1. Electric field on the equatorial line of a dipole

For a dipole with charges $-q$ and $+q$ separated by distance $2a$, the dipole moment is $p = q(2a).$

At a point on the equatorial line at distance $r$ from the center, when $r \gg 2a$, the electric field magnitude is $E_{\text{eq}} = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}.$

Hence the force on a charge $Q$ placed there is $F = QE_{\text{eq}} = \frac{1}{4\pi\varepsilon_0}\frac{Qp}{r^3}.$

2. Force at point $P$

Initially, $OP = y$, so $F = \frac{1}{4\pi\varepsilon_0}\frac{Qp}{y^3}.$

This is given as $F$.

3. Force at point $P'$

Now the charge is moved along the same equatorial line to $P'$ such that $OP' = \frac{y}{3}.$

Since $\frac{y}{3} \gg 2a$, the dipole approximation is still valid.

So the new force is $F' = \frac{1}{4\pi\varepsilon_0}\frac{Qp}{(y/3)^3}.$

Simplify: $F' = \frac{1}{4\pi\varepsilon_0}\frac{Qp}{y^3/27} = 27\left(\frac{1}{4\pi\varepsilon_0}\frac{Qp}{y^3}\right).$

Thus, $F' = 27F.$

4. Checking options

• A: $9F$ ❌
• B: $3F$ ❌
• C: $F/3$ ❌
• D: $27F$ ✅

So the correct answer is D.