Interpret the dipoles

Given dipole moments: $\vec d_A = -4qa\,\hat i, \qquad \vec d_B = -2qa\,\hat i$

Both dipoles lie on the $x$-axis, separated by distance $R$.

Let dipole $A$ be at $x=0$ and dipole $B$ be at $x=R$.

Potential due to a dipole on its axial line

The electric potential due to a dipole at a point located at position vector $\vec r$ is $V = \frac{1}{4\pi\varepsilon_0}\frac{\vec p\cdot \vec r}{r^3}$

On the axis of the dipole, this becomes $V = \frac{1}{4\pi\varepsilon_0}\frac{p}{x^2}$ with sign determined by the direction of $\vec p$ and the side of the point.

Assume the point lies to the right of both dipoles

Let the required point be at distance $x$ from dipole $A$, so its coordinate is $x$ with $x>R$.

Then its distance from dipole $B$ is $x-R$

Since both dipole moments are along $-\hat i$, and the point is to the right of each dipole, both potentials are negative. For equal potential values: $\frac{4qa}{x^2} = \frac{2qa}{(x-R)^2}$

Cancel $qa$: $\frac{4}{x^2} = \frac{2}{(x-R)^2}$

$2(x-R)^2 = x^2$

Solve the equation

$2(x^2 - 2Rx + R^2) = x^2$ $2x^2 - 4Rx + 2R^2 = x^2$ $x^2 - 4Rx + 2R^2 = 0$

Using quadratic formula: $x = \frac{4R \pm \sqrt{16R^2 - 8R^2}}{2}$ $x = \frac{4R \pm \sqrt{8}R}{2}$ $x = \frac{4R \pm 2\sqrt2 R}{2}$ $x = 2R \pm \sqrt2 R$

So, $x = R(2 \pm \sqrt2)$

Choose the physically relevant point

Since we assumed $x>R$, both values are possible: $x = R(2+\sqrt2), \qquad x = R(2-\sqrt2)$

Now rewrite them in option form:

$2+\sqrt2 = \frac{\sqrt2}{\sqrt2-1}$ because $\frac{\sqrt2}{\sqrt2-1} = \frac{\sqrt2(\sqrt2+1)}{2-1} = 2+\sqrt2$

Thus, $x = \frac{\sqrt2 R}{\sqrt2-1}$

The other root is $2-\sqrt2 = \frac{\sqrt2}{\sqrt2+1}$ which corresponds to $x = \frac{\sqrt2 R}{\sqrt2+1}$

From the standard figure-based interpretation for this question, the required point corresponds to the farther point on the right side, i.e. $x = \frac{\sqrt2 R}{\sqrt2-1}$

Final answer

$\boxed{\frac{\sqrt2 R}{\sqrt2-1}}$

So the correct option is C.