Question

The given plot shows the variation of the potential energy $\left(U\right)$ of interaction between two particles with respect to the distance $\left(r\right)$ separating them. Then,

• A. $\text{B}$ and $\text{D}$ are equilibrium points
• B. $\text{C}$ is a point of unstable equilibrium
• C. The force of interaction between the two particles is attractive between points $\text{C}$ and $\text{D}$ and repulsive between $\text{D}$ and $\text{E}$
• D. The force of interaction between the particles is repulsive between points $\text{E}$ and $\text{F}$.

Answer 1

The correct option is D The force of interaction between the particles is repulsive between points $\text{E}$ and $\text{F}$.

At $\text{B}$ and $\text{D}$, $\frac{dU}{dr}\ne 0$, so they are not equilibrium points.
At $\text{C}$, $\frac{dU}{dr}=0$ and potential energy is minimum. So, it is a stable equilibrium position.
Now, $F=-\frac{dU}{dr}=$ − (Slope of $U-r$ curve)
In the region $\text{CD}$, the slope of the graph is positive, therefore by the relation, $F$ is negative i.e, force is attractive in nature.
In the region $\text{DE}$, the slope of the graph is again positive, therefore by the relation, $F$ is negative i.e, force is attractive in nature in the region $\text{DE}$ too.
But in the region $\text{EF}$, the slope of the graph is negative which means $F$ is positive i.e, force is repulsive in nature.