Capillary force on an ‘inert’ colloid: a physical analogy to dielectrophoresis

“Inert” colloids are μm-scale particles that create no distortion when trapped at a planar fluid–fluid interface. When placed in a curved interface, however, such colloids can create interfacial distortions of quadrupolar symmetry – so-called “induced capillary quadrupoles.” The present work explores the analogy between capillary quadrupoles and electric dipoles, and the forces exerted on them by a symmetry-breaking gradient. In doing so, we weigh in on an outstanding debate as to whether a curvature gradient can induce a capillary force on an inert colloid. We argue that this force exists, for the opposite would imply that all dielectrophoretic forces vanish in two dimensions (2D). We justify our claim by solving 2D Laplace problems of electrostatics and capillary statics involving a single particle placed within a large circular shell with an imposed gradient. We show that the static boundary condition on the outer shell must be considered when applying the principle of virtual work to compute the force on the particle, as verified by a direct calculation of this force through integration of the particle stresses. Our investigation highlights some of the subtleties that emerge in virtual work calculations of capillary statics and electrostatics, thereby clarifying and extending previous results in the field. The broader implication of our results is that inert particles – including particles with planar, pinned contact lines and equilibrium contact angles – interact through interparticle capillary forces that scale quadratically with the deviatoric curvature of the host interface, contrary to recent claims made in the literature.


Introduction
In this supplemental document, we provide details for the solution of the Laplace problems for the electric potential ψ and interface height ζ described in the main text (the relevant geometry is sketched in Fig. 3). For both the electric and capillary problems, we solve the Laplace problem approximately using the method of reflections 1-4 ( §S.1 and §S.3) and validate our approximation using an exact solution in bipolar coordinates 3,5 ( §S. 2 and §S.4). Concurrently, we also calculate the (electric or capillary) force F and energy W. In the main text, results for F and W are presented in the "unbounded" limit as the shell radius R → ∞. By expanding to higher reflections (and verifying against the bipolar solution), we also determine the first finite-size correction to F and W in terms of the ratio a/R, where a is the particle radius. The resulting analytical expressions are summarized in Tables S.1 and S.2 for the electric and capillary problems, respectively.  1 Analytical expressions for the electric force and energy, including the first finite-size correction, for a particle held in Conductor

S.1.1 Conducting particle
Conducting particles are discussed in §3.1 in the main text. Below, we describe the mathematical details underlying the results presented in that section.

S.1.1.1 Governing equations
Laplace problem: Consider the boundary-value problem for a conducting particle described in the main text [Eqs. (3.8)-(3.11)]. The electric potential ψ obeys Laplace's equation, ∇ 2 ψ = 0 for r ≥ a and r ≤ R, (3.8) subject to boundary conditions on the shell, eithern · ∇ψ Q =n · ∇ψ ext at r = R (3.9a) or ψ V = ψ ext at r = R, (3.9b) and the particle, ψ = V at r = a, (3.10) Here, ψ ext is the externally applied potential given by Eq. (3.1), V is the constant potential on the particle boundary, and Q is the net charge.
Electric force: Once the potential ψ is known, the force F may be calculated using Eq. (3.7) from the main text: Substituting Eq. (S.1.2) into (3.7) yields the double-series expansion, where F (mn) depends upon the product of terms containing ψ (m) and ψ (n) . In Eq. (S.1.7), we have defined F (mn) such that m runs from 0 to n; thus, each nth reflection contributes n + 1 terms to the sum. E.g., for the zeroth reflection (n = 0), we need only compute one term: Thus, the self-coupling of the zeroth reflection ψ (0) (i.e., the external potential) contributes no force to the particle.
Electric energy: To calculate the insertion energy W, we must apply Eq. (3.5) from the main text: After integrating by parts and using ∇ 2 ψ = ∇ 2 ψ ext = 0 to eliminate the area integrals, we obtain which depends only upon line integrals over the particle and shell boundaries. Substituting Eq. (S.1.2) into (S.1.9) for ψ and setting ψ ext = ψ (0) yields the double-series expansion, where W (mn) depends upon the product of terms containing ψ (m) and ψ (n) . Using Eq. (S.1.4), we may calculate the part of the energy due solely to ψ (0) : i.e., the work done to eliminate the external field within the particle. Equation (S.1.11) does not account for the field disturbance created by the particle, which contributes like-ordered terms to the energy. These terms depend on higher reflections. Below, we compute the sequential approximations ψ (n) , V (n) , F (mn) , and W (mn) for the first few values of n and m = 0, . . . , n. We implicitly assume that the particle is much smaller than the characteristic dimension of the shell, a/R 1, and sufficiently far away from the shell boundary, ξ = o(R). Initially, we shall approximate the reflections such that only the leading-order contributions to the force and energy are retained, neglecting errors of O(a 2 /R 2 ). These leading-order contributions are exact in the limit as the shell becomes infinitely large. Subsequently, we evaluate the O(a 2 /R 2 ) correction to the force and energy, which accounts for the finite size of the shell.

S.1.1.2 First reflection
The particle-free potential ψ (0) violates the particle boundary conditions (3.10)-(3.11). Thus, we add the first reflection ψ (1) such that which remediates the conditions at r = a. To solve the boundary-value problem (S.1.12)-(S.1.14), it is convenient to first recast ψ (0) in terms of the particle-centered position r . Substituting r = r + ξ into Eq. (S.1.4) gives, where we have defined the external field and field gradient at the particle's center: Substituting Eq. (S.1.15) into (S.1.13)-(S.1.14) and integrating then yields the first reflection It is straightforward to verify that Eqs. (S.1.18)-(S.1.19) uniquely satisfy (S.1.12)-(S.1.14). Thus, the first reflection induces dipolar and quadrupolar disturbances to the potential. The sum ψ (0) + ψ (1) represents the potential in a semi-infinite dielectric medium, which is exact in the limit as the shell radius R → ∞. At this order of approximation, we may compute the following contributions to the force from Eq. (S.1.7): Thus, the first non-vanishing contribution to the force arises from the coupling between the particle-free potential ψ (0) and the leading-order ("unbounded") disturbance ψ (1) . Using Eq. (S.1.10), we obtain the following contributions to the energy: where terms of O(a 2 /R 2 ) been neglected in Eq. (S.1.23). The integrals over r = R in Eqs. (S.1.22)-(S.1.23) are evaluated by first expanding the decaying harmonics with respect to r (contained in ψ (1) ) as an infinite series of harmonics in r, using Eq. (S.A.4) in Appendix S.A. Then, each term in the series is integrated over r = R up to the desired level of accuracy. It should be remembered that the unit normaln = r /r and −r/r at r = a and r = R, respectively. The derived expression for W (11) [Eq. (S.1.23)] exactly cancels W (00) [Eq. (S.1.11)] in the limit as R → ∞; that is, W (00) + W (11) 0. This implies that truncating the approximation for the potential after the first reflection is insufficient to calculate the energy. The source of error can be traced to the fact that ψ (0) + ψ (1) does not satisfy the electrostatic condition at r = R. As we show in the next section, adding the second reflection ψ (2) results in a finite energy.

S.1.1.3 Second reflection
The addition of the first reflection ψ (1) perturbs the shell boundary condition (3.9) at r = R. Thus, we add a correction ψ (2) that satisfies Laplace's equation, subject to one of two conditions on the outer shell: To solve this problem, we must first rewrite ψ (1) in terms of the shell-centered position r. Substituting r = r − ξ into Eq.   (1) and integrating, we arrive at the expression, Thus, the second reflection induces an infinite series of growing harmonics in order to rectify the boundary condition at r = R.
For the purpose of calculating the leading-order contributions to the force and energy, it is sufficient to retain only the terms shown in Eq. (S.1.27). These terms satisfy the electrostatic condition at r = R up to O(a 3 /R 3 ) errors. The remaining terms in the series may be omitted, as these contribute corrections of similar magnitude to the higher reflections.
This expression is equivalent to Eqs. (3.17) and (3.18) in the main text. Since W (00) and W (11) exactly cancel each other in the limit as R → ∞, the only finite contribution is due to W (02) . This term results from the coupling between ψ (0) and ψ (2) . Finite-size corrections are of O(a 2 /R 2 ) and are discussed in §S.1.1.5.

S.1.1.5 Higher reflections: finite-size effects
The results (S.1.36) and (S.1.37) presented above are exact if the shell radius R is infinitely large compared to the particle radius a. In this limit, one need only compute the first reflection ψ (1) emitted from the particle to calculate the force F from Eq. (S.1.7); to compute the energy W from Eq. (S.1.10), one additionally needs the second reflection ψ (2) emitted from the outer shell. For infinitely large domains, the force is independent of the conditions used to establish the field, whereas the energy is not. The situation changes if the shell radius R is comparable to the particle radius a. For finite-sized shells, the force must depend on the condition used to establish the external field. The reason is the local field in the vicinity of the particle is now influenced by reflected modes from the outer shell. Higher odd-and even-numbered reflections may be computed by retracing the steps described in §S.1.1.2 and §S.1.1.3, respectively. The third reflection ψ (3) contributes an O(a 2 /R 2 ) correction to the force F by coupling to the ambient potential ψ (0) , adding a term F (03) to Eq. (S.1.7). This correction reverses sign depending upon whether the charges or potentials are held fixed on the outer shell. For the energy W, it is necessary to determine the potential up to the fourth reflection ψ (4) ; the coupling terms W (11) , W (02) , W (22) , W (13) , and W (04) all give O(a 2 /R 2 ) contributions to Eq. (S.1.10). Subsequent reflections contribute corrections to F and W in successive powers of a 2 /R 2 .
Calculating the third and fourth reflections (not presented here, for the sake of brevity) allows us to expand Eqs. (S.1.36) and (S.1.37) up to the O(a 2 /R 2 ) correction. Note that we must also expand the lower-order reflections up to O(a 2 /R 2 ) terms, are plotted against the relative particle position ξ = (ξ − ξ * )/a (left) and the radius ratio a/R (right), where ξ * is the fixed point as defined by Eq. (3.24) in the main text. For the numerical calculations in bipolar coordinates, we set the field amplitudes E 0 = E 1 = 1 and phase angles α whereas we had previously invoked the limit R → ∞. For finite-sized shells with a fixed charge distribution, we obtain the following approximations for the force and energy: and 1. Second, we find F Q F V and W Q −W V for shells of finite size, contrary to the limiting result for a semi-infinite medium. It remains true, however, that W Q and −W V are equivalent to the particle's potential energy when, respectively, the charges and potentials are held fixed on the outer shell. In the latter case, the external work done to maintain the shell potential is 2W V , as shown in the main text [cf. Eq. (3.22)].

S.1.2 Insulating particle
Insulating particles are discussed in §3.2 in the main text. For such particles, the Dirichlet condition (3.10) at r = a is replaced by the Neumann condition,n · ∇ψ = 0 at r = a. (3.25) Once again, the solution for the potential ψ can be obtained using the method of reflections. The solution procedure is essentially the same as the one described in §S.1.1. The key difference to recognize is that the insulating Neumann condition (3.25) induces reflected modes −(ψ (1) + ψ (2) ), −(ψ (5) + ψ (6) ), etc. of opposite sign compared to the conducting Dirichlet condition (3.10). We may immediately deduce this result by applying Eqs. (S.A.5)-(S.A.6) to the boundary condition at the particle. To avoid redundancy, we shall not describe this calculation in any detail. Instead, we shall simply present the final results for insulators that are analogous to Eqs. (S.1.38)-(S.1.41) for conductors. For fixed charges on the outer shell, we find and whereas, for fixed potentials, we find instead It is interesting to note that W Q and W V for insulators are, respectively, identical to W V and W Q for conductors. As anticipated, insulators are attracted to fixed charges and repelled by fixed potentials.
To summarize, we have described an approximate solution to the Laplace problem, both for conducting and insulating particles, using the method of reflections. We have validated this reflections solution using an exact solution in bipolar coordinates; the numerical data from the bipolar solution are plotted as markers in Figs. S.1-S.4. Details on the bipolar solution are described in the following section.

S.2 Electric problem: bipolar coordinates
The Laplace problem can be solved exactly using bipolar coordinates (σ, τ). Previously, Liu et al. 5 applied eigenfunction expansions in bipolar coordinates in order to solve the 2D electric Laplace problem for a fixed potential distribution on the outer shell. Here, we apply the same method and expand upon their analysis by additionally considering a fixed distribution of charges on the outer shell. Our main purpose in developing an exact solution in bipolar coordinates is to validate the analytical expressions derived for the electric force [Eqs.

S.2.1 Governing equations
The transformation from Cartesian coordinates (x, y) to bipolar coordinates (σ, τ) can be found in numerous texts; 3 for details, the reader is referred to Appendix S.B of this supplemental document. In bipolar coordinates, the Laplace problem for conducting particles becomes The problem for insulating particles is similar, except that Eq. (S.2.3) is replaced by which is equivalent to Eq. (3.25) from the main text. If the charge density on the outer shell is specified [Eq. (S.2.2a)], then we additionally require that the average potential on the outer shell vanishes: where we have used R = c/sinh τ R to eliminate the shell radius R. The last condition represents a choice of reference potential.
To calculate the force, equation (3.7) may be expressed in bipolar coordinates as Finally, to calculate the energy, equation (S.1.9) may be expressed as Below, we derive a solution for ψ by means of eigenfunction expansions.

S.2.2 Eigenfunction expansions
To solve Eqs. (S.2.1)-(S.2.6) above, we must first represent the external potential ψ ext and its normal derivative ∂ψ ext /∂τ at the outer shell τ = τ R in terms of orthogonal modes: where the coefficients a n , b n , c n , and d n are given by the inversion theorem, Then, we may expand the potential ψ in terms of eigenfunctions of Laplace's equation: where the coefficients ψ 0 , ψ + n , ψ − n , φ + n , and φ − n are to be determined from the boundary conditions. Note that the zero-charge condition (S.2.4) is automatically satisfied by the general solution (S.2.12). Substituting Eq. (S.2.12) into (S.2.6) gives the average potential on the outer shell, Equations (S.2.7)-(S.2.8) for the force F and energy W may also be written as double series in terms of the coefficients ψ + n , ψ − n , φ + n , and φ − n from the eigenfunction expansions. However, it is more expedient to simply apply a numerical quadrature to the integrals in (S.2.7)-(S.2.8).

S.2.3 Solution of the series coefficients
To solve for the series coefficients ψ 0 , ψ + n , ψ − n , φ + n , and φ − n , we first substitute the general solution (S.2.12) into the boundary conditions (S.2.2)-(S.2.5) and resolve the equations into orthogonal modes. This yields a system of linear equations that can be uniquely solved for ψ 0 , ψ + n , ψ − n , φ + n , and φ − n . For conducting particles, we obtain the system of equations, c n e nτ a e −n(τ a −τ R ) − e n(τ a −τ R ) , ψ − Vn = a n e nτ a e −n(τ a −τ R ) + e n(τ a −τ R ) , (S.2.18c) This completes the solution for the electric potential for conducting and insulating particles.

S.3.1 Particle with a pinned contact line
Below, we present supplemental details of the calculation for particles with pinned contact lines (discussed in §4.1 of the main text). For the most part, we focus on symmetrically pinned (i.e., non-undulated) contact lines; results for undulated contact lines are discussed in Appendix E of the main text.

S.3.1.1 Governing equations
Laplace problem: The boundary-value problem for a particle with a symmetrically pinned contact line is given by Eqs.
(4.8)-(4.12) in the main text (reproduced here for convenience): where either ζ U = ζ ext at r = R (4.9a) orn · ∇ζ P =n · ∇ζ ext at r = R, (4.9b) and ζ = U + Ω × · r at r = a, Here, ζ ext is the host interface height given by Eq. (4.1), U is the vertical translation, Ω = · Ω × is the horizontal rotation, P is the vertical force, and N = · N × is the horizontal torque. In addition to the zero-monopole condition (4.11), we also have the constraint (4.12) on the dipole moment. This is what distinguishes the capillary problem from the electric problem.
Interface height: As in the electric problem, we pursue a solution of Eqs. (4.8)-(4.12) by the method of reflections: The zeroth reflection is just the height of the host interface, [cf. Eq. (4.1)], and the ζ (n) , U (n) , and Ω (n) satisfy The even and odd reflections have the same meaning as in the electric problem.
Capillary energy: The capillary energy W is given by Eq. (4.5), Integrating by parts gives where we have used ∇ 2 ζ = ∇ 2 ζ ext = 0 to eliminate the area integrals. Substituting Eqs. (S.3.2) and (S.3.4) into (S.3.10) for ζ and Ω, respectively, and setting ζ ext = ζ (0) yields the double-series expansion, similar to (S.1.10) (it is implied that Ω (n) depends upon ζ (n) and Ω (n) = 0 for even values of n). Using Eqs. (S.3.5) and (S.3.11), we find the "self energy" due to ζ (0) is given by Adding the "trapping energy" −πγa 2 to W (00) gives the total work needed to eliminate a patch of interfacial area. We shall soon see that the boundary perturbation due to rotation of the particle out of the undeformed plane contributes a term to the energy that exactly cancels the last term in Eq. (S.3.12). The leading contribution to the particle rotation is determined by the first reflection, which we consider next.
which is equivalent to Eqs. (4.18) and (4.21) in the main text.

S.3.1.5 Higher reflections: finite-size effects
In §S.1.1.5, we computed the O(a 2 /R 2 ) corrections to the electric force and energy due to higher reflections. The magnitude of these corrections are characteristic of dipolar interactions, which decay like 1/r from the center of the particle. For quadrupolar interactions, the rate of decay goes like 1/r 2 and so the first correction to the capillary force and energy are of O(a 4 /R 4 ). Computing this correction is straightforward for the capillary force, which only requires integrals over the particle boundary. For the capillary energy, which includes integrals over the shell boundary, a large number of terms must be retained in the approximation for the interface height in order to acquire all O(a 4 /R 4 ) integral contributions. This very quickly becomes a tedious endeavor. A more practical approach is the one used to derive Eqs. (4.18) and (4.21) in the main text -that is, apply the wetting condition (4.9) to eliminate the integral over r = R in Eq. (S.3.10). This allows us to compute the energy solely in terms of integrals over the particle boundary. Without going into the details of the derivation and instead focusing on the results, we expand Eqs. (S.3.39) and (S.3.40) up to the O(a 4 /R 4 ) correction. For fixed heights on the outer shell, we obtain and where the terms neglected are of O(a 8 /R 8 ). For fixed slopes, we obtain instead and Note that the force now depends upon the wetting condition applied at the shell boundary through the O(a 4 /R 4 ) finite-size correction. A fixed-height condition repels particles with symmetrically pinned contact lines, whereas a fixed-slope condition attracts them. Although the two solutions agree very well when a/R is small, qualitative differences are observed as a/R → 1 when the outer height distribution is fixed. Whereas the reflections solution (S.3.41) predicts a negative finite-size correction of O(a 4 /R 4 ), the bipolar solution indicates that this correction switches sign (from negative to positive) as the particle fills the shell (see Fig. S.5, right). The same effect is not observed when the outer slope distribution is fixed (Fig. S.6). Further exploration of this finite-size effect is beyond the scope of the present study.
To conclude this section, it is worth discussing finite-size corrections to the external work W ext supplied to the system, which has not been mentioned up to this point. As was indicated near the beginning of this section, the results (S.3.42) and (S.3.44) for the energy W can be derived without having to evaluate integrals over the shell boundary, if one is clever about applying the wetting condition (4.9) before integration. This same trick cannot be applied to the Young-Dupré work of adhesion [Eq. (4.23) in the main text], which is defined solely as an integral over the shell boundary. This implies that, to compute the O(a 4 /R 4 ) correction to Eq. (4.23) (not presented here), one has to expand ζ up to a large number of terms. Fortunately, if there are no force sources in the domain, then the work of adhesion is always = 0 for fixed heights on the outer shell and = 2W P for fixed slopes, regardless of the value of a/R. This means that we need not compute the adhesion energy directly for particles with symmetrically pinned contact lines. On the other hand, consider a particle with an undulated contact line (discussed in Appendix E of the main text) trapped at an interface with a fixed distribution of slopes on the outer shell. In this case, the work of adhesion 2W P because the contact-line undulation acts as a source of force density. Thus, one is again faced with the problem of directly evaluating the adhesion energy through integrals over the shell boundary [cf. Eq. (E.5) in Appendix E]. Exactly the same problem occurs in electrostatics if a permanent electret is suspended in a dielectric medium and a field is applied through a fixed distribution of potentials.

S.3.2 Particle with an equilibrium contact angle
Particles with equilibrium contact angles are discussed in §4.2 in the main text. In keeping with the structure of that section, we divide our attention between cylindrical particles with 90 • contact angles ( §S.3.2.1) and spherical particles with unrestricted

S.3.2.2 Spherical particle
For spherical particles with arbitrary contact angles, the correct boundary condition iŝ Compared to the Neumann condition (4.26) for cylinders, the Robin boundary condition (4.30) for spheres alters the numerical prefactors of the particle reflections. Each particle reflection introduces an extra prefactor of 1 3 for terms multiplied by K ext ξ and a factor of 1 2 for those multiplied by (∇K ext ) ξ . The resulting expression for the interface height around a spherical particle with an arbitrary contact angle is ζ U = 1 2 K ext 0 : r r + 1 6 (∇K ext ) 0 (·) 3 r r r ζ (0) + 1 6 a 4 K ext ξ : r r r 4 + 1 12 a 6 (∇K ext ) ξ (·) 3 r r r r 6 4r r(r · ξ) − 2r 2 r ξ R 6 + · · · − 1 12 a 6 (∇K ext ) ξ (·) 3 r r r R 6 + · · · ζ (2) U + · · · , (S.3.49) for fixed outer heights, or ζ V = 1 2 K ext 0 : r r + 1 6 (∇K ext ) 0 (·) 3 r r r r r r 4 + 1 12 a 6 (∇K ext ) ξ (·) 3 r r r r 6 4r r(r · ξ) − 2r 2 r ξ R 6 + · · · + 1 12 a 6 (∇K ext ) ξ (·) 3 r r r R 6 + · · · ζ (2) P + · · · , (S.3.50) for fixed outer slopes. This pattern continues to higher reflections. Another important difference between cylinders and spheres is that the contact line is, in the latter case, non-circular. Thus, additional terms must be added to the energy to account for the contact-line distortion, as discussed in §4.2.2 in the main text. For the energy calculation, we must account for the distortion of the contact line needed to maintain a fixed contact angle α on a spherical particle. We denote by W 1 the energy given by Eq. (S.3.10), which accounts for interfacial distortions outside the particle. The additional work W 2 done to distort the contact line is , and Ω (n) , yielding a double series akin to (S.3.11) for W 1 . The total energy is then given by the sum W = W 1 + W 2 . The calculation is straightforward, and, for the sake of brevity, will not be delineated in any further detail here. Below, we present the final results for the force and energy. Up to the O(a 4 /R 4 ) correction, the force and energy for spherical particles with equilibrium contact angles are, respectively, and assuming a fixed distribution of heights on the outer shell. For fixed slopes, we obtain instead and

S.4 Capillary problem: bipolar coordinates
In §S.2, we developed an exact solution to the electric problem using bipolar coordinates. We apply the same method, below, for the capillary problem.

S.4.1 Governing equations
For details on the transformation from Cartesian coordinates (x, y) to bipolar coordinates (σ, τ), see Appendix S.B at the end of this supplemental document. In bipolar coordinates, the Laplace problem (4.8)-(4.12) for particles with pinned contact lines becomes and where the particle rotation Ω = · Ω × must be determined as part of the solution. If the particle has an equilibrium contact angle instead of a pinned contact line, then the boundary conditions at τ = τ a must be modified. For cylindrical particles with 90 • contact angles, we have Here, we have used the identity a = c/sinh τ a to eliminate the particle radius a. Finally, if the slope on the outer shell is specified [Eq. (S.4.2b)], then we must add another constraint, so that the reference height is taken to be the average height of the outer shell. Here, we have used R = c/sinh τ R to eliminate the shell radius R.
To calculate the force, equation (4.7) may be expressed in bipolar coordinates as To calculate the energy, equation (S.3.10) may be expressed as where the components Ω x , Ω y of the particle rotation must be determined simultaneously with the interface height ζ. For spherical particles with equilibrium contact angles, an additional term must be added to (S.4.10) that accounts for the work needed to distort the contact line. Following the same procedure as in §S.3.2.2, we denote by W 1 the energy given by Eq.
(S.4.10). The extra energy W 2 needed to deform the contact line is then given by Eq. (S.3.51), which is straightforwardly written in bipolar coordinates:

S.4.2 Eigenfunction expansions
To find a unique solution, we must first represent ζ ext and ∂ζ ext /∂τ on the outer boundary as a series of orthogonal modes: [a n cos (nσ) + b n sin (nσ)] , (S.4.12) where the coefficients a n , b n , c n , and d n are given by where the coefficients ζ 0 , ζ + n , ζ − n , χ + n , and χ − n are to be determined from the boundary conditions. Note that the force-free condition (S.4.4) is automatically satisfied by the general solution (S.4.15). Substituting Eq. (S.4.15) into (S.4.8) gives the average height of the outer shell: Equations (S.4.9)-(S.4.11) may also be written as double series in terms of the coefficients ζ + n , ζ − n , χ + n , and χ − n . For our purposes, we simply approximate the integrals in these expressions by numerical quadratures.
Particle rotation and torque: Since the particle rotation Ω must also be calculated, it behooves us to expand the right-hand side of Eqs. (S.4.3) and (S.4.6) in terms of orthogonal modes: The trick to obtaining these series representations is to substitute cosh τ = 1 2 (e τ + e −τ ) and sinh τ = 1 2 (e τ − e −τ ) and Taylor expand about the singular point e τ = ∞, which is excluded from the domain. Then, projecting the Taylor series onto the cos (nσ) and sin (nσ) modes directly gives the series coefficients in Eqs. (S.4.17)-(S.4.18).
The Cartesian components Ω x and Ω y of the particle rotation must be determined by applying the torque-free condition (S.4.5). After inserting the general solution (S.4.15) into (S.4.5) and exchanging the order of integration and summation, we obtain the following series representation for the lateral torque: Thus, all of the growing modes (with respect to τ) are coupled through the torque-free condition (S.4.5) such that the capillary dipole is annihilated.

S.B Transformation to bipolar coordinates
We choose the Cartesian coordinates (x, y) such that the center of the shell and particle are aligned at y = 0 and the particle is positioned to the left of the shell's center, with the x-axis directed from left to right. In this basis, we define r = xx + yŷ as the position measured from the shell's center and r = x x + y ŷ as the position measured from the particle's center, where the coordinates (x, y) and (x , y ) are related by x − x 0 = x − x 0 , y = y . (S.B.1) The particle's position relative to the shell's center is then given by ξ = r − r = ξx, where The bipolar-symmetry axis is defined at the horizontal position x = x 0 (in shell-centered coordinates) or x = x 0 (in particlecentered coordinates). The transformation from Cartesian coordinates (x, y) to bipolar coordinates (σ, τ) is defined as where 0 ≤ σ ≤ 2π, 0 ≤ τ < ∞, and c is related to a and R by The curves τ = τ a and τ = τ R (where 0 < τ R < τ a < ∞) define the particle and shell boundaries, respectively, and are given by • The isocontours of σ (blue curves in Fig. S.11) form a family of circular arcs centered at x = x 0 , y = c cot σ (the bipolar-symmetry axis) and passing through the limiting point x = x 0 + c, y = 0 (the center of the particle). For values of σ < π, the circular arc lies in the upper half plane; for values of σ > π, the arc lies in the lower half plane. The value σ = 0 (2π) generates a circular arc of infinite radius centered at y = ∞. When σ = π, the circles degenerate to lines on the x-axis connecting the bipolar-symmetry axis to the center of the particle. The value σ = π/2 (3π/2) is a semi-circle of radius c located in the upper (lower) half plane and whose center is located at the origin.
• The isocontours of τ (red curves in Fig. S.11) form a family of non-intersecting circles of radius c/sinh τ whose centers are located at x = x 0 + c coth τ, y = 0. For values of τ > 0, the circle lies to the right of x = x 0 . The value τ = 0 generates a circle of infinite radius centered at x = ∞, y = 0. When τ = ∞, the circles degenerate to points located at x = x 0 + c, y = 0 (the center of the particle).
Several other geometric quantities will be useful to define. These include the Cartesian basis vectorsx andŷ expressed in terms of the bipolar basis vectorsσ andτ: