Lecture 5 R 2R2R

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

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/2  R charge per unit length. The electric field at the origin is R x y (a) zero (b) (c)

Question 1 1.a 2.b 3.c

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/2  R charge per unit length. The electric field at the origin is R x y (a) zero (b) (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!!

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

Charge Densities How do we represent the charge “ Q ” on an extended object? total charge Q small pieces of charge dq Line of charge :  = charge per unit length dq = dx Surface of charge:   = charge per unit area dq =  dA Volume of Charge :   = charge per unit volume dq =  dV

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

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  : from  to  r E(r) = ? Any symmetries ? This may help for easy cancellations  x

Infinite Line of Charge  We need to add up the E -field components dE at the point r given by contributions from all segments dx along the line.   and x are not independent, choose to work in terms of   Write dE in terms of , r, and  Integrate  from -  /2 to +  /2 Charge density = x y dx r' r  dE

Infinite Line of Charge We use Coulomb’s Law to find dE : What is r’ in terms of r ? What is dq in terms of dx ? Therefore, x y dx r' r  dE What is dx in terms of  ?

Infinite Line of Charge Components: Integrate: ExEx x y dx r' r  dE  EyEy

Infinite Line of Charge Now The final result: ExEx x y dx r' r  dE  EyEy

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

Question 2 Examine the electric field lines produced by the charges in this figure. Which statement is true? (a) q 1 and q 2 have the same sign (b) q 1 and q 2 have the opposite signs and q 1 > q 2 (c) q 1 and q 2 have the opposite signs and q 1 < q 2 q1q1 q2q2

Question 2 1.a 2.b 3.c

Question 2 Field lines start from q 2 and terminate on q 1. This means q 2 is positive; q 1 is negative; so, … not (a) Now, which one is bigger ? Note that more field lines emerge from q 2 than end on q 1 This indicates that q 2 is greater than q 1 Examine the electric field lines produced by the charges in this figure. Which statement is true? (a) q 1 and q 2 have the same sign (b) q 1 and q 2 have the opposite signs and q 1 > q 2 (c) q 1 and q 2 have the opposite signs and q 1 < q 2 q1q1 q2q2

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

Electric Field Lines Electric Field Patterns Dipole ~ 1/ R 3 Point Charge ~ 1/ R 2 ~ 1/ R Infinite Line of Charge Distance dependence

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

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.

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 L 2

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:

The Story Thus Far We want to be able to calculate the electric fields from various charge arrangements. Two ways: 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. 1.Brute Force: Add up / integrate contribution from each charge. Often this is pretty difficult. Ack! Ex: electron cloud around nucleus

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