 previous lecture we were discussing about the electrical double layer potential distribution around a spherical particle. So, to summarize our findings we solved the Poisson Boltzmann equation with Debye-Hakl linearization and for that we made a substitution that like a new variable was introduced as equal to r into psi where psi is the ideal potential distribution and that gave rise to the general solution of the following form which is given in the form of the last equation of this slide. The total surface charge on the particle we evaluated as negative of the total charge in the ideal and that was evaluated by using the ideal potential distribution and then we got an expression for the total surface charge on the particle as a function of the zeta potential which is given in the last equation of the slide. Now we considered a case 1, so there are several limiting cases. So, let us come to the board to explain the cases. So, you have a spherical particle of radius r. The question is when would you consider the particle as a point mass or the particle of a finite extent whose geometrical features need to be considered. So, it depends on what is r by lambda, lambda is the Debye length and r is the radius of the particle. Now if r by lambda is small that means what that means that with respect to the characteristic ideal thickness the particle is like a point mass and that is the case that we have considered in the previous slide or in the in the previous lecture. Now we will quickly go to the slide and see that what was the consequence. So, as a consequence we wrote an expression for the stokes drag on the particle as a function of the surface charge and using the expression for surface charge we got an expression for the electrophoretic velocity and because r is much much less than lambda we got u equal to 2 by 3 epsilon zeta e by mu. Now it is something like that the it is 2 by 3 times the Helmholtz-Moluchowski velocity. Now typically in the Helmholtz-Moluchowski velocity you have minus epsilon zeta e by mu that minus sign is missing here and there is a reason behind this it depends on the reference frame that you are looking into that if you are sitting on a particle then of course the velocity of flow is in the opposite direction that is if the particle is stationary and the fluid is flowing past it the situation is opposite to the situation when the particle is stationary and you are observing a flow to pass around the particle and there is opposite situation when the particle is moving you are fixed on the particle and with respect to the moving particle you are observing the flow. So, it depends on from which reference frame you are looking into the flow. So, this equation is known as Huckel equation and we have discussed up to this in the previous lecture. Now we will continue with this today and we will consider the other case when lambda much much less than r that is called as thin debilure approximation when lambda much much less than r then the curvature of the particle cannot be neglected the because the radius of the particle is significantly large as compared to the characteristic electrical double layer thickness. So, you cannot neglect the curvature of the particle. So, if you cannot neglect the curvature of the particle then you can because of typically large radius of curvature you can consider each region of the particle as a localized flat plate as a localized flat plate this concept is very important. So, it is not that it is like a flat plate, but because the curvature is cannot be neglected that is the radius of curvature is large then you can consider the particle as a localized flat plate with the local parallel and perpendicular velocity components u parallel and u perpendicular. u parallel u perpendicular of course is not there because of no penetration this is 0 but u parallel is there and u parallel depends on what u parallel is a function of e parallel what is e parallel e parallel is the electric field applied electric field parallel to the direction that direction of the small elemental surface that you have considered on the sphere. So, u parallel is a function of e parallel and in fact you can write u parallel as epsilon zeta e parallel by mu because it is like a flat plate. So, you can use the Helmholtz-Moluchowski velocity plus or minus sign depends on from which reference frame you are looking into the flow. So, it could be plus sign or it could be minus sign it is a that is what I am saying it is as it as it is like a flat plate, but the question is what is this e parallel you have an electric field e now that you have an applied electric field e now the potential due to the applied electric field e is distributed over the sphere over the surface of the sphere such that at each location you have a different e parallel and of course you have e perpendicular now we will see that the e perpendicular would not be there on the surface of the sphere just like the no penetration boundary condition, but e parallel will be there and we should have a governing equation in the r theta coordinate system to solve for the potential distribution remember this is not the ideal potential distribution this is the potential distribution due to the applied electric field. So, let us go to the corresponding governing equation. So, as we can see that the corresponding governing equation for the ideal potential distribution we have already seen, but this is the potential distribution for the field due to the applied potential that field is given by phi. So, what is the corresponding governing equation? The corresponding governing equation is del square phi equal to 0 that is the Laplace equation remember as I have told several times, but I can again repeat that the induced electrical field due to the electrical double layer satisfies the Poisson equation, but the applied the field due to the applied potential satisfies the Laplace equation. So, here we are interested to solve for the applied field because if you recall the derivation of the electro osmotic velocity eventually it was the result was that minus epsilon zeta e external by mu when we consider the electrical body force and the osmotic and the body force due to the osmotic pressure. So, I am just trying to recapitulate because the fundamentals are important. So, the result of the electrical body force that is the body force due to the Maxwell's trace and the body force due to the osmotic pressure gradient these two together came out to be charge density times the external electric field and when we plucked that with the Navier-Stokes equation and solved it the velocity that came out was minus epsilon zeta e external by mu. So, this e parallel is the electrical field parallel to the element of the surface because of the applied potential and that because the applied potential satisfies the Laplace equation. So, here we have to appeal to the Laplace equation and not the Poisson equation in general. So, this is the form of the equation of course we will try to outline the solution of this equation in the board, but I am just giving you first the form of the equation with the boundary conditions. So, this is nothing but the Laplace equation in r theta coordinate system boundary conditions at small r equal to capital R there is no penetration boundary condition that is basically del phi del n equal to 0. So, here because of the spherical nature n will be r. So, it is del phi del r equal to 0. So, this is just like the no penetration boundary condition and at first stream phi equal to phi infinity. So, these are the two boundary conditions for this equation. Now, let us try to outline a solution of this equation. I will try to outline a solution to the extent possible in the board and then we will refer to the slides wherever we need to cut short. This is the equation that we want to solve. This equation is solvable by method of separation of variables because of the nature of the equation and the boundary condition. So, let us say. So, let me work it out a little bit because you may not be familiar with the solution of the Laplace equation in r theta system using the separation of variables. So, let me work it out a little bit to the extent possible. So, del phi del r is g dash into h r square del phi del r del del r of r square del phi del r. Now, the next equation or the next term. So, sin theta del phi del theta del phi del theta is h dash. So, you have h d dr of r square g dash plus g by sin theta d d theta of h dash. So, you have h d dr of r square g dash plus g by sin theta d d theta of h dash sin theta equal to 0. So, this becomes 1 by g d dr of r square g dash plus g by sin theta 1 by h sin theta d d theta of h dash sin theta equal to 0. So, as you know from your previous experience that why we write it in this form because eventually this is function of r only this is function of r only this is function of theta only. So, function of r only is equal to function of theta only that means each is equal to a constant. So, we can write that this is equal to minus of this is equal to k say. So, we will solve this equation. First we will solve 1 by g d dr of r square d g dr is equal to k. So, d dr of r square d g dr is equal to k g. A typical standard form of this equation is recovered in the last by replacing k with n into n plus 1. Just introduce a new variable n a new parameter rather not a variable new parameter n such that k is equal to n into n plus 1. Then what we do is we let g is equal to r to the power alpha that is a standard form of solution of this equation and you can see that it is possible because I mean the influence of r will be cancelled from both sides if we take it in this form. For instance what is d g dr is equal to alpha r to the power alpha minus 1 r square d g dr is equal to alpha r to the power alpha plus 1 d dr of r square d g dr is equal to alpha into alpha plus 1 into r to the power alpha. So, now we can write alpha into alpha plus 1 into r to the power alpha is equal to n into r to the power alpha is equal to n into n plus 1 g is equal to r to the power alpha. So, from this equation so you can see r to the power alpha effect is cancelled from both sides that is why this particular form is a particular solution of the equation. So, you can write alpha square plus alpha minus n square minus n is equal to 0. So, you can write alpha square plus n plus 1 minus n into alpha minus n into n plus 1 is equal to 0. So, alpha into alpha minus n plus 1 minus n into this there any problem alpha square plus n plus 1 minus n alpha alpha into alpha minus n plus 1 minus n plus 1 minus n plus 1 minus 1 is there any problem with plus minus which one this is alpha plus yes right minus n into alpha please tell whenever there is a plus minus mistake otherwise it waste time ok. So, then alpha is equal to minus of n plus 1 and n then we have the other equation minus 1 by h sin theta d d theta of sin theta d h d theta is equal to k k is n into n plus 1. So, d d theta of sin theta or 1 by sin theta d d theta of sin theta d h d theta plus n into n plus 1 h is equal to 0. Now, we make a substitution let us say cos theta equal to w. So, d h d theta is equal to d h d w into d w d theta. What is d w d theta that is minus sin theta? So, minus sin theta. So, sin theta d h d theta is equal to minus sin square theta d h d w. So, minus sin square theta means minus of 1 minus cos square theta cos square theta is w square d d theta of sin theta d h d theta is equal to d d w of this into d w d theta d w d theta is minus sin theta and 1 by sin theta of sin theta of sin theta d h d theta is equal to d d w of 1 minus w square d h d w that 2 minus sin gets cancelled. So, you have d d w of 1 minus w square d h d w plus n into n plus 1 h is equal to 0 this is your equation. So, we are interested to solve this equation d d w of 1 minus w square d h d w plus n into n plus 1 h equal to 0. This is nothing but the Legendres differential equation that is the solution of this equation can be expressed in terms of the Legendres polynomial. So, Legendres polynomial we commonly write as p n where different values of n will give different Legendres polynomials like p 0 x equal to 1 p 1 x equal to x like in this way you have the Legendres polynomials. So, let us write the general solution. So, recall that your phi is equal to 0. So, your solution was gr into h theta and this gr was r to the power alpha with alpha is equal to minus of n plus 1 and n. So, because it is a linear differential equation the resultant solution is a linear superposition of the solution for the 2 values of alpha. So, you can write phi is equal to summation of a n r to the power n p n cos theta plus summation of b n r to the power minus n plus 1 into p n cos theta because w is cos theta. So, we have written p n cos theta. Now, let us evaluate the first part let us evaluate the first part. Now, we will apply the boundary conditions. So, at r tends to infinity what is phi? Phi infinity is what minus e infinity into r cos theta because r cos theta is x e x is equal to minus del phi del x. So, phi is equal to minus e into x x is r cos theta. So, this is at r tends to infinity first stream where e infinity is the uniform electric field in the x direction that is applied. So, we are finding out that because of the applied uniform electric field along the x direction in the first stream what is the potential distribution on the surface of the sphere or what is the field on the surface of the sphere that is what we are finding out. So, if you use this then you can say that the other terms for r to the power n dependence should go away should not be there when r tends to infinity because that will blow up the solution. So, you can say safely that phi is equal to minus e infinity r cos theta plus summation of b n r to the power minus n plus 1 p n cos theta. Now, we can expand this. So, we can write this as minus e infinity r cos theta plus b 0 by r p 0 cos theta plus b 1 by r p 0 cos theta. Now, we will use the second boundary condition. This was the boundary condition for the potential at infinity. The second boundary condition is the no penetration that is on the surface of the sphere you have del phi del r equal to 0. So, what is del phi del r minus e infinity cos theta minus b 0 by r square. What is p 0 cos theta that is equal to 1 because p 0 x is equal to 1 then minus 2 b 1 by r cube b 1 cos theta is cos theta. This at small r is equal to capital R is equal to 0. So, you can find out the coefficients. So, from here it is straight forward. So, I will use the slides to give you the final compact form. So, let us go to the slides. So, the solution is a n r to the power n p n cos theta and b n by r to the power n plus 1 p n cos theta and summation of that linear superposition of that. So, the first boundary condition will indicate that the first term that is a n r to the power n p n cos theta will be nothing but minus e infinity r cos theta and for the second term we will evaluate equating like powers of cos theta. So, if you look into this expression let us come to the board for one final moment. So, you can you can look into this equation this equation del phi del r at small r is equal to capital R this must be equal to 0. So, b 0 is equal to 0 and if you equate the like powers of cos theta then you will get b 1 in terms of e infinity. There is no other term you can say that if the series has infinite terms but there is no other term which you can match with this cos theta. So, all other coefficients will be 0. So, only these two terms will match b 0 will be 0 and all other coefficients all other b n's will be 0. So, eventually let us come to the slide. So, you will get b 0 is equal to 0 and b 1 is equal to minus half e infinity r cube and b n is equal to 0 for n greater than equal to 2. So, this is your final solution phi is equal to minus e infinity cos theta into small r plus capital R cube by 2 r square these two terms together and e parallel is minus 1 by r del phi del theta at small r is equal to capital R. This comes out to be minus 3 by 2 e infinity sin theta just you differentiate this. You differentiate this and set small r is equal to capital R differentiate this with respect to theta. So, the cos theta becomes sin theta minus sin theta basically. So, and e perpendicular is 0 and that is what that is obvious because no penetration boundary condition was there to constrain this. So, the solution must satisfy that. So, you get a e parallel and e perpendicular. So, using e parallel now your local velocity is epsilon zeta e parallel by mu. So, if you know e parallel you can now work with the u parallel. So, to summarize if you have a thin electrical double layer as compared to the radius of the particle your electrical field if you apply a e infinity in the first stream then the electrical potential distribution due to that e infinity will be a function of theta and so this is minus 3 by 2 infinity sin theta. So, at different locations theta you will get different e parallel based on that you will get different u parallel and those that means the electrophoretic velocity will also be theta dependent and that comes from the largeness of the particle as compared to the characteristic electrical double layer thickness okay. Now we have discussed about several primary electro kinetic effects but I will discuss about some other effects without getting into much of the details. So, I mean I mean these effects are also important and but I mean within the scope of your curriculum we do not have the details of this but I will just give you a glimpse with the physical idea so that if you have more interest you can read a large volume of fundamental text and research materials on these topics. One is dielectrophoresis. So, dielectrophoresis relies on the difference in polarizability between the particle and a solution and non-uniformity in the electrical field. These are the two essential ingredients of the dielectrophoretic transport. So, one is difference in polarizability between the particle and the solution. So, there is a particle in a solution, a particle is of different polarizability than the solution and the electric field is non-uniform. If that is there then this gives rise to a net force acting on the particle. This is just like reminiscent of the force due to a dipole formation. In positive dielectrophoresis particles are there are two types of dielectrophoresis. One is positive dielectrophoresis another is negative dielectrophoresis. In positive dielectrophoresis particles are more polarizable than the solution and tend to move towards high field regions. So, you can have so why dielectrophoresis is important because you can have dielectrophoresis of like charged particles or I mean you can have dielectrophoresis of DNA and so it is possible to transport charged species or the charged particle in a solution by using dielectrophoretic principles. So, in positive dielectrophoresis particles are more polarizable than the solution and tend to move towards high field regions. And in negative dielectrophoresis particles are less polarizable than the solution and migrate towards low electrical field regions. So, this is the difference between positive and negative dielectrophoresis. Now the question is because the particle is moving relative to a solution there must be a force acting on it. So, the expression for the dielectrophoretic force it is time average is given by this formula. This is a very well established formula and there is a fundamental proof of this I mean we will not get into the details of this because this is beyond the scope of our curriculum. But I mean you can see that it depends on the permittivity of the medium and here however you can see that there is till day at the top of epsilon this is complex permittivity this is not normal permittivity. So, the complex permittivity is like if you have I mean let me write an expression it will contain a combination of permittivity and conductivity electrical conductivity. So, epsilon till day for example can be epsilon plus j sigma by omega where j square root of minus 1 and omega is the frequency of the electric field being applied. So, in that spirit if you look into the formula that we were discussing. So, you will get terms with the complex permittivity of the particle complex permittivity of the medium and the radius of the medium and you can see that the force essentially depends on the square of the gradient square of the electric field AC electro osmosis. Now in AC electro osmosis unlike the standard electro osmosis or DC electro osmosis you create a bias by using a alternating electric field. So, the idea is so you can see that there is a surface on the surface you create an induced charge by an electric field by an alternating electric field. So, one of the basic or a couple of the basic technological challenges which motivated researchers to use AC electro osmosis is that like in DC electro osmosis you have you face significant overheating problems due to joule heating and not only that there is a technological problem which is associated with bubble formation due to local electrolysis at the electrodes. So, many times because of that bubble formation due to local electrolysis it becomes very difficult to operate such systems. Now, electrical potential within the electrodes causes charges to accumulate on the electrode surface you can see the schematic which is given in the slide. So, electrical potential within the electrodes causes charges to accumulate on the electrode surface which changes charge density near the surface and forms the ideal. So, electrical double layer always keep one thing in mind when we discussed about electrical double layer we discussed about a spontaneous charging but it is not necessary that you have a spontaneously formed electrical double layer. You can give an external bias to form a charged layer and then if you have electro kinetic transport that is called as induced charge electro kinetics that is you induce an external charge instead of having a spontaneous charge on the surface. So, you have a formation of a charge layer near the surface this process is called as electrode polarization. In an alternating electric field sign of charges in the electrical double layer and the applied field change. So, if you see it is a very interesting thing recall what is the expression for the body force the charge density times the electrical field if there is no magnetic field. So, if you have an alternate polarity then the charge density changes its sign it flips its sign the electric field also flips its sign. So, the product the sign of the product remains the same. So, despite the alternating nature of the field you still have the same direction in which the body force is acting which is a very interesting facet of the AC electro osmosis. Now, here so how the next question is how does the AC electro osmosis velocity depend on the frequency, frequency of the alternating electric field. So, typical graph of course these graphs are for various parameters, but typical graph if you look into any of these these are some experimental graphs taken from literature you will see that at low frequency the velocity is low at high frequency also the velocity is low and at some intermediate frequency the velocity is maximum. So, if you want to design a system for high velocity you should go for the intermediate frequency. Now, the question is why at low frequency what happens see when you have large frequency what it essentially means is that you create a quick change to the system to which the system cannot relax so easily. So, at low frequency space charges can quickly respond to external electric field and most of the electrical potential drops within the electrical double layer that means you have low electro osmotic velocity at high frequency on the other hand the net charge itself in the ideal is very small because the net charge in the ideal is small then you have basically the velocity is very small. So, the velocity is small in the cases of small low frequency and high frequency for 2 different reasons, but the velocity is small for these 2 cases. So, at intermediate frequency you have the maximum velocity. The next concept that we will learn is electro thermal flow. Now, electro thermal flow the basic physics is like this as I have discussed that there is a joule heating in electro kinetic flows because of joule heating there is a change in temperature right because of joule effect there is heating and because of heating there is a change in temperature. Because of change in temperature there is a change in electrical properties and because of this gradient in electrical properties there can be a body force. So, you can so if you recall our derivation of the Maxwell stress and all these you will not get this term why because we assume the constant epsilon not we use the Maxwell's equation at free space. So, we use the constant epsilon not, but if you have a gradient of epsilon you have an extra term in the body force in addition to the charge density times electric field this is called as electrostriction term. So, this additional term is due to the for example, gradient in permittivity. So, permittivity being a function of temperature if the temperature changes you can see that there is so this I mean from our exposure to heat transfer I call it as something which is very much similar to natural convection that you have a temperature gradient and the temperature gradient can give rise to certain body forces. Those body forces are because of gradients in certain properties as a function of temperature. So, it is the temperature gradient induces local changes in conductivity, permittivity, viscosity and density. So, not only the permittivity you can have change you can have body forces due to the gradients in several other properties. So, for example, conductivity gradients produce free volume charge and coulomb forces where as permittivity gradient produces dielectric forces. So, you can see that heating effect in electro kinetics can give rise to flows because it can give rise to body forces very much analogous to natural convection in heat transfer. So, to summarize in continuation of the discussions in on electrical double layer and electro osmosis we discussed the other facets of electro kinetics. We have discussed the physical aspects behind the generation of flow induced streaming potential that we discussed in the previous lecture and we obtained an analytical expression for the streaming potential and finally in the previous lecture and in today's lecture we have derived the expressions for motion of a charge particle due to the application of an external electric field that is electrophoresis. For two limiting cases one is the electrical double layer thickness is much smaller than the radius of the particle and other in the electrical double layer thickness is much greater than the radius of the particle. So, we stop here for this lecture in the next lecture we will start with a different topic. Thank you very much.