Question

Two point charges $(Q$ each$)$ are placed at $(0,y)$ and $(0,–y)$. A point charge $q$ of the same polarity can move along $x-$axis. Then:

• A. The force on $q$ is maximum at $x=\pm \frac{y}{\sqrt{2}}$.
• B. The charge $q$ is in equilibrium at the origin.
  
• C. The charge $q$ performs an oscillatory motion about the origin.
  
• D. For any position of $q$ other than origin, the force is directed away from the origin.
  

Answer 1

Answer: (A), (B), (D)

For any position of $q$ other than origin, the force is directed away from the origin.



Net force on charge $q$,

$F_{\text{net}}=2F\cos \theta =2\times \frac{1}{4\pi {ϵ}_0}\frac{Qq}{(x^2+y^2)}\frac{x}{\sqrt{x^2+y^2}}$
$\Rightarrow F_{\text{net}}=\frac{Qq}{2\pi {ϵ}_0}\times \frac{x}{(x^2+y^2{)}^{3/2}}$

To maximize $F_{\text{net}}$,

$\frac{d{F}_{\text{net}}}{dx}=0$ or $x=\pm \frac{y}{\sqrt{2}}$

Therefore, force on $q$ will be maximum at
$x=\pm \frac{y}{\sqrt{2}}.$

The charge $q$ will be in equilibrium at the origin.
For any position of charge $q$ other than origin, the force will be directed away from the origin.
Therefore, charge $q$ will not perform an oscillatory motion about the origin.