Electric Fields 3

1. R 2R Lecture 5 Electric Fields 3

2. Lecture 5 • Yesterday we introduced electric field lines • Today we will cover some extra topics on Electric Fields before going on to Electric Flux and Gauss’ Law next week: • Continuous Charge Distributions • Infinite line of charge • Motion of a charge in an electric field

3. y + + + + + + + + + R + + + + x + + + + + + + + + Question 1 • Consider a circular ring with total charge +Q. The charge is spread uniformly around the ring, as shown, so there is λ = Q/2pR charge per unit length. • The electric field at the origin is (a) zero (b) (c)

4. 150 • of • 300 0 Question 1 • a • b • c

5. y + + + + + + + + + R + + + + x + + + + + + + + + Question 1 • Consider a circular ring with total charge +Q. The charge is spread uniformly around the ring, as shown, so there is λ = Q/2pR charge per unit length. • The electric field at the origin is (b) (a) zero (c) • The key thing to remember here is that the total field at the origin is given by the VECTOR SUM of the contributions from all bits of charge. • If the total field were given by the ALGEBRAIC SUM, then (b) would be correct. (exercise for the student). • Note that the electric field at the origin produced by one bit of charge is exactly cancelled by that produced by the bit of charge diametrically opposite!! • Therefore, the VECTOR SUM of all these contributions is ZERO!!

6. E(r) = ? r ++++++++++++++++++++++++++ Electric FieldsfromContinuous Charge Distributions • Examples: • line of charge • charged plates • electron cloud in atoms, … • Principles (Coulomb’s Law + Law of Superposition) remain the same. • Only change:

7. small piecesof chargedq total chargeQ • Line of charge: dq = ldx l=charge per unit length • Surface of charge: dq = sdA s=charge per unit area • Volume of Charge: • r=charge per unit volume dq = rdV Charge Densities • How do we represent the charge “Q” on an extended object?

8. How We Calculate (Uniform) Charge Densities: • Examples: • 10 coulombsdistributed over a 2-meter rod. Take total charge, divide by “size” 14 pC(pico = 10-12) distributed over the surface of a sphere of radius 1 μm. 14 pCdistributed over the volume of a sphere of radius 1 mm.

9. E(r) = ? r ++++++++++++++++++++++++++ q +++++++++++++++++++++++++++++ x Electric field from an infinite line charge Approach: “Add up the electric field contribution from each bit of charge, using superposition of the results to get the final field.” In practice: • Use Coulomb’s Law to find the E-field per segment of charge • Plan to integrate along the line… • x:from-¥to+¥OR q : from -p/2to+p/2 Any symmetries ? This may help for easy cancellations

10. dE y q r' r ++++++++++++++++ x dx Infinite Line of Charge Charge density = l • We need to add up the E-field componentsdEat the pointr given by contributions from all segmentsdx along the line. •  and x are not independent, choose to work in terms of  • Write dEin terms of , r,and  • Integrate from -/2 to +/2

11. dE y q r' r ++++++++++++++++ x dx Weuse Coulomb’s Law to find dE: Infinite Line of Charge What isdq in terms of dx? What isr’in terms of r ? Therefore, What is dxin terms of  ?

12. y dE q Ey q r' r Ex ++++++++++++++++ x dx Infinite Line of Charge • Components: • Integrate:

13. y dE q Ey q r' r Ex ++++++++++++++++ x dx Infinite Line of Charge • Now • The final result:

14. dE y q r' r ++++++++++++++++ x dx Infinite Line of Charge Conclusion: • The Electric Field produced by an infinite line of charge is: - everywhere perpendicular to the line (Ex=0) - is proportional to the charge density () - decreases as - Gauss’ Law makes this trivial!!

15. Examine the electric field • lines produced by the charges • in this figure. • Which statement is true? q1 q2 (a)q1 and q2 have the same sign (b)q1 and q2 have the opposite signs and q1> q2 (c)q1 and q2 have the opposite signs and q1 < q2 Question 2

16. 172 of 300 0 Question 2 • a • b • c

17. Examine the electric field • lines produced by the charges • in this figure. • Which statement is true? q1 q2 Field lines start fromq2and terminate on q1. This meansq2 is positive; q1 is negative; so, … not (a) (a)q1 and q2 have the same sign (b)q1 and q2 have the opposite signs and q1> q2 (c)q1 and q2 have the opposite signs and q1 < q2 Now, which one is bigger? Note that more field lines emerge from q2 than end on q1 This indicates that q2is greater than q1 Question 2

18. a)+q(blue) c a b)-q(red) c)midpoint (yellow) b Electric Dipole: Lines of Force Consider imaginary spheres centered on : • All lines leavea) • All lines enterb) • Equal amounts of leaving and entering lines for c)

19. Dipole ~ 1/R3 Point Charge ~ 1/R2 Infinite Line of Charge ~ 1/R Electric Field LinesElectric Field Patterns Distance dependence

20. Motion of a Charge in a Field • An electron passes between two charged plates (cathode ray tube in your television set) • While the electron is between the plates, it experiences an acceleration in the y-direction due to the electric field

21. Motion of a Charge in a Field The only force is in the y direction and the acceleration is: The initial velocity is in the x-direction, so the velocity as a function of time (t) is: The time (T) taken by the charge to traverse the plates is determined only by the initial velocity in the x direction.

22. Motion of a Charge in a Field • Therefore, the particle’s deflection in the y-direction is: • It exits the field making an angle θ with its original direction, where: • Exercise for the student; calculate where it hits the screen after a distance L2

23. The Story Thus Far Two types of electric charge: opposite charges attract, like charges repel Coulomb’s Law: Electric Fields • Charges respond to electric fields: • Charges produce electric fields:

24. The Story Thus Far • 1. Brute Force: Add up / integrate contribution from each charge. • Often this is pretty difficult. We want to be able to calculate the electric fields from various charge arrangements. Two ways: Ack! Ex: electron cloud around nucleus • 2. Gauss’ Law: The net electric flux through any closed surface is proportional to the charge enclosed by that surface. • In cases of symmetry, this will be MUCH EASIER than the brute force method.

25. Finished Coulomb’s Law • Next lecture: Electric field Flux and Gauss’ Law • Read Chapter 23 • Try Chapter 22, Problems 28,45,70