Physics 012 1st Edition Lecture 8Outline of Last Lecture I. Review: a. Two positive charges i. V will always be positive (sum of two positive numbers)ii. E will equal zero at point half way between the two points (the two cancelout)b. A positive and a negative chargei. V will equal zero at point half way between the two points (the two cancel out)ii. E will never be zeroII. Capacitors: two conductors near each other (not touching) carrying equal magnitude charge and opposite signsa. Capacitance (C): how much charge conductors can hold for a given potential differencei. C = Q/ΔV in farads (F) = 1 C/Vii. Depends on geometry of capacitorb. Parallel plate capacitor: consists of two parallel plates of opposite but equal (in magnitude) charge that, if they are close enough together, can be viewed mathematically as being infinitei. When connected to a battery (neg. plate to neg. side of battery and pos. plate to pos. side of battery), the voltage of the battery will equal the potential difference between the plates (0V at pos. plate and max potential at neg. plate)ii. C = (Aε0)/d when plates have an area of A and a distance of d between themiii. E = σ/ε0 with σ being the charge density (Q/A) of the platesc. Cylindrical capacitorThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.d. Spherical Capacitore. When starting with two parallel neutral plates, it takes work to remove a charge from one of the plates as, once the charge is removed, you are pulling the chargeaway from a plate with a net charge opposite of that which you are removing, thus requiring you to fight against the attractive force of the two charges.1. Requires more energy with each successive charge removed2. Energy = ½ QΔV = ½ CΔV23. For parallel plate capacitor, energy = ½ [(Aε0)/d] [Ed]2 = ½ Adε0E24. Energy density = energy/volume = energy/Ad = ½ ε0E2III. Dielectrics: insulators that can be placed between conductors of a capacitor to increase capacitancea. If dielectric completely fills space between conductors, it will increase capacitance by a factor of κ (dielectric constant, depends on dielectric material)i. C = κC0ii. E = E0/κIV. Problem: A parallel plate capacitor is charged and then disconnected from a battery. Then, the two plates are pulled together from d to 2d. What happens to the capacitance,potential difference, electric field, and energy of the system?a. C0 = (Aε0)/d ; Cf = (Aε0)/2d ; Cf = ½ C0b. Qf = Q0 ; Cf ΔVf = C0 ΔV0 ; ΔVf = (C0 ΔV0)/ Cf = 2 ΔV0c. E = Q/( Aε0) ; Ef = E0 (Q, A, and ε0 never changed)d. Energy0 = Q02/2C0 ; Energyf = Qf2/2Cf ; Energyf = 2 Energy0 (increase in energy comes from energy of pulling plates apart)Outline of Current Lecture V. Electric current: net flow of charge per unit time through some regiona. By convention, current flows in the direction that positive charge would move. Electrons (the charge carriers for metals) will flow in the opposite direction of electric current.b. Measured in ampheres, amps (A)i. 1 A = 1 C/sVI. Ohm’s Law: relationship between potential difference and currenta. Drift speed (vd): net speed an electron movesi. Path of motion is not actually linear due to collisions with other particles, which causes a certain amount of resistance as electrons move through a material.b. ΔV = IRi. R = resistance = (ρL)/A for a conductive wire1. ρ = resistivity (larger with more resistance, like in non-metals)2. L = length of conductive wire3. A = cross-sectional area of wire (πr2)c. Only applicable to ohmic materialsd. R = ΔV/I in ohms (Ω)i. 1 Ω = 1 V/Ae. ρ in Ω*mVII. Electrical power: rate of energy consumed by resistora. P = IΔV ( = I2R = V2/R )i. V = IR ; I = V/Rb. Electric circuits: i. Basic circuit- wire in picture (connecting battery and resistor) is an ideal conducting wire (no resistance)ii. In series- In general, Req. = R1 + R2 + … +
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