Question

The curve represents the distribution of potential along the straight line joining the two charges $Q_1$ and $Q_2$ (separated by a distance $r$) then which of the following statements are correct?


1. $|Q_1|>|Q_2|$
2. $Q_1$ is positive in nature​.
3. Points $\text{A}$ and $\text{B}$ are equilibrium points.
4. Point $\text{C}$ is a point of unstable equilibrium.

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

Answer 1

The correct option is C $1,2$ and $4$

The electric potential $V$ at a distance $r$ from a point charge $q$ is given by
$$V=\frac{kq}{r}$$ Since from the given graph, the electric potential around of charge $Q_1$ is positive, so $Q_1$ is positive.

Similarly, the electric potential around the charge $Q_2$ is negative, hence $Q_2$ is negative.

Also, at the point $\text{A}$, the net electric potential is zero so, $Q_1$ and $Q_2$ must have opposite sign.

Since, $V\propto Q$ and $V\propto \frac{1}{r}$

The point $\text{A}$ is closer to the charge $Q_2$.

So, we can conclude that, $|Q_1|>|Q_2|$

For equation $F_{net}=0$ and conservative force,

$\overrightarrow{F}=-\frac{dU}{dr}=-q\frac{dV}{dr}$

Where,

$U=qV=$Potential energy

$\Rightarrow |F|\propto \frac{dV}{dr}$

$F_{net}=0$, when the slope of $V-r$ graph becomes zero.

But at points $\text{A}$ and $\text{B}$, the slope is non-zero.

So, $\text{A}$ and $\text{B}$ are not equilibrium points.

At Point $\text{C}$, if the particle is displaced slightly, it does not regain its original position. Hence, the particle is said to be in unstable equilibrium at this point.

Thus, statements $1$, $2$ and $4$ are correct.

Hence, option (c) is the correct answer.