Chapter 24. QUIZ 6 January 26, Example: Three Point Charges. Example: Electrostatic Potential Energy 1/30/12 1

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1 QUIZ 6 January 26, 2012 An electron moves a distance of 1.5 m through a region where the electric field E is constant and parallel to the displacement. The electron s potential energy increases by J. What is the magnitude of E? a) 1.3 N/C b) N/C c) 2.0 N/C d) N/C e) N/C e = C k = Nm 2 /C 2 Chapter 24 Electrostatic Potential Energy of a system of fixed point charges is equal to the work that must be done by an external agent to assemble the system, bringing each charge in from an infinite distance. Point 1 Point 2 1/30/12 1 ε 0 = C 2 /(Nm 2 ) 1/30/12 2 Example: Electrostatic Potential Energy Example: Three Point Charges Point 1 Point 2 q 2 q 3 Point 3 r 2,3 r 1,3 Point 1 r 1,2 q 2 Point 2 If & q 2 have the same sign, U is positive because positive work by an external agent must be done to push against their mutual repulsion. If & q 2 have opposite signs, U is negative because negative work by an external agent must be done to work against their mutual attraction. 1/30/12 3 1/30/12 4 1

2 Electrostatic Potential Energy We can conclude that the total work required to assemble the three charges is the electrostatic potential energy U of the system of three point charges: kq kq q U = Wtotal = W2 W3 = r1,2 r1,3 q kq3q r 2,3 2 Calculate Electric Field from the Potential Electric field always points in the direction of steepest descent of V (steepest slope) and tis magnitude is the slope. Potential from a Negative Point Charge Potential from a Positve Point Charge V(r ) The electrostatic potential energy of a system of point charges is the work needed to bring the charges from an infinite separation to their final position 1/30/12 5 x y y -V(r ) 1/30/12 6 x Calculating the Electric Field from the Potential Field E = V = V x î V y ĵ V z ˆk Example: Calculating the Electric Field from the Potential Field What is the electric field at any point on the central axis of a uniformly charged disk given the potential? E x = V x, E y = V y, and E z = V z 1/30/12 7 1/30/12 8 2

3 Potential due to a Group of Point Charges Potential from a Continuous Charge Distribution r 1 X r 2 r 3 r 4 q 2 q3 q4 N V(r) = V n (r) = 1 n=1 4πε o N q n n=1 r n 1/30/12 9 1/30/12 10 Charge Densities total charge Q small pieces of charge dq Line of charge: appleapple= charge per unit length [C/m] dq = apple dx Calculate Potential on the central axis of a charged ring Surface of charge: apple apple = charge per unit area [C/m 2 ] dq =apple da Cylinder: dq = σrdθdz Sphere: dq = σr 2 sinθdθdφ Volume of Charge: apple = charge per unit volume [C/m 3 ] dq = appledv Cylinder: dq = ρrdrdθdz Sphere: dq = ρr 2 dr sinθdθdφ 1/30/ /30/

4 Calculate Potential on the central axis of a charged disk Calculate Potential on the central axis of a charged disk (another way) apple apple 1/30/ /30/12 14 Calculate Potential due to an infinite sheet 1/30/12 15 E due to an infinite line charge Corona discharge around a high voltage power line, which roughly indicates the electric field lines. 1/30/

5 Equipotentials Definition: locus of points with the same potential. Equipotentials: Examples General Property: The electric field is always perpendicular to an equipotential surface. Point charge infinite positive charge sheet electric dipole 1/30/ /30/12 18 Equipotential Lines on a Metal Surface Locally E = σ ε 0 Gauss: E = 0 at electrostatic equilibrium in electrostatic equilibrium all of this metal is an equipotential; i.e., it is all at the same voltage 1/30/12 19 Potential inside & outside a conducting sphere V ref = 0 at r =. Common mistake: thinking that potential must be zero inside because electric field is zero inside. 1/30/

6 Summary If you know the functional behavior of the potential V at any point, you can calculate the electric field. The electric potential for a continuous charge distribution can be calculated by breaking the distribution into tiny pieces of dq and then integrating over the whole distribution. Finally no work needs to be done if you move a charge on an equipotential, since it would be moving perpendicular to the electric field. Quiz 7 Jaunary 31, 2012 The graph gives the electric potential V as a function of x. ank the five regions according to the magnitude of the x-component of the electric field within them, greatest first. (A) 4 > 2 > 3 > 1 > 5 V (B) 3 > 1 > 5 > 2 = 4 (C) 1 = 2 > 4 = 3 > 5 (D) 2 > 4 > 1 = 3 = 5 The charge concentrates on a conductor on surfaces with smallest radius of curvature x 1/30/ /30/12 22 Another Method: Electrostatic Potential Energy Leave the charges in place and add all the electrostatic energies of each charge and divide it by 2. where: q i -- is the charge at point i V i -- is the potential at location i due to all of the other charges. 1/30/ /30/

7 Another Method: Electrostatic Potential Energy q 3 Point 3 r 2,3 r 1,3 Summary: Electrostatic Potential Energy for a Collection of Point 1. Bringing charge by charge from infinity Point 1 U = 1 2 q kq 2 1 kq 3 r 1,2 r 1,3 q k 2 kq 3 r 1,2 r 2,3 q k 3 kq 2 r 1,3 r 2,3 q 2 Point 2 U = kq 2 2 kq 3 2 kq 2q 3 = kq 2 kq 3 kq 2q 3 r 1,2 r 1,3 r 2,3 r 1,2 r 1,3 1/30/12 25 r 1,2 n 2. Or using the equation: U 1 = q V i i 2 and trying i= 1 not to forget the factor of ½ in front. Note: The second way, could involve more calculations for point charges. However, it becomes very handy whenever you try to calculate the electrostatic potential energy of a continuous distribution of charge, the sum becomes an integral. r 2,3 1/30/12 26 V due to an infinite line charge emember Gauss s Law line charge with charge density E da = EndA = s EndA E dabarrel s E A L Q ε inside s o = E da Endcaps Barrel λl = E (2π L) = ε 1/30/12 27 o Q inside = λl 0 E Gaussian cylinder λ = or 2π ε o 2 kλ E = V due to an infinite line charge (continued) Where we define V = 0 at = ref dv = E dl = E ˆ dl = E d V V E d P d = 2kλ ref P = 2kλ ln ref Problem : you can' t choose ref = 0 or ref = because ln0 = and ln =. 0 < ref < and ref V = 2kλln 1/30/12 28 P ref = P ref 7

8 Work and Equipotential Lines Which of the following statements is true about the work done by the electric field in moving a positve charge along the paths? (A) III = IV > 0, I = II = 0 (B) I = II = 0, III > IV (C) III = IV < 0, I = II = 0 electric field V 1 = 100 V lines V 2 = 80 V V 3 = 60 V V 4 = 40 V 1/30/