Question

Two charges each equal to $q$, are kept at $x=a$and $x=-a$ on the $x$-axis. A particle of mass $m$ and charge $q_0=\frac{q}{2}$ is placed at the origin. If a charge $q_0$ is given, a small displacement ($y<<a$) along the $y$-axis, the net force acting on the particle is proportional to

• A. $y$
• B. $-y$
• C. $\frac{1}{y}$
• D. $-\frac{1}{y}$

Answer 1

The correct option is A

$y$

Step 1: Given data

$Mass=m$

$\text{Charge}=q$

Step 2: To find

The net force acting on the particle is proportional to.

Step 3: Calculate the net force

We are given that two charges, each having a charge $q$ are kept opposite to each other along the $x$-axis at a distance of $a$ from the origin. Another particle of charge ${\displaystyle \frac{q}{2}}$ is placed along the y-axis at a distance $Y$ from the origin as shown in the figure. The charges $q$ each exert a repulsive force directing outwards on the particle as shown.
Let the magnitude of the force exerted by each charge $q$ on the particle of charge ${\displaystyle \frac{q}{2}}$ be $F$. From Coulomb’s law, this will be equivalent to:
$F={\displaystyle \frac{kq\left({\displaystyle \frac{q}{2}}\right)}{r^2}}$
where $k$ is the Coulomb’s constant, and $r$ is the distance between the charges $q$ and ${\displaystyle \frac{q}{2}}$
The distance between $q$ and ${\displaystyle \frac{q}{2}}$ can be found by applying Pythagora's theorem to one of the right angles formed, as shown in the figure.
$$\begin{gathered} \Rightarrow r^2=y^2+a^2 \\ \Rightarrow r=\sqrt{y^2+a^2} \end{gathered}$$
Therefore,

$F=\frac{k{q}^2}{2\left(y^2+a^2\right)}$
To find the net force acting on the particle, we resolve the force that they exert on the particle into their respective horizontal and vertical components. Let $\theta$ be the angle that the force exerted by the charges makes with the y-axis.
The net force along the horizontal direction will be:
$F_H=Fsin\theta +(-Fsin\theta )=Fsin\theta –Fsin\theta =0$
Therefore, no force acts on the particle of charge ${\displaystyle \frac{q}{2}}$ in the horizontal direction.
The net force acting along the vertical direction will be:
$F_V=Fcos\theta +Fcos\theta =2Fcos\theta$
Therefore, the total force acting on the particle of charge ${\displaystyle \frac{q}{2}}$ will be the sum of the forces acting on it in the horizontal and vertical directions, i.e.,

$$\begin{gathered} F_{net}=F_H+F_v \\ =0+2F\cos \theta \\ =2F\cos \theta \end{gathered}$$
But

$F={\displaystyle \frac{k{q}^2}{2\left(y^2+a^2\right)}}$
Therefore, $F_{net}={\displaystyle \frac{2k{q}^2}{2\left(y^2+a^2\right)}}cos\theta ={\displaystyle \frac{k{q}^2}{\left(y^2+a^2\right)}}cos\theta$
From the diagram, taking any one of the right-angled triangles, we have:
$cos\theta ={\displaystyle \frac{y}{r}}={\displaystyle \frac{y}{\sqrt{\left(y^2+a^2\right)}}}$
Plugging this into our net force equation we get:
$F_{net}={\displaystyle \frac{2k{q}^2}{2\left(y^2+a^2\right)}}cos\theta ={\displaystyle \frac{k{q}^2}{\left(y^2+a^2\right)}}.{\displaystyle \frac{y}{\sqrt{\left(y^2+a^2\right)}}}$
$\Rightarrow F_{net}={\displaystyle \frac{k{q}^{2}y}{{\left(y^2+a^2\right)}^{3/2}}}$
But we are given that $y<<a$ which means that we can approximate y2+a2≈a2$y^2+a^2\approx a^2$, so the above equation becomes:
$F_{net}={\displaystyle \frac{k{q}^{2}y}{{\left(a^2\right)}^{3/2}}}={\displaystyle \frac{k{q}^{2}y}{\left(a^2\right)}}$
$F_{net}\propto y$

Hence, option(A) is correct.