r 4 r 2 q 2 r 3 V () r En dln (r) 1 q 1 Negative sign from integration of E field cancels the negative sign from the original equation.

Transcription

1 Question from last class Potential due to a Group of Point Charges rr V () r E dl r r 1 q 1 X rr N V () r E dl r n n N V(r) V n (r) 1 n1 4 o 1/30/ q 2 N r r N r r q n V () r En dln dr 2 n n r n r 4 orn N n r N q n qn 4 orn n 4 r n o n N q n n1 r n r 2 r 3 r 4 q3 q4 Negative sign from integration of E field cancels the negative sign from the original equation.

2 Chapter 23 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 q 1 1/30/2018 3

3 Example: Electrostatic Potential Energy Point 1 q 1 Point 2 q 2 If q 1 & 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 1 & q 2 have opposite signs, U is negative because negative work by an external agent must be done to work against their mutual attraction. 1/30/2018 4

4 Example: Three Point Charges q 3 Point 3 r 2,3 r 1,3 Point 1 q 1 r 1,2 q 2 Point 2 1/30/2018 5

5 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: U W total W 12 W 13 W 23 kq 2 q 1 r 1,2 kq 3 q 1 r 1,3 kq 3 q 2 r 2,3 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 1/30/2018 6

6 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. 1/30/2018 7

7 Another Method: Electrostatic Potential Energy r 1,3 q 3 Point 3 r 2,3 Point 1 q 1 r 1,2 q 2 Point 2 U 1 2 q 1 kq 2 r 1,2 kq 3 r 1,3 q 2 kq 1 r 1,2 kq 3 r 2,3 q 3 kq 1 r 1,3 kq 2 r 2,3 U kq q kq q kq q 2 3 kq q 1 2 kq q 3 1 kq q 2 3 r 1,2 r 1,3 r 2,3 r 1,2 r 1,3 r 2,3 1/30/2018 8

8 Summary: Electrostatic Potential Energy for a Collection of Point 1. Bringing charge by charge from infinity n U Or using the equation: 2 and trying i not to forget the factor of ½ in front. q i V i 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. 1/30/2018 9

9 Calculate Electric Field from the Potential Electric field always points in the direction of steepest descent of V (steepest slope) and its magnitude is the slope. Potential from a Negative Point Charge Potential from a Positve Point Charge V(r ) x y -V(r ) y x 1/30/

10 Calculating the Electric Field from the Potential Field If we can get the potential by integrating the electric field: b Va b E dl We should be able to get the electric field by differentiating the potential. E V In Cartesian coordinates: V E x dx V E y dy V E z dz a In the direction of steepest descent 1/30/

11 Calculating the Electric Field from the Potential Field V V V E V iˆ ˆj kˆ x y z V V V E, E, and E x y z x y z 1/30/

12 Potential from a Continuous Charge Distribution 1/30/

13 Charge Densities total charge Q small pieces of charge dq Line of charge: = charge per unit length [C/m] dq = dx Surface of charge: = charge per unit area [C/m 2 ] dq = da Cylinder: dq rddz Sphere: Volume of Charge: = charge per unit volume [C/m 3 ] dq = dv Cylinder: dq rdrddz Sphere: dq r 2 sindd dq r 2 drsindd 1/30/

14 Calculate Potential on the central axis of a charged ring 1/30/

15 Calculate Potential on the central axis of a charged disk dq A 2a da V k dq r 1/30/

16 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? Given: V 2 0 ( z 2 2 z) E x E y E z E V 1/30/

17 Calculate Potential on the central axis of a charged disk (another way) V E dl From Lecture 3: E z 2 o 1 z z 2 2 1/30/

18 Calculate Potential due to an infinite sheet dv E dl dv 2 k iˆ ( dx iˆ dx ˆj dx kˆ ) V 2 k dx 2 k x C 1/30/

19 E due to an infinite line charge Corona discharge around a high voltage power line, which roughly indicates the electric field lines. August 26, 2014: Four Kentucky Firefighters Electrocuted When Participating in ALS Ice Bucket Challenge 1/30/

20 V due to an infinite line charge emember Gauss s Law Gaussian cylinder s E line charge with charge density s n E E da da E A Barrel s E da n da barrel Q E E (2 L) L inside o da L 1/30/ o Endcaps Q inside L 0 E E 2 2k o or

21 V due to an infinite line charge (continued) dv E dl E ˆ dl E d 2k 2k ln Problem : you can' t choose 0 or because ln0 and ln. 0 and 1/30/ ref Where we define V = 0 at = ref V P V V ref ref ref P ref 2kln P ref d P E ref ref d

22 Equipotentials Definition: locus of points with the same potential. General Property: The electric field is always perpendicular to an equipotential surface. 1/30/

23 Equipotentials: Examples Point charge V(r) k q r infinite positive charge sheet V(x) 2kx V o electric dipole 1/30/

24 Equipotential Lines on a Metal Surface Locally E 0 E 0 Gauss: at electrostatic equilibrium in electrostatic equilibrium all of this metal is an equipotential; i.e., it is all at the same voltage 1/30/

25 Potential inside & outside a conducting sphere V ref 0 at r. The electric field is zero inside a conductor. The electric potential is constant inside a conductor. 1/30/

26 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. The charge concentrates on a conductor on surfaces with smallest radius of curvature. 1/30/