Magnitude of a dipole moment [closed]

Question

I am now learning Electromagnetism, and I am doing the practice question. However I am stuck in the following question.

A uniform electric field has a dipole placed in it. If you rotate the dipole you find that the maximum torque it experiences is $7.9 × 10^{-26} N·m$. If the electric field is generated by two parallel plates separated by $11.8cm$ subject to a potential difference of $33.1 V$, what is the magnitude of the dipole moment $p$.

After checking lecture notes, I know the following (same as the one in my lecture notes) Would someone explain to me if the figure above is related to my question or not.

p is pretty sure stands for moment, F is force, q is the charge, E is electric field. I am not sure if the question said 11.8 cm is the distance of the particles or not.

Answer 1

As $d\sin\phi$ is maximized when $\phi=\frac{\pi}{2}$ we find that the torque between the two particles is maximized when $\phi=\frac{\pi}{2}$ where: \begin{equation} \begin{aligned} \tau_{\textrm{max}}&=\frac{F_+d}{2}+\frac{F_-d}{2}\\ &=Eqd\\ &=Ep \end{aligned} \end{equation} where in our last step we used the definition of the dipole moment. In addition, as the magnitude of the electric field in a parallel plate capacitor is given by $E=\frac{V}{l}$ we find that: \begin{equation} \begin{aligned} \tau_{\textrm{max}}&=Ep\\ &=\frac{Vp}{l}\\ \implies p&=\frac{\tau_{\textrm{max}}l}{V} \end{aligned} \end{equation} Substituting our given values we therefore find that: \begin{equation} \begin{aligned} p&=\frac{\tau_{\textrm{max}}l}{V}\\ &=\frac{7.9\times 10^{-26}\times 0.118}{33.1}\textrm{ C m}\\ &\approx 2.82\times 10^{-28}\textrm{ C m} \end{aligned} \end{equation}

Answer 2

Torque = P*E ( P & E are vectors, also * means cross product)

Since the torque is maximum P is perpendicular to E so we can forget about the cross product(sin90 = 1) Torque given here = PE

In a capacitor E = potentia/distance between the plates

Therefore P = max torque/(V/D)