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 Homework Statement

I want to evaluate the potential energy of a configuration where nonpolar particle 1, of charge ##q## and polarizability ##\alpha_1## is at distance ##r## from nonpolar particle 2, of zero charge and polarizability ##\alpha_2##.
I am neglecting gravitational forces and dispersion forces, net charge can be thought as pointlike and polarizabilites are considered scalar quantities. I'll be working in units where ##4\pi\epsilon_0\epsilon=\epsilon## for brevity.
I could not manage to find references to check my work.
 Homework Equations

Electric field generated by a pointcharge: ## E_q(r) = \dfrac{q}{\epsilon r^2} ##
Electric field generated by a pointdipole: ## E_\mu(r) = \dfrac{\mu\sqrt{1+3\cos^2\theta}}{\epsilon r^3} ##
Dipole moment induced by electric field: ## \mu = \alpha E##
I tried an approach where I build the interactions one by one and then add them all to find the total potential energy.
The electric field of particle 1 induces a dipole moment on particle 2 given by ## \mu_2 = \alpha_2 E_q(2) ##(by ##E(2)## I mean the field evaluated at the position of particle 2); this induced dipole originates a reaction field which acts on particle 1
$$ E_\mu(1) = \dfrac{\mu_2\sqrt{1+3\cos^2\theta}}{\epsilon r^3} = \dfrac{2\mu_2}{\epsilon r^3} = \dfrac{2\alpha_2E_q(2)}{\epsilon r^3} $$
where I used the fact that ##\theta## must be ##0## or ##\pi## because ##E_q(2)## acts along the line between the particles, hence the dipole is parallel to the field if ##q>0##, antiparallel if ##q<0##; this also explains the minus sign since this interaction will always be attractive.
Now I can evaluate the force that is acting on particle 1 as
$$ F(1) = qE_\mu(1) = \dfrac{2q\alpha_2E_q(2)}{\epsilon r^3} = \dfrac{2q^2\alpha_2}{\epsilon^2 r^5}$$
The potential energy of this configuration can then be found as the work needed to bring the charge from infinity up to a distance ##r## from the dipole, which is equal to the work needed for a charge ##q## to create a dipole ##\mu_2##:
$$ v'(r) = \int_\infty^r \text{d}s F(s) = \dfrac12\alpha_2 E_\mu^2(1) = \dfrac{1}{\epsilon^2}\dfrac{\alpha_2 q^2}{2r^4} $$
I am pretty sure this chargedipole contribution to the potential energy is correct; now is where I start having doubts.
The induced dipole on particle 2 induces in turn a dipole on particle 1, such that ##\mu_1 = \alpha_1E_\mu(1)##; this process should physically occur "simultaneously" to the induction of ##\mu_1## so that every infinitesimal increment ##\text{d}\mu_2## induces an infinitesimal ##\text{d}\mu_1##, but I think the result should not be affected by evaluating the 2 processes one after the other, since I only care for the equilibrium state and not for the transient.
The ##\mu_1## dipole is aligned to the ##\mu_2## dipole, thus the work needed to induce the dipole can be written as
$$ v''(r) = \int_0^{\mu_1}\text{d}\mu E_\mu(1) = \dfrac{1}{2}\alpha_1\int_0^{E_\mu(1)} \text{d}E E = \dfrac{1}{2}\alpha_1 E_\mu^2(1)
= \dfrac{2\mu_2^2\alpha_1}{\epsilon^2 r^6} $$
since the interaction energy between a permanent dipole and an electric field has the form ##v = \vec{E}\cdot\vec{\mu}##.
At this point what I have is
$$ v(r) = v'(r) + v''(r) = \dfrac{1}{\epsilon^2}\left[ \dfrac{\alpha_2 q^2}{2r^4} + \dfrac{2\mu_2^2\alpha_1}{r^6} \right] $$
Now I should be taking into account the fact that the induced dipole on particle 1 alters the field on particle 2, which adds another contribution the induced dipole of particle 2 and so on.. but the spatial decays add up pretty fast: the second term in ##v(r)## is already ##\sim r^{10}##, so I don't think I want to go further unless, for some reason, it adds essential variations to the result.
Are my approach and result correct?
If not, what is wrong and why?
What if I wanted to consider 2 polar particles instead, i.e., each of them possess a permanent dipole ##\mu_i^{(0)}##? Will the permanent dipole contributions just add on top on the nonpolar interaction(I believe this should work fine)(also reminding that in this case I would have an orientational contribution arising in the polarizibilites), or should I rework it from the beginning(if so, why)?
I'm doing this just 'for fun', no constraints, so feel free to add whatever you think is relevant. Thank you.
EDIT: I'm not neglecting dispersion forces because I think they are not relevant, I'm just temporarily pretending they don't exist.
The electric field of particle 1 induces a dipole moment on particle 2 given by ## \mu_2 = \alpha_2 E_q(2) ##(by ##E(2)## I mean the field evaluated at the position of particle 2); this induced dipole originates a reaction field which acts on particle 1
$$ E_\mu(1) = \dfrac{\mu_2\sqrt{1+3\cos^2\theta}}{\epsilon r^3} = \dfrac{2\mu_2}{\epsilon r^3} = \dfrac{2\alpha_2E_q(2)}{\epsilon r^3} $$
where I used the fact that ##\theta## must be ##0## or ##\pi## because ##E_q(2)## acts along the line between the particles, hence the dipole is parallel to the field if ##q>0##, antiparallel if ##q<0##; this also explains the minus sign since this interaction will always be attractive.
Now I can evaluate the force that is acting on particle 1 as
$$ F(1) = qE_\mu(1) = \dfrac{2q\alpha_2E_q(2)}{\epsilon r^3} = \dfrac{2q^2\alpha_2}{\epsilon^2 r^5}$$
The potential energy of this configuration can then be found as the work needed to bring the charge from infinity up to a distance ##r## from the dipole, which is equal to the work needed for a charge ##q## to create a dipole ##\mu_2##:
$$ v'(r) = \int_\infty^r \text{d}s F(s) = \dfrac12\alpha_2 E_\mu^2(1) = \dfrac{1}{\epsilon^2}\dfrac{\alpha_2 q^2}{2r^4} $$
I am pretty sure this chargedipole contribution to the potential energy is correct; now is where I start having doubts.
The induced dipole on particle 2 induces in turn a dipole on particle 1, such that ##\mu_1 = \alpha_1E_\mu(1)##; this process should physically occur "simultaneously" to the induction of ##\mu_1## so that every infinitesimal increment ##\text{d}\mu_2## induces an infinitesimal ##\text{d}\mu_1##, but I think the result should not be affected by evaluating the 2 processes one after the other, since I only care for the equilibrium state and not for the transient.
The ##\mu_1## dipole is aligned to the ##\mu_2## dipole, thus the work needed to induce the dipole can be written as
$$ v''(r) = \int_0^{\mu_1}\text{d}\mu E_\mu(1) = \dfrac{1}{2}\alpha_1\int_0^{E_\mu(1)} \text{d}E E = \dfrac{1}{2}\alpha_1 E_\mu^2(1)
= \dfrac{2\mu_2^2\alpha_1}{\epsilon^2 r^6} $$
since the interaction energy between a permanent dipole and an electric field has the form ##v = \vec{E}\cdot\vec{\mu}##.
At this point what I have is
$$ v(r) = v'(r) + v''(r) = \dfrac{1}{\epsilon^2}\left[ \dfrac{\alpha_2 q^2}{2r^4} + \dfrac{2\mu_2^2\alpha_1}{r^6} \right] $$
Now I should be taking into account the fact that the induced dipole on particle 1 alters the field on particle 2, which adds another contribution the induced dipole of particle 2 and so on.. but the spatial decays add up pretty fast: the second term in ##v(r)## is already ##\sim r^{10}##, so I don't think I want to go further unless, for some reason, it adds essential variations to the result.
Are my approach and result correct?
If not, what is wrong and why?
What if I wanted to consider 2 polar particles instead, i.e., each of them possess a permanent dipole ##\mu_i^{(0)}##? Will the permanent dipole contributions just add on top on the nonpolar interaction(I believe this should work fine)(also reminding that in this case I would have an orientational contribution arising in the polarizibilites), or should I rework it from the beginning(if so, why)?
I'm doing this just 'for fun', no constraints, so feel free to add whatever you think is relevant. Thank you.
EDIT: I'm not neglecting dispersion forces because I think they are not relevant, I'm just temporarily pretending they don't exist.
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