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 electric dipole of dipole moment: -q - q +

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

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

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/r2. D) Electric fields describe a conservative force field.

SUMMARY: FIND THE ELECTRIC FIELD GIVEN THE CHARGES 1) GEOMETRY FOR qi or dqi 2) DISCRETE CHARGES qi 3) CONTINUOUS CHARGES dqi 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

Continuous Charge Distribution 1 Charged Line At a point P on axis: Ex = k λ (1/r1 – 1/r2 ) ( Eqn 22-2a) Ex = k λ (1/( XP – L/2 ) - 1/( XP + L/2 ) Ex = k ( Q/L) L [ XP2 – (L/2)2 ] -1 For XP2 >> (L/2)2 Ex = k Q / XP2 For XP = 0 Ex = 0

Again: Continuous Charge Distribution 1: Charged Line At a point P on perpendicular axis:  x

Physics 241 –Warm-up quiz 2 The rod is uniformly charged with a positive charge density . 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. to the right to the left up down lower right  p

General location of P: Charged Line At a point P off axis:

Continuous Charge Distribution 2: Charged Ring At point P on axis of ring: ds Use symmetry!  Ex = k Q x ( x2 + a2 )-3

Continuous Charge Distribution 3: Charged Disk At a point P on axis: Use the ring with radius a EX value dEx = k dq x ( x2 + a2 ) -3 Integrate rings from 0 to R Ex = -2πσkx ( 1/( x2 + R2 )1/2 – 1/x ) Superposition of rings! E = σ/2εo <= Independent of x

Continuous Charge Distribution 4: Charged Sheets E=const in each region Superposition! Capacitor geometry

MULTIPLE CHARGE SHEET EXAMPLE DOCCAM 2

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.

General definition of electric flux: To state Gauss’s Law in a quantitative form, we first need to define Electric Flux. # of field lines N = density of field lines x “area” where “area” = A2 x cos   Sum over surface General definition of electric flux: (must specify sense, i.e., which way)

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 - E proportional to # field lines coming through outward

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.

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.

Physics 241 – 10:30 Quiz 3 The left half of a rod is uniformly charged with a positive charge density , whereas the right half is uniformly charged with a charge density of . What is the direction of the electric field at a point on the perpendicular bisector and above the rod as shown? to the right to the left up down E is zero.  

Physics 241 – 11:30 Quiz 3 The upper half of a ring is uniformly charged with a positive charge density , whereas the lower half is uniformly charged with a charge density of . What is the direction of the electric field at a point on the perpendicular axis and to the left of the ring as shown?  to the right to the left up down E is zero. 

Gauss’ Law - Examples Shell Theorem Outside shell: E is as if Q at center Inside: E is zero

Proof of the Shell Theorem Electric Field Outside By symmetry, the electric field must only depend on r and is along a radial line everywhere. Apply Gauss’s law to the blue surface , we get

Uniformly charged thin shell: Inside E = 0 inside By symmetry, the electric field must only depend on r and is along a radial line everywhere. Apply Gauss’s law to the blue surface , we get E = 0. Equal and opposite contributions from charges on diagonally opposite surface elements. Discontinuity in E

Electric Field of a Uniformly Charged Sphere Apply Gauss’s Law directly or use superposition of the shell results

Physics 241 –Quiz 2a Two identical point charges are each placed inside a large cube. One is at the center while the other is close to the surface. Which statement about the net electric flux through the surface of the cube is true? The flux is larger when the charge is at the center. The flux is the same (and not zero). The flux is larger when the charge is near the surface. Not enough information to tell. The flux is zero in both cases. +Q +Q

Physics 241 –Quiz 2b Two identical point charges are each placed inside a large sphere. One is at the center while the other is close to the surface. Which statement about the net electric flux through the surface of the sphere is true? The flux is larger when the charge is at the center. The flux is larger when the charge is near the surface. The magnitude of the flux is the same (and not zero). Not enough information to tell. The flux is zero in both cases. +Q +Q

Physics 241 –Quiz 2c Two identical point charges are placed, at the center of a large sphere in one case, and outside an identical sphere in the other case. Which statement about the net electric flux through the surface of the sphere is true? The flux is larger when the charge is inside. The flux is larger when the charge is outside. The flux is the same (and not zero). Not enough information to tell. The flux is zero in both cases. Q>0 Q>0