EELE 3331 Electomagnetic I Chapte 4 Electostatic fields Islamic Univesity of Gaza Electical Engineeing Depatment D. Talal Skaik 212 1
Electic Potential The Gavitational Analogy Moving an object upwad against the gavitational field inceases its gavitational potential enegy. An object moving downwad within the gavitational field would lose gavitational potential enegy. When you move a chage in an electic field its potential enegy changes. This is like moving a mass in a gavitational field. 2
Electic Potential In ode to bing two like chages nea each othe wok must be done. In ode to sepaate two opposite chages, wok must be done. As the monkey does wok on the positive chage, he inceases the enegy of that chage. The close he bings it, the moe electical potential enegy it has. The extenal foce F against the E-field inceases the potential enegy. 3
4.7 Electic Potential To move a point chage a distance dl, the wok done is the poduct of the foce and distance : F l= E l dw d d Displacement of point chage in an electostatic field E. Negative sign indicates that wok is done by an extenal agent. (The foce we apply to move the chage against the electic field is equal and opposite to the foce due the field). If we want to move the chage in the diection of the field, we do not do the wok, the field does. 4
Electic Potential The total wok done in moving fom A and B, is: Define potential diffeence V as the wok done (by extenal foce) in moving a unit positive chage fom one point to anothe in an Electic Field. Potential Diffeence B W E dl A final point (Joules/coulomb) B W =V AB= E dl o (J/C) A o (V) initial point 5
Electic Potential Notes: V E dl AB If V AB is negative, thee is loss in potential enegy in moving fom A to B Wok is don by the field. If V AB is positive, thee is gain in potential enegy in moving fom A to B Extenal agent pefoms the wok. V AB is independent of the path taken. B A 6
Electic Potential If E is due to a point chage located at the oigin, E 4 2 dl= d a d a sin d a B AB 2 4 A a B V E dl a d a B 1 1 V AB d 2 4 4 B o A V V V AB B A A A 7
Electic Potential If we choose a zeo efeence fo potential, Let V= at infinity. Thus if V = as V V -V V AB B A B zeo 4 Geneally, the potential at any point distant fom a point chage at the oigin is: V 4 V B A A at infinity is efeence The potential at any point is the potential diffeence between that point and a chosen point (o efeence point) at which the potential is zeo. V AB 4 B A 1 1 8
If the point chage is located at a point with position vecto ', V() becomes: V 4 ' 1 2 at infinity is efeence Fo n point chages,,..., located at points with position vectos V o V 1 2 1 2 ( )= 4 1 4 2 4 V ( )= n 4 k1 k n,,...,, the potential at is: n Electic Potential k n n 9
Fo continuous chage distibution, the potential at becomes: 1 L( ') dl ' V( )= (Line Chage) 4 ' 1 S ( ') ds ' V( )= (Suface Chage) 4 ' V( )= 1 v( ') dv ' (Volume Chage) 4 ' L S v Electic Potential pimed coodinates fo souce point locations. 1
Electic Potential The diagam shows some values of the electic potential at points in the electic field of a positively chaged sphee. When one coulomb of chage moves fom A to B it gains 15 Joules of potential enegy. 11
Notes: Fo E due to a point chage : Assume V = as Zeo efeence at infinity A V V -V V A AB B A B zeo 4 If any point othe than infinity is chosen as efeence with zeo potential, Let / 4 be constant=c V V V V C V C AB B A B B 4B 4 zeo Geneally, the potential at any point distant fom a point chage at the oigin is: V B B 1 1 VAB E dl 4 A B A A 4 + C B Geneally, fo any E: V= E dl C 12
Example 4.1 Two point chages -4μC and 5μC ae located at (2,-1,3) and (,4,-2). Find the potential at (1,,1), assuming zeo potential at infinity. Let 4C, 5C V()= 1 2 1 2 1 2 4 4 If V( )=, C 1 2 (1,,1) (2, 1,3) ( 1,1, 2) 6 (1,,1) (, 4, 2) (1, 4,3) 26 Hence V(1,,1)= 4 36 C 6 1 4 5 9 1 6 26 5.872 KV 13
Example 4.11 A point chage of 5nC is located at (-3,4,), while line y=1, z=1 caies unifom chage 2 nc/m. (a) If V=V at O(,,), find V at A(5,,1) (b) If V=1V at B(1,2,1), find V at C(-2,5,3) (c) If V=-5V at O(,,), find V BC. Let V=V +V L V E dl= a d a C 2 1 4 4 VL E dl= a d a ln C 2 2 V L ln C 2 4 L L 2 14
V L ln C 2 4 Example 4.11 - continued (a) V= at O(,,), V at A(5,,1)=?? V V V L AO O A ln 2 A 4 known,,,1,1 2 5,,1 5,1,1 1 A,, 3, 4, 5 5,,1 3, 4, 9 A 1 1 A V AO 2 1 9 9 2 (1 / 36 ) 9 2 51 1 1 ln 9 1 4 (1 / 36 ) 5 9 V V 36ln 2 45(1/ 5 1/ 9) V 8.477 V AO A A 15
C B Example 4.11 - continued (b) If V=1 at B(1,2,1), V at C( 2,5,3)=?? V -V 2,5,3 2,1,1 2 C 1, 2,1 1,1,1 1 B 2,5,3 3, 4, 11 C 1, 2,1 3, 4, 21 B L C 1 1 ln 2 B 4 C B 2 1 1 VC 1 36ln 45 5.175 V 1 11 21 V 49.825 V C (C) V V V 49.825 1 5.175 V BC C B We don't need a potential efeence if a common efeence is assumed. 16
V V V V E dl BA AB BA AB E dl integal fom L No net wok is done in moving a chage along a closed path in an electostatic field. Applying Stoke's theoem: L 4.8 Relationship between E and V Maxwell s Equation L E dl E ds= E= (consevative, iotational) S E= 2 nd Maxwell s equation fo static electic field. (diffeential fom) 17
Relationship between E and V E= E V Electic Field Intensity E is the gadient of V. Negative sign shows that the diection of E is opposite to the diection in which V inceases; E is diected fom highe to lowe levels of V. V 18
Example 4.12 Given the potential V=(1/ 2 )sinθ cosφ. a) Find the electic flux density D at (2,π/2,). b) Calculate the wok done in moving a 1µC chage fom point A(1,3,12 ) to B(4,9,6 ). V 1 V 1 V D= E, E= V a a a sin 2 1 1 E= sin cos a cos cos a sin a 3 3 3 At (2, /2,) D= E 2 = a a a 8 =2.5 a C/m 22.1 a pc/m 2 2 19
2 1 1 E= sin cos a cos cos a sin a 3 3 3 (b) Method 1: W W E dl o E dl L Beak up path (to make integation easie) W E dl AA' A' B' B' B AA' dl d a A' B ' dl d a Example 4.12 - continued (b) Calculate the wok done in moving a 1µC chage fom point A(1,3,12 ) to B(4,9,6 ). L B ' B dl sin d a 2
AA' dl d a, A' B ' dl d a, B ' B dl sin d a 4 9 W 2sin cos 1cos cos d d 3 1 3 6 12 3 3 =12 1sin + sin d 3 4 =9 W 75 5 1 45 W 28.125 J 32 32 16 16 Method 2: Example 4.12 continued W VAB W V 1 1 W VB VA 16 1 W 28.125 J AB 4 =12 6 1 1 sin 9 cos6 sin 3 cos12 21
The potential at point P (,, ) is: V 4 1 2 2 1 4 1 2 If d : d cos 2 1 4.9 An Electic Dipole and Flux Lines An electic dipole is fomed when two point chages of equal magnitude but opposite sign ae sepaated by a small distance. 1 1 2 21 V dcos 2 4 22
V 4 dcos 2 V 1V 1 V E V a a a sin d cos d sin E a a 2 4 o An Electic Dipole and Flux Lines 3 3 d E= 3 2 cos a sin a 4 23
V 4 dcos 2 Define dipole moment P=d, d (P is diected fom to ) P a P V=, a V= 4 4 2 3 If the dipole cente is at ': P ( ') V( )= 4 ' An Electic Dipole and Flux Lines 3 cos =P a 24
Equipotential Lines EUIPOTENTIAL LINES ae lines along which each point is at the same potential. On an equipotential suface, each point on the suface is at the same potential. The equipotential line o suface is pependicula to the diection of the electic field lines at evey point. Thus, if the electic field patten is known, it is possible to detemine the patten of equipotential lines o sufaces, and vice vesa. 25
Equipotential Lines In the following diagams, the dashed lines epesent equipotential lines and the solid lines the electic field lines. 26
An Electic Dipole and Flux Lines Equipotential sufaces fo (a) a point chage and (b) an electic dipole. 27
An Electic Dipole and Flux Lines Notes: An electic flux line is an imaginay path o line and its diection at any point is the diection of the electic field at that point. Equipotential suface: potential is the same at any point. E is always nomal to the equipotential sufaces. Equipotential line : is intesection of equipotential suface and a plane. 28
Example 4.13 Two dipoles with dipole moments -5a z nc.m and 9a z nc.m ae located at points (,,-2) and (,,3). Find the potential at the oigin. 1 P P V= 4 whee 1 1 1 2 3 3 1 2 P 51 a, (,,) (,, 2) 2a 9 1 z 1 P 9 1 a, (,,) (,,3) 3a 9 2 z 2 Hence V 1 1 27 1 2 3 4 36 9 = 1 2.25 9 3 3 z z 29
4.1 Enegy density in electostatic fields Conside a egion fee of electic fields. Let thee be thee point chages 1, 2, 3 at infinity To detemine the enegy pesent in the assembly of chages, we have to detemine the amount of wok necessay to assemble them. No wok is equied to tansfe 1 fom infinity to P 1 because the space is initially chage fee (No electic field) 3
Enegy density in electostatic fields The wok done in tansfeing 2 fom infinity to P 2 is equal to the poduct of 2 and the potential V 21 at P 2 due to 1. The wok done in positioning 3 at P 3 is equal to 3 (V 32 +V 31 ). The total wok in positioning the thee chages is: W W W W E 1 2 3 V V V (1) 2 21 3 31 32 V 21 Potential at point 2 due to 1. V 31 Potential at point 3 due to 1. V 32 Potential at point 3 due to 2. If chages wee positioned in evese ode: W W W W E 3 2 1 V V V (2) 2 23 1 12 13 31
Enegy density in electostatic fields Adding equations (1) and (2) 2W V V V V V V E 1 12 13 2 21 23 3 31 32 1 2 WE= V + V V o W V + V V 2 1 1 2 2 3 3 E 1 1 2 2 3 3 V 1, V 2, V 3 Total potentials at points 1,2,3 espectively. If thee ae n point chages: W E k k 2 k1 In case of continuous chage distibution, summation becomes integation: 1 WE LVdl (Line Chage) 2 W W E E 1 2 1 2 L S v VdS S Vdv v 1 n V (Suface Chage) (Volume Chage) 32
Enegy density in electostatic fields Define the electostatic enegy density w E (in J/m 3 ) w E so, W dw dv 1 1 2 D 2 2 2 E 2 D E = E E v E 1 1 2 2 2 WE D E dv E dv v w dv v (Enegy Stoed) 33
Example 4.14 The point chages -1nC, 4nC, 3nC ae located at (,,), (,,1), and (1,,). Find the enegy in the system. 34
Example 4.15 (section 4.6 D) (The enegy stoed) 35
Example 4.15 36
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