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This is a classic problem in electrostatics with a twist, so I thought it could be useful for others here. I did have a problem with the integration at the end, but on general I think the idea can help others.

The idea is that a conductor, chargeless sphere of radius R is floating half-submerged on a liquid with dielectric constant (the upper half is in air, with dielectric constant equal 1). Now a differential of charge $\delta Q$ is added to the surface and the sphere moves, and we want to find how much it displaces.

Question

Find the displacement of the sphere in terms of the differential charge

My attempt at a solution

While I don't know if it's really useful, if we apply mechanical equilibrium before adding the charge, it's straightforward to find that $\rho_s=\rho_l$, where $\rho_s$ and $\rho_l$ are the volumetric densities of the sphere and the liquid, respectively.

After adding the charge, we can suppose that the sphere moves upwards (the direction doesn't matter, since at the end the sign will determine the correct one). Assuming the charge distributes uniformily over the surface, we can apply Gauss Law to find that:

$$\int_{air} \vec{D}\cdot d\vec{A}+\int_{liquid}\vec{D}\cdot d\vec{A}=\delta Q$$

Letting $\delta y$ be the vertical distance that the sphere moves up (the area is a small ring of perimeter $2\pi R$ and height $\delta y$), and since the electric field (and thus the displacement) of a submerged sphere is radial:

$$D_{air}(2\pi R^2+2\pi \delta y)+D_{liquid}(2\pi R^2-2\pi \delta y)=\delta Q$$

Figure 1: Sphere after adding charge

Assuming an isotropic, homogenous, linear dielectric, we can apply the constitutive relationship $\vec{D}=\epsilon \vec{E}$ to find that:

$$\vec{E}=\frac{1}{2\pi a}\frac{\delta Q}{(a+\delta y)+\epsilon(a-\delta y)}$$

And thus:

$$\vec{D_{air}}=\frac{1}{2\pi a}\frac{\delta Q}{(a+\delta y)+\epsilon(a-\delta y)}$$

$$\vec{D_{liquid}}=\frac{\epsilon}{2\pi a}\frac{\delta Q}{(a+\delta y)+\epsilon(a-\delta y)}$$

Now, for these kinds of materials, the energy of the fields is:

$$U_E=\int d^3 r \vec{E}\cdot\vec{D}$$

Substituting:

$$U_E=\frac{(\delta Q)^2}{4\pi^2 a^2}\left ( \int_{air} \frac{d^3r}{((a+\delta y)+\epsilon(a-\delta y))^2} + \epsilon\int_{liquid} \frac{d^3r}{((a+\delta y)+\epsilon(a-\delta y))^2} \right )$$

To first order, we can simplify the denominator since $\delta y$ is small:

$$((a+\delta y)+\epsilon(a-\delta y))^2=a^2(1+3\epsilon)+2a\delta y(1-\epsilon^2)$$

Therefore:

$$U_E=\frac{(\delta Q)^2}{4\pi^2 a^2}\left ( \int_{air} \frac{d^3r}{a^2(1+3\epsilon)+2a\delta y(1-\epsilon^2)} + \epsilon\int_{liquid} \frac{d^3r}{a^2(1+3\epsilon)+2a\delta y(1-\epsilon^2)} \right )$$

Now, if we could just express $\delta y$ as a function of spherical coordinates we could perform the integral, so I got stuck here.

However, assuming we can do the above integral, we can easily find the force as:

$$F_E=-\left ( \frac{\partial U_E}{\partial R} \right )_{\delta Q}$$

And perform a new mechanical equilibrium equation:

$$\rho_l V_{submerged} g - \rho_s V_{total} g +F_E=0$$

And theoretically find the submerged volume to find the percentage that the sphere moved after adding the charge (don't forget we calculated $\rho_s=\rho_l/2$)

If I'm allowed to ask: is my intuition about finding the displacement correct? Or how can I find it if I consider a virtual displacement aroound the equilibrium position instead?

I have received the suggestion to consider the equilibrium state and then apply a first order perturbation in $\delta Q$, which should be easier but I can't find a way to do it this way.

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  • $\begingroup$ cant we use electrostatic pressure instead of this complex inegration? $\endgroup$ – Aditya Garg Feb 24 at 6:45
  • $\begingroup$ How would that be? I've never used electrostatic pressure to solve problems. $\endgroup$ – Charlie Feb 24 at 17:00
  • $\begingroup$ like see its just an idea imagine that when the charge is given it would distribute on the surfaces as τ₁ and τ₂ and this would lead to an excess pressure downwards and force due to this will be pressure times projected area. the only problem i encountered is that i need two equations between τ₁ and τ₂ one is from charge conservation other i think would follow from gauss law but haven't tried yet $\endgroup$ – Aditya Garg Feb 24 at 17:16
  • $\begingroup$ Yeah, I checked the topic on Zangwill and you could theoretically calculate the pressures from the electric field. However, as you stated you need to find the two equations for the charges on each side. Maybe one can calculate the surface charges (which follow straight from the field $\vec{D}$), but I don't see how you can relate it to the virtual displacement of the problem. $\endgroup$ – Charlie Feb 24 at 22:10
  • $\begingroup$ They also told me one could start from the equilibrium position and then consider a virtual displacement, I'll ask for a bounty in a couple of hours to study this case. $\endgroup$ – Charlie Feb 25 at 18:03

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