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# Block03Notes.pdf PHYS 242

DB
NC A&T
GPA 3.8

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These notes cover topics ranging from Charges stored, dielectric constants, capacitance, to polarization, bound charges, permittivity, parallel-plate capacitors, and more.
COURSE
General Physics II
PROF.
Prof. Sandin
TYPE
Class Notes
PAGES
2
WORDS
CONCEPTS
General Physics, Physics, General Education, Engineering
KARMA
25 ?

## Popular in Physics 2

This 2 page Class Notes was uploaded by DB on Tuesday September 27, 2016. The Class Notes belongs to PHYS 242 at North Carolina A&T State University taught by Prof. Sandin in Fall 2016. Since its upload, it has received 2 views. For similar materials see General Physics II in Physics 2 at North Carolina A&T State University.

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Date Created: 09/27/16
PHYS 242 BLOCK 3 NOTES Sections 24.1, 24.3 to 24.6 Suppose we have two conductors of any shape that are insulated from one another. Conductor a has a positive excess charge Q (in C) on it. Conductor b has a negative excess charge –Q on it. This Q is called the charge stored. We consider only cases in which the potential differeace V b V =ab (in V) is directly proportional toQ. Then the ratio ofQ tis a constant and the equation C ≡  defines the capacitance C ab V ab coulomb C (in F = farad = volt = V ). (Don’t confuse the capacitance C with the charge unit C = coulomb.) The capacitanceC is never negative. Also, the quQ andV abe never negative in the above defining equation. A capacitor is a circuit element that mainly provides capacitance.Example 24.3 could be improved by ⌠ 1 q showing Vab= V a V b ⌡E cos φ dl with E = 4πε 0r2 and φ = 0 gives theab used. Example 24.4 could be a λ similarly improved by showingE2πε r andφ = 0 gives theabused (as we did forb and a in class in Block 2). 0 For varying values, let’s use υ abrand q for Q. Then C = Q solves to υ =q . In moving an Vab C q infinitesimal chargedq from plate to plate to charge a capacitor, we do an adW = υ dq =rC dq. This U 1 Q 1 Q 2 work is stored as electric potential energyU. Integrating,W = dW = U – 0q dq = ( – 0) . That is, C C 2 0 0 Q2 1 Q 1 U = 2C . SubstitutingQ = CV gives U 2  CV . Finally, substitutinVC gives U =2 QV , whereV is short for the potential differanceb – Vab V (in V). ThisU is the electric potential energy (in J),which is stored in the capacitor’s electric field. See if you can work Example 24.7. The word dielectric is a synonym for insulator. Thedielectric strength E is the maximum electricfield m magnitude the dielectric can withstand without breaking down and conducting. Suppose we charge a capacitor (with vacuum betweeslates and capacitanc0C ) so that it has a charge stored Q, a potential differe0ce V , and—at some point between the plates—an electric field0magnitudeE . If we thencompletelyfill the volume between the plates of the capaa dielectricand Q stays constant,we would V0 E0 Q find that the potential difference and electric fieeecreaseto V =K and E = K . Since C =V , the capacitanceincreasesto K0 . The dielectric constantK has no unit and is greater than or equal to one (K ≥ 1). E 0 Why does E = K ? Step 1. Thefree chargeson theconductingcapacitor plates give an applied electric field.This field exerts a torque on the electric dipoles in the dielectric, tending to line them up, and giving induced bound charges on the insulatingdielectric surfaces. (See Fig. 24.19.) (The dielectric’s resuldipole moment per volumeis called its polarization.) Step 2. The induced bound charges give an opposing electric field, thus decreasing the resultant electric field magnitude fromE0to E0/K. (See Fig. 24.20.) 2 We’ve been usingε = 8.854 × 1012 C or F , calling it theelectricconstant., but it has another name: 0 N·m2 m ε is the permittivity of a dielectric an0 thenε is the permittivity of vacuum. The relation betw0en ε and ε is ε = Kε . Therefore,in vacuum, ε ≡ ε by definition,so K = 1 there. In air, ε is slightly greater than ε and K is 0 0 0 C 2 F slightly greater than 1 (for dry air at 1 atm pressure and 20˚C,ε = 8.8592 or m and K = 1.00059). N·m An integral form of Gauss’s lawfor vacuum∫ E ·d A = Q encl, which gave us electricfield relations ε0 for high symmetry cases. The comparable integral form of Gauss’s law when we have one or more dielectrics → → Q encl-free → → present is∫K E ·d A  =  ε0 or ∫ ε E ·d A = Q encl-freehereQencl-free the total free(not bound) charge enclosed by the Gaussian surface. Thus, if the dielectric(s) keep sufficient symmetry, we can take our previous results for E and V in vacuum and simply replace ε with either ε or Kε . The table below ab 0 0 gives some examples for the three symmetries: Symmetry Spherical Cylindrical Flat Vacuum 1 |q| λ b σ E =4πε 0r2 Vab 2πε0 la E =ε0 1 |q| λ b σ E = 4πε 2 Vab 2πε lna E = ε Dielectric r σ 1 |q| or V = λ ln or E = or E =4πKε0 r2 ab 2πKε 0 a Kε0 A parallel-plate capacitor has two parallel conducting plates. Each plate has an area A. The plates are separated by a distanced. We assume the dimension(s) of the plates that give us the area are much larger than d. Then E is constant between the plates (except for the small “fringing” region near the edges, which we can accurately ignore). (See Fig. 24.2). Integrating alongan electricfield line from positiveplate a to negativeplate b, b d we findVab V –aV =b⌡E cos φ dl = E cos 0 dl = E(1)d. That is, ab = Ed . The quantitiesab , E, and d are a   0 never negative in this equation. Between the parallel conductsabis the potential difference (in V),E is the V N magnitude of theconstantelectric fielm (orC ), andd is the distance (in m). σ Q Q/A From the table above,Eε=(whereσ = A . We substitute E = ε into ab= Ed and that result foabV into C = Q to obtain C = εA . Remembering that ε = Kε , we find C  = ε   = Kε for parallel-plate Vab d 0 d d capacitors only. Cover up the solutionsand carefully work Examples 24.1 (in vacuum)and 24.2, as well as most of Example 24.10 (don’t worry abouti , σ, aidσ ). Now we find an expressionfor the electricfield’s energy density u, which is its electricpotentialenergy U per volume. For an idealized parallel-plate capacitor (no electric-field fringu) volumes , whereU = 1 A A 2 CV , C = Kε0d = εd , andV = Ed. Also the volume betweenthe plates (the location of the electricfield) equals Ad. When we make all these substitutions, we find that the parallel-plate capacitor’s A and d cancel out, and we J 1 1 have a more general expression for the energy densimy3 ): u =2 K0 E  =2 εE . Work vacuum Examples 24.8 and 24.9, as well as Example 24.11 (replg withvol in its EVALUATE part).

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