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Ch – 30 Potential and Field Learning Objectives – Ch 30 • To establish the relationship between and V. • To learn more about the properties of a conductor in electrostatic equilibrium. • To introduce batteries as a practical source of potential difference. • To find the connection between current and potential difference for a conductor. • To find the connection between charge and potential difference for a capacitor. • To analyze simple capacitor circuits. Finding potential (V) from field (E) -ΔU = ΔV = ΔU/q, E =F/q , therefore If you want to find a value for Vf , instead of ∆V, you must specify a position, si , where Vi =0. This position is often at infinity. Finding E field from potential (V) Es = - dV/ds Given this graph of V, make a graph of Ex Assume the equation of the parabola is of the form y = Kx2 where K is a constant Answer E = -dV/dx 1500 1000 500 0 0 E (V/m) for parabola K = 50,000 V/m2 E0-2cm =(-)100,000x E2-4cm = 0 E4-6cm = (-)-20V/.02m =1000V/m 1 2 3 4 -500 5 6 7 E (V/m) -1000 -1500 -2000 -2500 x (cm) Given this graph of Ex , make a graph of V (V = 0 at x = 0) Answer ∆V = area under the curve (V0 = 0 at x=0) ∆ V1-3 = (-)-200V x .02m = +4 V ∆ V3-4 = ∆ V1-3 = +4V 5 4 4 3 3 V V ∆ V4-6 = (-) 200V x .02m = -4 V 2 .. 2 1 1 0 0 1 2 3 4 x (cm) 5 6 7 A point charge of + 5/9nC is located at the origin • Determine the values of x at which the potential is 100, 200, 300, 400, 500V. • Graph V vs x along an x axis with the charge at the origin. • Describe E, the electric field on both sides of the point charge (e.g.positive, negative, constant, increasing, decreasing). Graph of V vs x 600 500 400 V V x(cm) 100 ± 5.0 200 ± 2.5 300 ± 1.67 400 ± 1.25 500 ± 1.0 300 200 100 0 -6 -4 -2 0 2 4 6 x (cm) For values of x > 0, E is positive and decreasing with increasing value of x For values of x < 0, E is negative and decreasing with increasing values of |x| Geometry of Potential and Field • The direction of the electric field is perpendicular to the equipotential surfaces • E always points in the direction of decreasing potential • Field strength (magnitude) of E is inversely proportional to the spacing ∆s between equipotential surfaces Which set of equipotential surfaces is valid for the electric field shown? Answer • Answer is c • Field strength (magnitude) of E is inversely proportional to the spacing ∆s between equipotential surfaces Kirchoff’s Loop Law • The sum of all potential differences encountered while moving around closed path is zero • This is a result of the conservation of energy for a conservative force Conductor in electrostatic equilibrium • For a sphere of charge Q, outside the conductor (r >R), E = kQ/r2 with the maximum field strength at the surface of the sphere E = kQ/R2 • Inside the sphere, E=0. Conductor in electrostatic equilibrium • To find ∆V between 2 points outside the sphere, integrate E along a line between the two points. • At the surface of the sphere, there is a nonzero potential. Conductor in electrostatic equilibrium • Inside, since E = 0, ∆V =0, which means the potential inside is constant • When a conductor is in electrostatic equilibrium, the entire conductor has the same potential, not necessarily the same charge Conductor in electrostatic equilibrium • At the surface of the conductor • Same charge and same potential here . • Same potential, but different charge and different electric field here Connecting Potential and Current • I = AJ = A σ E • In the battery shown: Ewire = |∆Vwire | L We can derive: I = ∆Vwire R Where R = L / A σ R = ρL / A R is a property of a specific wire, depending on the material, length and area. Current Ranking Task • Rank the currents I1–I5 at the five labeled points in this figure, from greatest to least. Explain. Fat and Skinny • Which current is greater? Both wires are made of the same material. • Explain. 2 conductive rods • Two conductive rods have been connected to a 6 V battery for “a long time.” What are the values of: • ∆V12 _______ • ∆ V34 _______ • ∆ V23_______ Equipotential Map • Compare the field strengths Ea and Eb. Are they equal, or is one larger than the other? Explain. • Compare the field strengths Ec and Ed. Are they equal, or is one larger than the other? Explain. • Draw the electric field vectors at points a through e. Answers • Electric field strength is the gradient of the potential • Ea > Eb • Ec >Ed E field vectors and potential • Is the potential at point 1 greater than, less than or equal to the potential at point 2? Explain. • Determine a value of ∆V12 (1 is initial position, 2 is final position). • Draw a series of equipotential surfaces spaced every 5 V. Answer • E points in direction of decreasing potential so V1 > V2 • |∆V12 |= 200 V/m x 10 cm = 20 V and since it is decreasing: ∆V12 = -20V • Arbitrary assumption that V2 = 0 for equipotentials 0V 10V 20V What is the Value for E? • Assume V is a linear function as in a capacitor. • Draw an arrow on the figure showing direction. Answer • E = 1000 V/m Answer