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 $(-q')$ is placed at a large distance $x$ from $O$.
Here, $F_{1}=F_{2}=\frac{K q q}{\left(a^{2}+x^{2}\right)}$ ...(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 $q'$ is
${\left[A C=B C=\sqrt{a^{2}+x^{2}} \text { and } \cos \theta=\frac{x}{\sqrt{a^{2}+x^{2}}}\right]}$
$F=F_{1} \cos \theta+ F_{2} \cos \theta=2 F \cos \theta$
$\left[\because F_{1}=F_{2}\right]$
$\Rightarrow F=\frac{2 K q q}{\left(a^{2}+x^{2}\right)} \cdot \frac{x}{\sqrt{a^{2}+x^{2}}}$
$=\frac{2 K q q x}{\left(a^{2}+x^{2}\right)^{3 / 2}}$ ...(ii)
But $F=m a\left[m=\right.$ mass of the charge $q'$ and $a=$ acceleration]
$\Rightarrow m a=\frac{2 K q q x}{\left(a^{2}+x^{2}\right)^{3 / 2}}$
or $a=\frac{2 K q q}{m} \cdot \frac{x}{\left(a^{2}+x^{2}\right)^{3 / 2}}$ ...(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