Download Electrostatics: Force on Dielectric Slabs and Uniformly Polarized Spheres - Prof. Phillip and more Study notes Physics in PDF only on Docsity! PHY481 - Lecture 21 Chapter 6 of PS A. Force pulling a dielectric slab between capacitor plates To find the force on the slab, we have to solve the problem when the slab partially penetrates the slab. It is easiest to see the effect in the case of an isolated (fixed charge) systems. Consider an l × w slab where the plates are penetrated by an area x × w of the slab, while an area (l − x) × w remains filled with air. The voltage on the plates are at equipotentials (they are metal), so the electric field in both regions is the same. However the capacitance in the two regions is different. The total capacitance can be found by adding the charge on the two parts of the plate, Exw + 0E(l − x)w = Q (1) Using V = Ed, we then find, C = Q V = xw d + 0(l − x)w d . (2) Note that we could have used the law of addition of capacitors to get this result. Now we can find the force on the slab by using F U0 = −dU(x) dx = − d dx Q2d 2(xw + 0(l − x)w) = Q2dw(− 0) 2(xw + 0(l − x)w)2 (3) In the case of a fixed voltage system, we have to take into account the work done by the battery or generator. dU = 1 2 dCV 2 = (dQ)V − Fdx = dCV 2 − Fdx (4) so that, F = 1 2 dC dx V 2 = V 2w − 0 2d (5) B. Uniformly polarized sphere (ρf = 0, no applied field) If we have a sphere where there is a uniform density of polarization, ~P = P0k̂, then at the surface of a sphere, the bound charge is σb = r̂ · P0k̂ = P0cosθ. (6) 1 We can construct an electrostatic potential corresponding to this charge distribution, by trying the l = 1 spherical solution, in which case, Vint(r, θ) = C1rcosθ; Vext(r, θ) = C2 r2 cosθ (7) The potential must be continuous across at r = a, which implies that C1 = C2/a 3. Using Er = −∂V/∂r, the radial component of the electric field obeys, Eextr (a, θ) − Eintr (a, θ) = σb 0 = 2C2 a3 cosθ + C1cosθ = 2C2 a3 cosθ + C2 a3 cosθ = P0cosθ 0 (8) From this we find, C2 = P0a 3/30, C1 = P0/30 so that, Vint = P0r 30 cosθ = P0z 30 ; Vext = P0a 3 30r2 cosθ (9) The electric field inside the sphere is given by, ~Eint = − ∂Vint ∂z k̂ = −P0k̂ 30 (10) The electric field is thus uniform and is in the opposite direction to the polarization. The potential outside the sphere is like that of a dipole, and the dipole moment is found by equating the general expression for a dipole with the potential Vext, 1 4π0 ~p · r̂ r2 = P0a 3 30r2 cosθ (11) which shows that the magnitude of the dipole moment of the uniformly polarized sphere is given by, p = 4π 3 a3P0. As expected, the magnitude of the dipole moment of the sphere is the volume of the sphere times the uniform polarization of the sphere. C. Relation between atomic dipoles and polarization At the beginning of our discussion we noted that there is often a linear relation between the polarization of an atom or molecule and the electric field that is applied to it, ~p = α~E. Now we want to connect this behavior with the behavior of the polarization density used in Maxwell’s equations ~P = 0χe ~E. Since ~p is for one atom or molecule while ~P is the polarization per unit volume, a naive assumption might be to write ~P = n~p where n is the number of atoms or molecules per unit volume in the dielectric material. This assumption is valid if the electric field is not altered by the dielectric material, which from the discussion above is rarely true. Nevertheless it is 2
