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# 228602.ppt

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Electrostatics 3
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# 228602.ppt

the divergence and curl of electric field

the divergence and curl of electric field

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### 228602.ppt

1. 1. The divergence of E If the charge fills a volume, with charge per unit volume . ' dq d    Where d is an element of volume. For a volume charge: ' ' 2 0 1 ( ) ˆ ( ) 4 v r E R rd r      R
2. 2. Thus: 3 ' ' ' 0 1 . 4 ( ) ( ) 4 v E r r r d         0 1 . ( ) E r     Gauss’s law in differential form. 3 2 ˆ . 4 ( ) r r r            ' ' 2 0 ˆ 1 . . ( ) 4 v r E r d r               
3. 3. Spherical polar coordinates (r, , ) r: the distance from the origin : the angle down from the z axis is called polar angle : angle around from the x axis is called the azimuthal angle sin cos x r    sin sin y r    cos z r   ˆ ˆ ˆ ˆ sin cos sin sin cos r x y z         ˆ ˆ ˆ ˆ cos cos cos sin sin x y z          ˆ ˆ ˆ sin cos x y       ˆ ˆ ˆ sin dl drr rd r d      
4. 4. The Curl of E For a point charge situated at origin: 2 0 1 ˆ 4 q E r r   Line integral of the field from some point a to some other point b: In spherical polar coordinates, ˆ ˆ ˆ sin dl drr rd r d       2 0 1 . 4 q E dl dr r  
5. 5. 2 0 1 . 4 b b a a q E dl dr r     0 1 4 a b q q r r          True for electrostatic field. . 0 E dl   Apply stokes theorem: 0 E   The integral around a closed path:
6. 6. Electric Potential Basic concept: The absence of closed lines is the property of vector field whose curl is zero. E is such a vector whose curl is zero. Using this special kind of it’s property we can reduce a vector problem: using V, we can get E very easily. Vector whose curl is zero, is equal to the gradient of some scalar function E=0  the line integral of E around any closed loop is zero (due to Stokes' theorem). E V  
7. 7. otherwise you could go out along path (i) and return along path (ii) and Because the line integral is independent of path, we can define a function O is some standard reference point. . 0 E dl   Therefore the line integral of E from point a to point b is the same for all paths. . 0 E dl   ( ) . r o V r E dl   is called electric potential
8. 8. The potential difference between two points a and b: ( ) ( ) . . b a o o V b V a E dl E dl       . . b o o a E dl E dl      . b a E dl   Using fundamental theorem for gradients: ( ) ( ) ( ). b a V b V a V dl     So ( ). . b b a a V dl E dl      E V   Electric field is the gradient of a scalar potential.
9. 9. Electric Potential at an arbitrary point •Electric potential at a point is given as the work done in moving the unit test charge (q0) from infinity (where potential is taken as zero) to that point. • Electric potential at any point P is Note that Vp represents the potential difference dV between the point P and a point at infinity. S.I. unit J/C defined as a volt (V) and 1 V/m = 1 N/C . p p V E ds    0 p W V q 
10. 10. Potential Difference in Uniform E field • Electric field lines always point in the direction of decreasing electric potential. Example: Uniform field along –y axis (E parallel to dl) . B B B A A A V V V E dl Edl          B A V Edl Ed       • When the electric field E is directed downward, point B is at a lower electric potential than point A. A positive test charge that moves from point A to point B loses electric potential energy.
11. 11. Potential Diff. in Uniform E field Charged particle moves from A to B in uniform E field.  . b a V E ds  
12. 12. Potential Diff. In Uniform E field (Path independence) Show that the potential difference between point A and B by moving through path (1) and (2) are the same as expected for a conservative force field. By path (1), . cos B A V E dl El       
13. 13. path (2) = 0 since E is  to dl . cos C A V E dl Eh El          . . C B A C V E dl E dl       
14. 14. Equipotential Surfaces (Contours) VC = VB ( same potential) In fact, points along this line has the same potential. We have an equipotential line. . 0 B C V E dl      No work is done in moving a test charge between any two points on an equipotential surface. The equipotential surfaces of a uniform electric field consist of a family of planes that are all perpendicular to the field.
15. 15. Equipotential Surface Equipotential Surfaces (dashed blue lines) and electric field lines (orange lines) for (a) a uniform electric field produced by infinite sheet of charge, (b) a point charge, and (c) an electric dipole. In all cases, the equipotential surfaces are perpendicular to the electric field lines at every point.
16. 16. 16 Electrostatic Potential of a Point Charge at the Origin Q P r       ' 2 ' 0 2 ' 0 0 4 4 4 r r r Q V r E dl dr r Q dr Q r r                 
17. 17. 17 Electrostatic Potential Resulting from Multiple Point Charges Q1 P(R,,) r 1 R 1 r O Q2 2 r   1 0 4 n k k k Q V r R     2 R
18. 18. 18 Electrostatic Potential Resulting from Continuous Charge Distributions       0 0 0 1 4 1 4 1 4 L S V dl V r R ds V r R dv V r R              line charge  surface charge  volume charge
19. 19. 19 Charge Dipole • An electric charge dipole consists of a pair of equal and opposite point charges separated by a small distance (i.e., much smaller than the distance at which we observe the resulting field). d +Q -Q
20. 20. Dipole Moment • Dipole moment p is a measure of the strength of the dipole and has its direction. p Qd  +Q -Q d p is in the direction from the negative point charge to the positive point charge
21. 21. 21 Electrostatic Potential Due to Charge Dipole observation point d/2 +Q -Q z d/2  P ˆz p a Qd  R R r
22. 22. 22 d/2 d/2    cos ) 2 / ( cos ) 2 / ( 2 2 2 2 rd d r R rd d r R         R r P     0 0 , 4 4 Q Q V r V r R R         R
23. 23. • first order approximation from geometry:   cos 2 cos 2 d r R d r R       d/2 d/2  lines approximately parallel R R r
24. 24. 24 • Taylor series approximation:                                         cos 2 1 1 1 cos 2 1 1 cos 2 1 1 cos 2 1 1 1 r d r R r d r r d r d r R   1 , 1 1 : Recall     x nx x n   2 0 0 4 cos 2 cos 1 2 cos 1 4 , r Qd r d r d r Q r V                             
25. 25. 25 • In terms of the dipole moment: 2 0 ˆ 4 1 r a p V r   
26. 26. Electric Potential Energy of a System of Point Charges 1 0 1 4 q V r   2 ( ) W q V r  1 3 2 3 1 2 12 13 23 0 12 13 23 1 ( ) 4 q q q q q q W W W W r r r        2 W F r q E r     q1 q2 2 b b a a W F dl q E dl        2[ ( ) ( )] W q V b V a   2[ ( ) ( )] W q V r V    and we know
27. 27. The Energy of a Continuous Charge Distribution For a volume charge density p, 1 2 W Vd     0 .E     Using Gauss’s Law: 0 ( . ) 2 W E Vd      So: 0 .( ) . 2 W E V d VE da             By doing integration by part: and so, V E    2 0 . 2 v s W E d VE da             If we take integral over all space: 2 0 2 allspace W E d    
28. 28. Poisson’s and Laplace’s Equation E V   The fundamental equations for E: 0 . ; E     0 E   2 . .( ) E V V       Gauss’s law then says that: 2 0 V      This is known as Poisson’s equation. In regions where there is no charge: 0   Poisson’s equation reduces to Laplace’s equation. 2 0 V   This is known as Laplace’s equation.