Phys 202 1nd Edition Lecture 6 Outline of Last Lecture I. Electric potential and capacitorsOutline of Current Lecture II. Topic ReviewIII. Isolated CapacitorIV. DielectricV. True or FalseCurrent LectureTopic review:Electric potential is the work done(U) to bring a charge from r to ∞ through an electric field. For a point charge, electric potential can be represented as V=kq/r volts. The electrical energy between two point charges(U) can be represented as U=QV where U is the change in energy from +1 coulomb to q which is rdistance away.From the definition of electric potential, we can use the equation ΔV=ΔER to derive the equation; E=-ΔV/ΔR (in units of Volts/meter or Newtons/Coulomb). The negative sign in the derived equation simply signifies that like charges will repel.Capacitor: A capacitor is a system of conductors with two parallel plates. The total charge stored in a capacitor is Q=CV. V is the potential and can be described equationally as V=Ed with a unit of volts. C is the capacitance and can be described with the equation C=(A/d)E0 with a unit of Farads.Isolated Capacitor: Within a capacitor, the area of each plate is A, and the separation between plates is d. One plate is positive, one is negative. When it is isolated, the charge Q is always conserved. There is no source or syncof charge. Whatever charge you put on the capacitor will always remain the same.Dielectric: A dielectric is an insulator between the plates of a capacitor, with the ability to become slightly polarized. When calculating capacitance(C), C=AE/d. The E in this equation is the permittivity of the dielectric. Kappa(K) is denoted as the dielectric constant. K is always greater than or equal to 1. If the space between the plates of a capacitor is empty, it is considered a vacuum and K is equal to 1. For all other Dielectric materials, K is greater than 1.These 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.If a dielectric is inserted between the plates of a capacitor, the potential difference between the plates will decrease (from V0 to V). Within an isolated capacitor, the charge Q is constant. So, if we consider the equation Q=CV, than in this case, Q0=Q, so C0V0=CV. If V0 decreases, then C0 has to increase to keep the equation in balance. So if K defines the ratios of capacitance, the K=C0/C. So then; C0V0=CV V=V0C0/C V=V0/KTrue or False:1. In an isolated capacitor, the distance(d) between the plates becomes d/2. This implies that Q=Q/2. For an isolated system, Q will always stay the same. Therefore, the answer is false.2. In an isolated capacitor, the distance(d) between the plates becomes d/2. This implies that the voltage will increase.V0C0=VCV=EdIf d is halved, then the value of V decreases. Therefore the statement is false.3. In an isolated capacitor, the distance(d) between the plates becomes d/2. This implies that U willincrease.U=QVV=EdOnce again, if d is halved, V decreases. If this case, if the value of V decreases, then the value of U also decreases. Therefore, the statement is
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