Question

Two fixed $4Q$ (positive) and $Q$ (negative) are located at $A$ and $B$, the distance $AB$ being $3m$.

Choose the correct alternative.

• A. The point $P$ where the resultant field due to both is zero lies outside the line joining $AB$
• B. The point P where the resultant field due to both is zero lies inside the line joining $AB$
• C. If a positive charge is placed at P and displaced slightly along $AB$ it will execute oscillations
• D. If a negative charge is placed at P and displaced slightly along $AB$ it will execute oscillations

Answer 1

Answer: (A), (D)

The point $P$ where the resultant field due to both is zero lies outside the line joining $AB$
If a negative charge is placed at P and displaced slightly along $AB$ it will execute oscillations

Since charges are of opposite sign so the field direction at any point in between them will be in the same direction, shown in the figure. So they can n't cancel each other. So to cancel the field the point P has to be outside AB, i.e. option A).

If a positive charge is placed at P, it will be in unstable equilibrium hence it will not execute oscillation. But if a negative charge is placed P, it will execute oscillatory motion as it will be in stable equilibrium.