Lecture Notes on Electric Field Near Two Charged Metal Sheet | PHY 481, Study notes of Physics

Download Lecture Notes on Electric Field Near Two Charged Metal Sheet | PHY 481 and more Study notes Physics in PDF only on Docsity! PHY481 - Lecture 11 Sections 4.2, 4.3 of PS A. Electric fields near two charged metal sheets The two sheets have the same area L2 and have total charge Q on the top sheet and q on the bottom sheet. The two metal sheets have thickness a and are separated by a distance d << L so we can treat the sheets as infinite. The sheets have normals oriented along the z-axis. Let the charge on the top of the top sheet be Q2u and the charge on the bottom of the top sheet be Q2l. The charge at the top of the bottom sheet is Q1u and the charge at the bottom of the lower sheet is Q1l. We need to find these charges, the electric field between the sheets and the voltage difference between the two sheets. The sheets are isolated so charge conservation requires that, Q1u + Q1l = q; and Q2u + Q2l = Q (1) In addition we require that the electric field inside the upper metal sheet and inside the lower metal sheet are zero. Using superposition to calculate these electric fields, we find (note that Gauss’s law does not help us here), Q2l 2ǫ0A − Q2u 2ǫ0A + Q1l 2ǫ0A + Q1u 2ǫ0A = 0 (2) and − Q2u 2ǫ0A − Q2l 2ǫ0A − Q1u 2ǫ0A + Q1l 2ǫ0A = 0 (3) Combining the second of equations (1) with Eq. (2), we find Q2u = (Q + q)/2. Similarly we find that Q2l = (Q−q)/2 and Q1u = −(Q−q)/2 and Q1l = (Q+ q)/2. The outer surfaces of the two sheets have the same charge and inner surfaces have charge of the same magnitude by opposite sign. This is consistent with the results found for the case Q = Q, q = −Q where there is no charge on the outer surfaces. Notice that if the charges of the two plates are the same, then all of the charge is on the outer surfaces of the metal sheets. The electric field is found by superposition or by using Gauss’s law. Using Gauss’s law is easier, but it is good to check it using superposition. We find, Ebetween = σ ǫ0 = −(Q − q) 2Aǫ0 ; Eupper = −Elower = (Q + q) 2Aǫ0 (4) 1 Finally the voltage between the plates is Ed = V , with the positively charged sheet at the higher potential. B. Definition of capacitance Capacitance measures the ability of a system to store charge. It is assumed that if the applied voltage is zero, no (net) charge is stored, and that when a voltage is applied charge starts to be stored. The basic geometry is set up by attaching one piece of metal to the positive lead of a battery, while another piece of metal is attached to the other electrode of a battery. The fundamental relation is then, Q = CV (5) where V is the voltage of the battery and Q is the charge stored on the pieces of metal, +Q on the piece of metal attached to the positive electrode of the battery and −Q on the piece of metal attached to the negative electrode. From the calculation above for two metal sheets we have, Q = σA; E = σ ǫ0 ; and V = Ed = Qd Aǫ (6) Which can be written as, Q = ǫ0A d V = CV (7) so that the capacitance of a parallel plate capacitor is Cplate = ǫ0A/d as you know. However the definition of capacitance of Eq. (11) is true for any linear dielectric, which applies to most materials in the low voltage limit. Example 1. A coaxial cable Consider two coaxial cylindrical metal shells of radii a and b, where b > a and d = b− a. A voltage V is applied to the outer cylinder and the inner cylinder is grounded so the total charge on the two shells is Q on the outer and −Q on the inner shell. The electric field for r > b and for r < a is zero by Gauss’s theorem. The electric field for a < r < b is found from Gauss’s law, E2πrL = −Q/ǫ0 so that E(r) = − Q 2πrLǫ0 (8) Now we find the potential difference by integration, Vb = − ∫ b a ~Ed~l = ∫ b a Q 2πrLǫ0 dr = Q 2πLǫ0 log(b/a) (9) 2