1. Ch – 30 Potential and Field

2. 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.

3. Finding potential (V) from field (E) -ΔU = ΔV = ΔU/q, E =F/q, therefore If you want to find a value for V f, instead of ∆V, you must specify a position, s i, where V i =0. This position is often at infinity.

4. Finding E field from potential (V) E s = - dV/ds

5. Given this graph of V, make a graph of E x Assume the equation of the parabola is of the form y = Kx 2 where K is a constant

6. Answer E = -dV/dx for parabola K = 50,000 V/m 2 E 0-2cm =(-)100,000x E 2-4cm = 0 E 4-6cm = (-)-20V/.02m =1000V/m

7. Given this graph of Ex, make a graph of V (V = 0 at x = 0)

8. Answer ∆V = area under the curve (V 0 = 0 at x=0) ∆ V 1-3 = (-)-200V x.02m = +4 V ∆ V 3-4 = ∆ V 1-3 = +4V ∆ V 4-6 = (-) 200V x.02m = -4 V

9. 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).

10. Graph of V vs x Vx(cm) 100± 5.0 200 ± 2.5 300 ± 1.67 400 ± 1.25 500 ± 1.0 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|

11. 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

12. Which set of equipotential surfaces is valid for the electric field shown?

13. Answer Answer is c Field strength (magnitude) of E is inversely proportional to the spacing ∆s between equipotential surfaces

14. 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

15. Conductor in electrostatic equilibrium For a sphere of charge Q, outside the conductor (r >R), E = kQ/r 2 with the maximum field strength at the surface of the sphere E = kQ/R 2 Inside the sphere, E=0.

16. 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.

17. 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

18. 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 

19. Connecting Potential and Current I = AJ = A σ E In the battery shown: E wire = |∆V wire | L We can derive: I = ∆V wire R Where R = L / A σ R = ρL / A R is a property of a specific wire, depending on the material, length and area.

20. Current Ranking Task Rank the currents I1–I5 at the five labeled points in this figure, from greatest to least. Explain.

21. Fat and Skinny Which current is greater? Both wires are made of the same material. Explain.

22. 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_______

23. 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.

24. Answers Electric field strength is the gradient of the potential E a > E b E c >E d

25. 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 ∆V 12 (1 is initial position, 2 is final position). Draw a series of equipotential surfaces spaced every 5 V.

26. Answer E points in direction of decreasing potential so V 1 > V 2 |∆V 12 |= 200 V/m x 10 cm = 20 V and since it is decreasing: ∆V 12 = -20V Arbitrary assumption that V 2 = 0 for equipotentials 0V 10V 20V

27. What is the Value for E? Assume V is a linear function as in a capacitor. Draw an arrow on the figure showing direction.

28. Answer E = 1000 V/m

29. Answer