Question

Two identical positive charges are fixed on the $y$-axis at equal distances from the origin $O$. A negatively charged particle starts on the $x$-axis, at a large distance from $O$, moves along the $x$-axis, passes through $O$ and moves far away from $O$. Its acceleration a is taken as positive along its direction of motion. The best graph between the particle’s acceleration and its $x$-coordinate is represented by

• A.
• B.
• C.
• D.

Answer 1

Answer: (C)

Let two identical positive charges $(+q)$ are placed on both sides at a distance $a$ from the origin $O$ and a negative charge $(-q^{\prime })$ is placed at a large distance $x$ from $O$.
Here, $F_1=F_2=\frac{Kqq}{\left(a^2+x^2\right)}$ ...(i)
The situation is as shown in figure. Here the vertical components $F_1\sin \theta$ and $F_2\sin \theta$ are equal and opposite. So, they cancel each other.
The resultant force on the charge $q^{\prime }$ is
$\left[AC=BC=\sqrt{a^2+x^2}\text{ and }\cos \theta =\frac{x}{\sqrt{a^2+x^2}}\right]$
$F=F_1\cos \theta +F_2\cos \theta =2F\cos \theta$
$\left[\because F_1=F_2\right]$
$\Rightarrow F=\frac{2Kqq}{\left(a^2+x^2\right)}\cdot \frac{x}{\sqrt{a^2+x^2}}$
$=\frac{2Kqqx}{{\left(a^2+x^2\right)}^{3/2}}$ ...(ii)
But $F=ma[m=$ mass of the charge $q^{\prime }$ and $a=$ acceleration]
$\Rightarrow ma=\frac{2Kqqx}{{\left(a^2+x^2\right)}^{3/2}}$
or $a=\frac{2Kqq}{m}\cdot \frac{x}{{\left(a^2+x^2\right)}^{3/2}}$ ...(iii)
When the particle starts to move, its acceleration is zero and after then the acceleration of the particle is given by Eq. (iii). So it is clear, when $x=0$, then $a=0$ and when $x=\pm \frac{a}{\sqrt{2}}$ then the acceleration is maximum. Hence, the graph between acceleration and its $x$-coordinate is given as