1. Lecture 3-1 Electric Field Define electric field, which is independent of the test charge, q, and depends only on position in space: dipole One is > 0, the other < 0 -qq electric dipole of dipole moment: - +

2. Lecture 3-2 Dipole in uniform electric fields No net force. The electrostatic forces on the constituent point charges are of the same magnitude but along opposite directions. So, there is no net force on the dipole and thus its center of mass should not accelerate. Net torque! There is clearly a net torque acting on the dipole with respect to its center of mass, since the forces are not aligned. http://qbx6.ltu.edu/s_schneider/physlets/main/dipole_torque.shtml

3. Lecture 3-3 Electric Field from Coulomb’s Law + + + + + - - - + - - Bunch of Charges (volume charge) (surface charge) (line charge) Continuous Charge Distribution dq P Summation over discrete charges Integral over continuous charge distribution P k http://www.falstad.com/vector3de/

4. Lecture 3-4 Reading Quiz 1 Which one of the following statements is incorrect ? A) Electric fields leave positive charges and end on negative charges B) Electric field lines can intersect at some points in space. C) Electric field field lines from a dipole fall off faster than 1/r 2. D) Electric fields describe a conservative force field.

5. Lecture 3-5 SUMMARY: FIND THE ELECTRIC FIELD GIVEN THE CHARGES 1) GEOMETRY FOR q i or dq i 2) DISCRETE CHARGES q i 3) CONTINUOUS CHARGES dq i line charge density λ (x) surface charge density σ (x.y) volume charge density ρ (x,y,z) Geometry may suggest other coordinate systems, R,θ,Φ or R,θ,Z

6. Lecture 3-6 Continuous Charge Distribution 1 Charged Line At a point P on axis: E x = k λ / (1/r 1 –1/r 2 ) = E x kλ/ (1/x 1 - 1/x 2 ) Eqn 22-2a E x = k λ {1/( X P + L/2 ) - 1/( X P - L/2)} E x = k ( Q/L) L [ X P 2 – (L/2) 2 ] -1 For X P 2 >> (L/2) 2 E x = k Q / X P 2 For X P = 0 E x = 0

7. Lecture 3-7 Again: Continuous Charge Distribution 1: Charged Line At a point P on perpendicular axis: θ x

8. Lecture 3-8 Physics 241 –Warm-up quiz 2 The rod is uniformly charged with a positive charge density σ. What is the direction of the electric field at a point P on a line perpendicular to the rod? Note that the line and the rod are in the same plane. a) to the right b) to the left c) up d) down e) lower right p

9. Lecture 3-9 Continuous Charge Distribution 2: Charged Ring Use symmetry! At point P on axis of ring: ds θ E x = k Q x ( x 2 + a 2 ) -3

10. Lecture 3-10 Continuous Charge Distribution 3: Charged Disk <= Independent of x Superposition of rings! At a point P on axis: Use the ring with radius a E X value dE x = k dq x ( x 2 + a 2 ) -3 Integrate rings from 0 to R E x = -2πσkx ( 1/( x 2 + R 2 ) 1/2 – 1/x ) k = 1/4πε o E = σ/2ε o

11. Lecture 3-11 Continuous Charge Distribution 4: Charged Sheets E=const in each regionSuperposition! Capacitor geometry

12. Lecture 3-12 MULTIPLE CHARGE SHEET EXAMPLE DOCCAM 2 (SKETCH)

13. Lecture 3-13 Gauss’s Law: Qualitative Statement  Form any closed surface around charges  Count the number of electric field lines coming through the surface, those outward as positive and inward as negative.  Then the net number of lines is proportional to the net charges enclosed in the surface.

14. Lecture 3-14 Electric flux # of field lines N = density of field lines x “area” where “area” = A 2 x cos θ General definition of electric flux: (must specify sense, i.e., which way) To state Gauss’s Law in a quantitative form, we first need to define Electric Flux. Sum over surface

15. Lecture 3-15 Electric Flux through Closed Surface The integral is over a CLOSED surface. Since is a scalar product, the electric flux is a SCALAR quantity The integration element is a vector normal to the surface and points OUTWARD from the surface. Out is +, In is - Φ E proportional to # field lines coming through outward

16. Lecture 3-16 Why are we interested in electric flux? is closely related to the charge(s) which cause it. Consider Point charge Q If we now turn to our previous discussion and use the analogy to the number of field lines, then the flux should be the same even when the surface is deformed. Thus should only depend on Q enclosed.

17. Lecture 3-17 Gauss’s Law: Quantitative Statement The net electric flux through any closed surface equals the net charge enclosed by that surface divided by  0. How do we use this equation?? The above equation is TRUE always but it doesn’t look easy to use. BUT - It is very useful in finding E when the physical situation exhibits a lot of SYMMETRY.

18. Lecture 3-18 Physics 241 – 10:30 Quiz 3, August 30, 2011 The left half of a rod is uniformly charged with a positive charge density σ, whereas the right half is uniformly charged with a charge density of -σ. What is the direction of the electric field at a point on the perpendicular bisector and above the rod as shown? a) to the right b) to the left c) up d) down e) E is zero. -σ +σ

19. Lecture 3-19 Physics 241 – 11:30 Quiz 3, September 1, 2011 The upper half of a ring is uniformly charged with a positive charge density σ, whereas the lower half is uniformly charged with a charge density of -σ. What is the direction of the electric field at a point on the perpendicular axis and to the left of the ring as shown? a) to the right b) to the left c) up d) down e) E is zero. +σ -σ

20. Lecture 3-20 Physics 241 – 11:30 Quiz 3, January 18, 2011 The upper half of a ring is uniformly charged with a negative charge density -σ, whereas the lower half is uniformly charged with a charge density of +σ. What is the direction of the electric field at a point on the perpendicular axis and to the left of the ring as shown? a) to the right b) to the left c) up d) down e) E is zero. -σ +σ