Welcome to Physics 272
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1 Welcome to Physics 7 Bob Mose mose@phys.hawaii.edu To do: Sign into Masteing Physics phys-7 webpage Registe i-clickes (you i-clicke ID to you name on class-list)
2 Review fo a point chage q: q = 4πε lectic Field Calculations ˆ q ˆ P What if moe than one chage? = q 4πε 1 = 4 πε q 1 ˆ 1 + ˆ 1 4πε i q i i ˆ i Use supeposition. q 1 q ˆ 1ˆ P 1 1 Can do fo any numbe of chages.
3 Chage Densities How do we epesent the chage Q on an etended object? total chage Q small pieces of chage dq Line of chage: λ = chage pe unit length Suface of chage: σ = chage pe unit aea dq = λ d dq = σ da Volume of Chage: ρ = chage pe unit volume dq = ρ dv
4 Review: Continuous Chage Distibutions Fun with calculus. What if we have a distibution of chage? Q chage of distibution. dq element of chage. d contibution to due to dq. Can wite dq = ρ dv; ρ is the chage density. Could also be a linea (λ dl) o aea chage density (σ da) depending on the eample. Q dq ˆ P dv d = 1 4πε qi 1 dq ˆ i ˆ = 4πε i i V 1 4πε V ρdv ˆ
5 Review: Continuous Chage Distibutions ample: Field of a line of chage on y ais. y at P? Use symmety. a Result: 1 Q = 4 πε + What if a? What if a? (Long staight wie) = 1 4πε 1 πε λ iˆ Q + a a iˆ = iˆ 1 4πε λ = Q/(a) = linea chage density a Q -a Q ( / a) + 1 Moe geneally: ˆ P wie iˆ = 1 πε P λ ˆ Integal woked out in 1.11 Impotant esult: Infinite line of chage
6 d Review: Continuous Chage Distibutions ample: Field on ais of a ing of chage. at P? Hee use symmety. z Q Details: 1 dq 1 dq = cosα = 4 ( + a ) 4 ( + a ) ( + a ) πε a y πε 1/ 1 dq 1 dq = d = = 4 πε ( + a ) 4 πε ( + a ) 3/ 3/ P 1 Q = 4 πε ( + a ) 3/ iˆ
7 ample: Field on ais of a disk of chage. Details of the calculation. y Integate ove ings of chage. Use symmety. z Q R d P dq = σ da da = d( π ) = π d dq = σ π d d 1 dq 1 ( πσd) = = 4πε 4 πε ( + ) 3/
8 Review: Continuous Chage Distibutions ample: Field on ais of a disk of chage. y at P? How to do? Integate ove ings of chage. Result: Result fo R? = σ ε 1 z Q R d σ 1 1 ( R / ) 1 = ε + ( R 1 / ) + 1 iˆ σ ε iˆ iˆ P σ = suface chage density = chage/aea 1. Pependicula. Independent of d. 3. Away/towads. Impotant esult: lectic field of an infinite plane of chage.
9 ample: Field on ais of a disk of chage. Details cont d Integate ove ings of chage. Use symmety. d z Q R y d 1 dq 1 ( πσd) = = 4πε 4 πε ( + ) 3/ 1 ( πσd) σ d = d = = 4 πε ( + ) ε ( + ) 3/ 3/ P Substitute v = + ; dv = d
10 ample: Field on ais of a disk of chage. Finale: Integate ove ings of chage. Use symmety. z Q R y d P σ 1 1 = + ε + R σ 1 σ = 1 ε / R + 1 ε when R
11 Hints fo HWK -σ σ Region I Region II Region III What is the field in Regions I and III?
12 Review: Continuous Chage Distibutions Review: 1. Have shown that we can use calculus to detemine electic fields fo a few special chage distibutions.. Method impotant. Know how to do. 3. Solutions fo infinite line of chage and infinite plane impotant. We will see these again.
13 Ways to Visualize the Field Conside the -field of a positive point chage at the oigin vecto map field lines + chg + chg + + Intoduced by Michael Faaday ( )
14 Rules fo Vecto Maps + chg + Diection of aows indicates the diection of the field at each point in space Length of aows is popotional to the magnitude of the field at each point in space
15 Rules fo Field Lines gaphical tick + fo visualizing - fields Lines leave (+) chages and etun to (-) chages Numbe of lines leaving/enteing chage amount of chage Field lines neve coss Tangent of line = diection of at each point Local density of field lines ~ magnitude of at each point Applets:
16 CLICKR QUSTION Which of the following coectly depicts the field lines fom an infinite unifomly negatively chaged sheet? Note that the sheet is being viewed edge-on in all pictues
17 CLICKR QUSTION Which of the following shows the coect electic field lines fo an electic dipole?
18 Gauss s Law and lectic Flu Peview Gauss s Law elates numbe of field lines enteing and leaving a suface to the net chage inside the suface. Conside imaginay sphees centeed on: a.) +q (blue) b.) -q (ed) c.) midpoint (yellow) c a Numbe of lines eiting: a.) (blue) 4 b.) -q (ed) -4 c.) midpoint (yellow) b Moe on Gauss s Law late.
19 Flu: lectic Flu Let s quantify pevious discussion about fieldline counting Define: electic flu Φ Ε suface S though the closed Φ S da S is suface of the bo da S
20 lectic Flu Φ S da What does this new quantity mean? The integal is ove a CLOSD SURFAC Since da is a SCALAR poduct, the electic flu is a SCALAR quantity The integation vecto da is nomal to the suface and points OUT of the suface. da is intepeted as the component of which is NORMAL to the SURFAC Theefoe, the electic flu though a closed suface is the sum of the nomal components of the electic field all ove the suface. The sign mattes!! Pay attention to the diection of the nomal component as it penetates the suface is it out of o into the suface? Out of is + into is -
21 lectic Flu Special case: unifom pependicula to plane suface A. Φ da = da cos θ = da = S S S A da constant =1 Anothe: unifom at angle to plane suface A. constants Φ da = da cos θ = cos θ da = S S S θ da A cos θ
22 CLICKR QUSTION A flat disk with adius =.1m, is oiented with its nomal unit vecto elative the a constant field as shown above. What is the electic flu though the disk aea? o A) B) C) π (.1) D) π (.1) cos( 3 ) π π ( ) ( o.1 cos 3 ) ( ) ( o.1 sin 3 ) ) π ( ) ( o.1 sin 3 )
23 How to think about flu We will be inteested in net flu in o out of a closed suface like this bo This is the sum of the flu though each side of the bo conside each side sepaately Let -field point in y-diection then and A ae paallel and z y A = w A is suface of the bo A w suface aea vecto: A = = aea w yˆ yˆ Look at this fom on top down the z-ais A
24 UI3PF3: 6) A cube is placed in a unifom electic field. Find the flu though the bottom suface of the cube. a)ф bottom < b)ф bottom = c)ф bottom >
25 ecise UI3A Imagine a cube of side a positioned in a egion of constant electic field as shown Which of the following statements about the net electic flu though the suface of this cube is tue? (a) φ (b) (c) φ =6a = φ =a a a
26 ecise UI3A Imagine a cube of side a positioned in a egion of constant electic field as shown Which of the following statements about the net electic flu t φ =a though the suface of this cube is tue? (a) φ = (b) φ =a (c) φ =6a a a The electic flu though the suface is defined by: Φ da da is ZRO on the fou sides that ae paallel to the electic field. on the bottom face is negative. (da is out; is in) da da on the top face is positive. (da is out; is out) Theefoe, the total flu though the cube is: Φ ds A = Φ a a top sides +Φ bottom +Φ = + =
27 c a b da = Φ = q enclosed ε
28 Fundamental Law of lectostatics Coulomb s Law Foce between two point chages OR Gauss Law Relationship between lectic Flu and chages
29 Gauss Law Gauss Law (a FUNDAMNTAL LAW): The net electic flu though any closed suface is popotional to the chage enclosed by that suface. qenclosed da = Φ = ε How do we use this equation?? The above equation is ALWAYS TRU but it may not be easy to use. It is vey useful in finding when the physical situation ehibits massive SYMMTRY.
30 UI4PF4: da da 1 ) A positive chage is contained inside a spheical shell. How does the electic flu dф though the suface element da change when the chage is moved fom position 1 to position? a) dф inceases b) dф deceases c) dф doesn t change
31 da da 1 3) A positive chage is contained inside a spheical shell. How does the flu Ф though the entie suface change when the chage is moved fom position 1 to position? a) Ф inceases b)ф deceases c)ф doesn t change
32 CLICKR QUSTION We have a sphee of adius and a negative chage -q, outside the sphee, away, using Gauss Law, what is the total electic flu though the sphee?? -q a) b) 1 q 4πε 1 q 4πε ( ) d) e) q ε zeo c) q ε
33 Gauss Law made easy Φ To solve the above equation fo, you have to be able to CHOOS A CLOSD SURFAC such that the integal is TRIVIAL. (1) Diection: suface must be chosen such that is known to be eithe paallel o pependicula to each piece of the suface; da da = If then If then da = enclosed ε () Magnitude: suface must be chosen such that has the same value at all points on the suface when is pependicula to the suface. q da = da da =
34 Gauss Law made easy With these two conditions we can bing outside of the integal and: da Φ = A da = da = da = Note that is just the aea of the Gaussian suface ove which we ae integating. Gauss Law now takes the fom: da = da = q enclosed This equation can now be solved fo (at the suface) if we know q enclosed (o fo q enclosed if we know ). ε q enclosed ε
35 Gauss Coulomb 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! da = da = da R Theefoe, = 4π! Gauss Law ε 4π R = Q = 1 4πε Q R +Q R We ae fee to choose the suface in such poblems we call this a Gaussian suface
36 Summay Gauss Law: lectic field flu though a closed suface is popotional to the net chage enclosed q enclosed Φ = da = Gauss Law is eact and always tue. Gauss Law makes solving fo -field easy when the symmety is sufficient spheical, cylindical, plana ε
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