ is stable equilibrium point , $\text{E}$ is unstable equilibrium point and $\text{F}$ is neutral equilibrium point.