Electrostatics unplugged

I like the direct integral route taken by @Karan Chatrath . I took a sort of empirical approach based on torques.

$|\vec{\tau}| = |\vec{p} \times \vec{E}| = |\vec{p}| |\vec{E}| \sin \theta$

In the above expression, $\theta$ is the angle between the dipole and an applied ambient field $\vec{E}$ . Suppose we do an experiment where we apply an ambient field $\vec{E}_1$ (along the $x$ axis) of unit magnitude and measure the torque on the dipole. Then we take away $\vec{E}_1$ and apply another field $\vec{E}_2$ (along the $y$ axis) of unit magnitude, and measure the torque again. The result is:

$|\vec{\tau_1}| = |\vec{p}| \sin \theta \\ |\vec{\tau_2}| = |\vec{p}| \cos \theta \\ |\vec{\tau_1}|^2 + |\vec{\tau_2}|^2 = |\vec{p}|^2$

Of course, the calculation of the torques requires some integration, which I have omitted for the sake of brevity

Nice problem. Consider the parabolic charged arc. The total length of the arc is:

$L = \int_{0}^{1} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx$ $\implies L = \int_{0}^{1} \sqrt{1 +4x^2} \ dx$

The charge per unit length on the wire is therefore:

$\lambda = \frac{Q}{L}$ The charge in the vicinity of a point $(x,y)$ on the parabola is:

$dq = \lambda \ \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx$ $\implies dq = \lambda \ \sqrt{1 +4x^2} \ dx$

So, having found the elementary charge in the vicinity of a general point on the arc, the elementary dipole moment due to this elementary charge can be computed using the definition of dipole moment as follows. Bear in mind that the point charge is treated as the sum of all the elementary charges as computed above.

$d\vec{p} = dq \ \left(x \ \hat{i} + \left(x^2-1\right) \ \hat{j}\right)$

$\implies d\vec{p} = \lambda \ \sqrt{1 +4x^2} \ \left(x \ \hat{i} + \left(x^2-1\right) \ \hat{j}\right) dx$

Finally, the total dipole moment is:

$\vec{p} = \int_{0}^{1} \lambda \ \sqrt{1 +4x^2} \ \left(x \ \hat{i} + \left(x^2-1\right) \ \hat{j}\right) dx$

$\vec{p} =\left(\int_{0}^{1} \frac{Qx}{L} \sqrt{1 +4x^2} \ dx\right) \ \hat{i} + \left(\int_{0}^{1} \frac{Q(x^2-1)}{L} \sqrt{1 +4x^2} \ dx\right) \ \hat{j}$ $\vec{p} =Q\left( A \ \hat{i} + B \hat{j}\right)$

Where $A$ and $B$ are the values of the integrals. The magnitude is:

$\lvert \vec{p} \rvert = Q \sqrt{A^2+B^2}$ $\boxed{\lvert \vec{p} \rvert = 0.8229Q}$