Lecture Notes on Planar Capacitor Containing a Dielectric | PHY 481, Study notes of Physics

Download Lecture Notes on Planar Capacitor Containing a Dielectric | PHY 481 and more Study notes Physics in PDF only on Docsity! PHY481 - Lecture 20 Chapter 6 of PS A. Planar capacitor containing a dielectric (Examples 3,4,5 of PS). Consider a parallel plate capacitor of area A and plate separation d that has dielectric material with dielectric permittivity  between the places. (i) Find the electric field between the plates when a charge Q is placed on the capacitor. Find the voltage across the capacitor and the energy stored in the capacitor. (ii) Now consider connecting a battery to the capacitor, with voltage V on the top plate and the lower plate at ground. Find the charge on the capacitor plates and the energy stored in the capacitor. In both cases compare your results to the case where there is no dielectric between the plates. That is, in the two cases of fixed charge and fixed voltage, does the stored energy go up or down when the dielectric is added? Also, if the dielectric slab is free to move, is it drawn into the space between the capacitor plates or is it pushed out. Find the force in both cases (i) and (ii). (This is a typical electro-mechanical actuator, used in doorbells amongst other things). Solution to (i) When a charge Q is placed on the capacitor, we can use Gauss’s law (or superposition), to find the diplacement field, ~D = −σk̂; so that ~E = −σ  k̂ (1) The voltage across the plates is given by V = |E|d = σd  . The capacitance is found from Q = CV , and using, V = σd  = Qd A so that C = A d (2) and the energy stored in the capacitor is U = Q2/2C. If there is no dielectric between the plates, the capacitance is C0 = 0A/d, and hence, U U0 = ( Q2 2C )/( Q2 2C0 ) = C0 C = 1/κ for isolated capacitor (3) so that the energy is reduced when the dielectric is between the plates. Note that from this calculation we deduce that the energy density by writing, U = Q2 2C = σ2A2d 2A = uV with u = 1 2 0κE 2 (4) where u is the energy density, V = Ad is volume and we used |E| = σ/. Now note that, u = 1 2 0κE 2 = 1 2 ~D · ~E. (5) 1 The result u = 1 2 ~D · ~E is a general result for linear isotropic dielectrics. Solution to (ii) Now consider the case where the voltage across the plates is fixed to V . In that case U/U0 = 1 2 CV 2 1 2 C0V 2 = C/C0 = κ for fixed voltage across capacitor (6) The case of a fixed voltage applies to energy storage, as capacitors are usually charged at fixed voltage, e.g. from a generator or solar cell. High energy storage materials require high values of κ (highly polarizable material) and high V (very good insulator with low leakage). B. An electric field applied to a uniform dielectric sphere We consider an uncharged uniform dielectric sphere of radius a and dielectric constant , in a constant applied field ~E0 = E0k̂, so that V = −E0z = rcosθ. (7) Since the dielectric sphere is uniform and there is no free charge, we have, ~∇ · ~D = 0 = ~∇ · ( ~E) = ~∇ · ~E = 0. uniform  (8) Using ~E = −~∇V we then find that V still obeys Laplace’s equation, so we try the solutions, Vint = −C1rcosθ, Vext = −E0rcosθ + C2a 3cosθ r2 (9) We impose continuity of V , and the condition 0E ext n (a, θ) = E int n (a, θ), to find, −E0 + C2 = −C1; −E0 − 2C2 = −  0 C1 (10) which lead to, C1 = E0 3 κ+ 2 ; C2 = E0 κ− 1 κ+ 2 (11) where κ = /0. Using the fact that Vint = −C1z, we find that the electric field inside the sphere is uniform ~Eint = − ∂Vint ∂z = E0 3 κ+ 2 k̂ (12) and that the dipole of the sphere is ~psphere = C2a 3 k = 4π0E0a 3κ− 1 κ+ 2 (13) 2