Question

Two equal positive charges $+2q$ are placed at a distance $a$ from the origin on either side of the x-axis. A negative charge placed at the origin is

• A. in stable equilibrium along the x-axis
• B. in unstable equilibrium along the y-axis
• C. in stable equilibrium along the y-axis
• D. in unstable equilibrium along the x-axis

Answer 1

Answer: (C), (D)

The correct options are
C in stable equilibrium along the y-axis
D in unstable equilibrium along the x-axis

If the charge particle is moved in x direction like x

force at A =$\frac{K2Q}{{(a-x)}^2}-\frac{K2Q}{{(a+x)}^2}=\frac{K2Q}{{(a-x)}^2}\frac{4ax}{{(a+x)}^2}$

F is always positive for increase in x, hence the particle will move and is in stable equilibrium.

now let y be the small movement

$F_{net}=\frac{2KqQ(-y)}{{(y^2+a^2)}^{3/2}}$

In the y force change its direct and is always acting in resistance to the motion.So in y direction the particle will oscilate.