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 charge $$q_0$$ is given a small displacement ($$y << a$$) along the y-axis, the net force acting on the particle is proportional to :

We have two identical point charges, each of magnitude $$q$$, fixed on the x-axis at the points $$(-a,\,0)$$ and $$(\,a,\,0)$$.

The test particle carries charge $$q_0=-\dfrac{q}{2}$$ and mass $$m$$. Initially this particle is at the origin, but we now give it a very small displacement $$y \;(\;y\ll a\;)$$ along the positive y-axis, so its new co-ordinates become $$(0,\,y)$$.

The electrostatic force between two point charges is given by Coulomb’s law: $$\vec F = k\,\dfrac{q_1q_2}{r^2}\,\hat r,$$ where $$k=\dfrac{1}{4\pi\varepsilon_0}$$, $$r$$ is the distance between the charges and $$\hat r$$ is the unit vector directed from the source charge to the test charge.

Let us find the force exerted on the displaced particle by each fixed charge.

Force due to the charge at $$(-a,0):$$ The position vector of the particle relative to this charge is $$\vec r_1 = (0-(-a))\,\hat i + (y-0)\,\hat j = a\,\hat i + y\,\hat j.$$ Its magnitude is $$r_1 = \sqrt{a^2+y^2}.$$ Therefore, using Coulomb’s law,

$$ \vec F_1 = k\,\dfrac{q\,q_0}{(a^2+y^2)}\,\dfrac{a\,\hat i + y\,\hat j}{\sqrt{a^2+y^2}} = k\,\dfrac{q\,q_0}{(a^2+y^2)^{3/2}}\,(a\,\hat i + y\,\hat j). $$

Force due to the charge at $$(+a,0):$$ Here the position vector of the particle relative to the charge is $$\vec r_2 = (0-a)\,\hat i + (y-0)\,\hat j = -a\,\hat i + y\,\hat j,$$ with the same magnitude $$r_2=\sqrt{a^2+y^2}$$. Hence,

$$ \vec F_2 = k\,\dfrac{q\,q_0}{(a^2+y^2)^{3/2}}\;(-a\,\hat i + y\,\hat j). $$

Adding the two forces:

$$ \vec F_{\text{net}} = \vec F_1 + \vec F_2 = k\,\dfrac{q\,q_0}{(a^2+y^2)^{3/2}}\Big[(a-a)\,\hat i + (y+y)\,\hat j\Big] = k\,\dfrac{q\,q_0}{(a^2+y^2)^{3/2}}\,(2y)\,\hat j. $$

So only a y-component survives: $$ F_y = 2k\,\dfrac{q\,q_0\,y}{(a^2+y^2)^{3/2}}. $$

Substituting $$q_0=-\dfrac{q}{2}$$, we obtain

$$ F_y = 2k\,\dfrac{q\left(-\dfrac{q}{2}\right)\,y}{(a^2+y^2)^{3/2}} = -k\,\dfrac{q^{\,2}\,y}{(a^2+y^2)^{3/2}}. $$

Because the displacement is very small compared with $$a$$  ($$y\ll a$$), we can approximate

$$ (a^2+y^2)^{3/2} = a^3\left(1+\dfrac{y^2}{a^2}\right)^{3/2} \approx a^3 \quad(\text{keeping only the leading term}). $$

Substituting this approximation,

$$ F_y \approx -k\,\dfrac{q^{\,2}\,y}{a^3}. $$

The constant factor $$-k\,\dfrac{q^{\,2}}{a^3}$$ is independent of $$y$$, so

$$ F_y \propto -y. $$

Thus the net restoring force is directly proportional to $$-y$$.

Hence, the correct answer is Option C.