Question

The given plot shows the variation of $U$ , the potential energy of interaction between two particles with the distance separating them $r$ .

1. $B$ and math $D$ are equilibrium points
2. $C$ is a point of stable equilibrium
3. The force of interaction between the two particles is attractive between points $C$ and $D$ and repulsive between $D$ and $E$
4. The force of interaction between particles is repulsive between points $E$ and $F$ . 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

Answer: (C)

Conservative force is equal to negative gradient of potential energy function, so $\vec{F}=-\frac{d U}{d r}$
At point of equilibrium, net force is zero.
So, $\frac{d U}{d r}=0$ i.e at point of equilibrium Slope of $(U-r)$
curve must be zero.
At $C$, P.E. is minimum so $C$ is a point of stable equilibrium.
Between $E$ and $F$ slope of potential energy curve is negative.
So, by $F=-\frac{d U}{d r}$, Force is positive i.e repulsive.