Today s Plan. Electric Dipoles. More on Gauss Law. Comment on PDF copies of Lectures. Final iclicker roll-call

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Gilbert King
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1 Today s Plan lectic Dipoles Moe on Gauss Law Comment on PDF copies of Lectues Final iclicke oll-call

2 lectic Dipoles A positive (q) and negative chage (-q) sepaated by a small distance d. lectic dipole moment vecto with length pq d, fom the negative chage to the positive chage. xamples include TV antennas and on the micoscopic scale (H 2 O).

3 lectic Field Lines of a Dipole

4 Toque and Foces on a Dipole N.B. No net foce on the dipole

5 Popeties of electic dipoles Potential enegy Toque U p τ p U p cos( θ ) τ p sin( θ ) In a unifom electic field, the field exets a toque (but no net foce) on the dipole to minimize its potential enegy and align it with the field.

6 xpeimental setup fo toque on a dipole

7 Toque on dipoles (wate molecules) povides heating in micowave ovens Altenating electic field

8 c a b da Φ q enclosed ε

9 BB Clicke Pob adius TAK s TO B RADIUS! Φ 1 2Φ 2 Φ 1 Φ 2 (A) (B) Φ 1 1/2Φ 2 (C) none (D) 25

10 xplanation Definition of Flux: Φ suface da constant on bael of cylinde pependicula to bael suface ( paallel to da) Case 1 Φ da A λ 2πε s A (2 s) L bael bael 1 Φ1 1 π λl ε A 2 Φ 1 2Φ 2 Φ 1 Φ 2 (A) (B) Case 2 λ 2πε (2s) (2π (2s)) L / 2 2πsL 2 2 Φ 1 1/2Φ 2 (C) none (D) RSULT: GAUSS LAW Φ popotional to chage enclosed! Φ λ( L / 2) ε 25

11 Review: Gauss Law Coulomb s Law We now illustate this fo the field of the point chage and pove that Gauss Law implies Coulomb s Law. Symmety -field of point chage is adial and spheically symmetic Daw a sphee of adius R centeed on the chage. Why? nomal to evey point on the suface da da has same value at evey point on the suface can take outside of the integal! 2 da da da R Theefoe, 4π! Gauss Law ε 4π R 2 Q We ae fee to choose the suface in such poblems we call this a Gaussian suface 1 4πε Q R Q 2 R

a ρ Outside sphee: (>a) We have spheical symmety centeed on the cente of the sphee of

12 Unifom chaged sphee What is the magnitude of the electic field due to a solid sphee of adius a with unifom chage density ρ (C/m 3 )? a ρ Outside sphee: (>a) We have spheical symmety centeed on the cente of the sphee of chage Theefoe, choose Gaussian suface hollow sphee of adius π 2 da 4 q 4 π a 3 ρ 3 q ε Gauss Law ρa ε 4π ε 1 q same as point chage!

13 Outside sphee: ( > a) Inside sphee: ( < a) Unifom chaged sphee ρa 3ε We still have spheical symmety centeed on the cente of the sphee of chage. Theefoe, choose Gaussian suface sphee of adius 3 2 a ρ π 2 da 4 q ε Gauss Law But, q 4 π 3 3 ρ Thus: ρ 3ε a

14 Infinite Line of Chage Symmety -field must be to line and can only depend on distance fom line Theefoe, CHOOS Gaussian suface to be a cylinde of adius and length h aligned with the x- axis. y h 2 x Apply Gauss Law: On the ends, da On the bael, da 2πh AND q λh λ 2πε NOT: we have obtained hee the same esult as we did last lectue using Coulomb s Law. The symmety makes today s deivation easie.

15 UI4PF4: 5) Given an infinite sheet of chage as shown in the figue. You need to use Gauss' Law to calculate the electic field nea the sheet of chage. Which of the following Gaussian sufaces ae best suited fo this pupose? Note: you may choose moe than one answe a) a cylinde with its axis along the plane b) a cylinde with its axis pependicula to the plane c) a cube d) a sphee

16 Infinite sheet of chage, suface chage density σ Symmety: diection of x-axis σ Theefoe, CHOOS Gaussian suface to be a cylinde whose axis is aligned with the x-axis. A x Apply Gauss' Law: On the bael, On the ends, da da 2 A The chage enclosed σ A ( 2A) A Theefoe, Gauss Law ε σ σ 2ε Conclusion: An infinite plane sheet of chage ceates a CONSTANT electic field. Same as integating chage distibution.

17 Two Infinite Sheets (into the sceen) Field outside must be zeo. Two ways to see: Supeposition Gaussian suface encloses zeo chage Field inside is NOT zeo: Supeposition Gaussian suface encloses non-zeo chage Q σa da A outside A inside σ ε - σ σ A A - - -

18 Gauss Law: Help fo the Homewok Poblems Gauss Law is ALWAYS VALID! ε da q enclosed What Can You Do With This? If you have (a) spheical, (b) cylindical, o (c) plana symmety AND: If you know the chage (RHS), you can calculate the electic field (LHS) If you know the field (LHS, usually because inside conducto), you can calculate the chage (RHS). Spheical Symmety: Gaussian suface sphee of adius LHS: ε da 4πε 2 1 4πε RHS: q ALL chage inside adius Cylindical symmety: Gaussian suface cylinde of adius LHS: ε da ε 2πL λ RHS: q ALL chage inside adius, length L Plana Symmety: Gaussian suface cylinde of aea A LHS: ε da 2 A q 2 2πε ε σ RHS: q ALL chage inside cylinde σ A 2ε

19 Gauss Law and Conductos We know that inside a conducto (othewise the chages would move). lectostatics! But since da Q. inside Conducto suface S Chages on a conducto only eside on the suface(s)! Gaussian Suface just inside S. Conducting sphee

20 Fields at suface of conductos Conducting sphee: chage distibutes unifomly. outside just like point chage Q. Moe geneal shape: 1. is to suface since thee can be no component tangent. 2. Flux on cyl. suface is zeo. 3. Flux on inside is zeo. 4. Theefoe da Φ A cos θ q ε σ A ε A Q Conducting sphee Gaussian pillbox with plane sufaces paallel. A at suface of conducto is nomal and σ/ε.

21 UL4PF4: A B A blue sphee A is contained within a ed spheical shell B. Thee is a chage Q A on the blue sphee and chage Q B on the ed spheical shell. 7) The electic field in the egion between the sphees is completely independent of Q B the chage on the ed spheical shell. Tue False

22 UI4ACT1 Conside the following two topologies: A) A solid non-conducting sphee caies a total chage Q -3 µc distibuted evenly thoughout. It is suounded by an unchaged conducting spheical shell. - Q σ 1 σ 2 B) Same as (A) but conducting shell emoved 1A Compae the electic field at point X in cases A and B: (a) A < B (b) A B (c) A > B 1B What is the suface chage density σ 1 on the inne suface of the conducting shell in case A? (a) σ 1 < (b) σ 1 (c) σ 1 >

23 UI4ACT1 Conside the following two topologies: σ 2 A) A solid non-conducting sphee caies a total chage Q -3 µc distibuted evenly thoughout. It is suounded by an unchaged conducting spheical shell. - Q σ 1 1A Compae the electic field at point X in cases A and B: (a) A < B (b) A B (c) A > B Select a sphee passing though the point X as the Gaussian suface. How much chage does it enclose? Answe: - Q, whethe o not the unchaged shell is pesent. (The field at point X is detemined only by the objects with NT CHARG.)

24 UI4ACT1 Conside the following two topologies: σ 2 A solid non-conducting sphee caies a total chage Q -3 µc and is suounded by an unchaged conducting spheical shell. - Q σ 1 B) Same as (A) but conducting shell emoved 1B What is the suface chage density σ 1 on the inne suface of the conducting shell in case A? (a) σ 1 < (b) σ 1 (c) σ 1 > Inside the conducto, we know the field Select a Gaussian suface inside the conducto Since on this suface, the total enclosed chage must be Theefoe, σ 1 must be positive, to cancel the chage - Q By the way, to calculate the actual value: σ 1 -Q / (4 π 12 )

25 Testing Gauss s Law Faaday s ice pail expeiment. If Gauss s Law coect, we will neve detect any chage on the inside. Application: Van de Gaaf geneato

26 Coaxial cable: signal is shielded fom extenal noise by Gauss Law

Gauss Law. Physics 231 Lecture 2-1

Gauss Law. Physics 231 Lecture 2-1 Gauss Law Physics 31 Lectue -1 lectic Field Lines The numbe of field lines, also known as lines of foce, ae elated to stength of the electic field Moe appopiately it is the numbe of field lines cossing