PHYS 012 1st Edition Lecture 7 Outline of Last Lecture I. Work done by a conservative force is equal to the negative change in potential energy (injoules).a. Wc = -ΔEPEb.II. Electric Potential (V)a. V = EPE/q0 = kq/rb. Scalar quantityc. Different way to describe an electric fieldd. ΔVAB = ΔEPEAB/q0 = -Edq0/q0 = -Ede. ΔVBC = 0 ; VB = VCf. ΔVAC = -Ed = ΔVABg. Equipotential line: line on which every point has the same electric potential (ex. line BC on diagram above); perpendicular to electric field linesh. Equipotential surface: surface on which every point has the same electric potentialThese 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.i. A positive charge that moves with the electric field will have a decreasing electricpotential energy. A negative charge that moves with the electric field will have anincreasing electric potential energy.j. Electric potential decreases further from the source charge.Outline of Current Lecture III. 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 zeroIV. 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 capacitord. 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.- Requires more energy with each successive charge removed- Energy = ½ QΔV = ½ CΔV2- For parallel plate capacitor, energy = ½ [(Aε0)/d] [Ed]2 = ½ Adε0E2- Energy density = energy/volume = energy/Ad = ½ ε0E2V. 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/κVI. 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
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