PHY481: Electromagnetism

Transcription

1 PHY48: Electomagnetism HW5 Lectue Cal Bombeg - Pof. of Physics

2 Bounday condition ( ) = C n cos n + V x, y ( ) n= ( ) = V cos x V x,± a 5.3 V = V cos x a & ' at y = ± a Geneal solution (fo even bounday conditions) ( ) x cosh ( n + ) y a &' a &' a & ' ( = C n cos ( n + ) )* x a +,- ( cosh ( n + ) )* Squae pipe coss section Only non-zeo tem +,- n =, C to be detemined Apply bounday condition to detemine coefficient ( ) = C cos x V x,± a C = V cosh &' a &' cosh &' = V cos x a &' Potential inside pipe V ( x, y) = V cos( x a)cosh( y a) cosh( ) Lectue Cal Bombeg - Pof. of Physics

3 Phase factos Pove: Asinh( x + ) = Bsinh(x) + C cosh(x) A ex+ ( x+ ) e & ( = B ex e x & ( + C ex + e x & ( ' ' ' A e e x e e x & ( = ( C + B) ex & ( + ( C B) e x & ( ' ' ' C + B = Ae C B = Ae C = A e e ; B = A e + e sinh x + ( ) = cosh()sinh(x) + sinh()cosh(x) Lectue Cal Bombeg - Pof. of Physics

4 X (x) = B n cos [(n + ) ] x a Y y V (x, y) = 4V n= 5.7 ( ) = C n sinh[ ( n + ) y a] + D n cosh[ ( n + ) y a] instead use phase: n = C n sinh ( n + ) y a + n bounday conditions: X ( ± a ) = V (x, a ) = = g n sinh [( n + ) + n ]cos[(n + ) x a] bc: Y a n= n = ( n + ) V (x,+ a ) = V = g n sinh ( n + ) a +a n= f ( x ) cos (n + ) x a a V (x, y) = g n sinh[ ( n + ) y a + n ]cos (n + ) x a n= Be caeful to beak integal into pieces cos (n + ) x a = g n sinh n + [( ) ] [ ( )] () n sinh ( n + ) y a + ( n + ) sinh[ ( n + ) ] ( ) = sets phase bc: Y ( + a ) = V sets coef. g n g n = 4V ( ) n ( n + ) sinh ( n + ) cos [(n + ) x a] Lectue Cal Bombeg - Pof. of Physics 3

5 5. See lectues 8- fo deivations of potential fo a conducting sphee in an extenal field V (, ) = + E a3 cos Dipole potential E cos V (, ) = pcos 4 p = 4 E a 3 Polaizability p = E = 4 a 3 ( ) V = E ( a, ) = =a E = a 3 cos + E cos = 3 E cos a 3 Chage density Compae / potentials ( ) = 3 E cos Dipole moment analogous to p = qd p = ()ds = acos ˆk + asin ˆ = a 3 ( 3 E cos )cos sind ˆk p z = 6a 3 E = 6a 3 E u du Dipole moment cos sind p z = 4a 3 E Lectue Cal Bombeg - Pof. of Physics 4

6 Linea quadupole exact potential V (,) = q 4 & '( + a a cos Appoximation Expansions + = = + a ± a cos Monopole and dipole tems cancel 5.3 ) * + a + a cos +( + + acos ± + 3a cos a ' + ( & ) q Linea quadupole +q +a a +q V (,) = qa ' 3cos )' & + * 4 (' 3 +' = qa P (cos) Lectue Cal Bombeg - Pof. of Physics 5

7 5.5 (see lectue 9) Half cylindes, adius R. V = +V on uppe half, and V = -V on lowe half Find potential inside and out. (odd function of φ -> no cosine tems) V (,) = Aln + B + A n n + B n n n= B.C. at = and = V Int (,) = A n n sin n n= V Ext (,) = B n n sin n n= Be caeful to beak Fouie integal into pieces Use Othogonality cos n cos m = nm ( ) C n cos n + D n sin n ( ) B.C. at = R V Int (R,) = V Ext (R,) A n R n = B n R n = c n A n = c n ; R n Fouie Integal at = R B n = c n Rn V (R,)sin md = c n sin n sin m d n= &V cos m + V cos m = cn ' mn c m = 4V m m =,3,5... Complete solution V Int (,) = 4V m odd m m sin m m R V Ext (,) = 4V m odd m R m sin m m Lectue Cal Bombeg - Pof. of Physics 6

8 Geneal solution in cylindical coodinates V (,) = Aln + B + A n n + B n n n= 5.6 ( ) C n cos n + D n sin n ( ) E (R ext ) E (R int ) = Bounday condition cossing the bounday V ( int,) = A cos V ( ext,) = B cos E n ( int,) = A cos E (,) = + B cos n ext ext Matching tems Apply bounday conditions B cos ext B R + A = Solution inside and out + A cos = cos V int (R,) = V ext (R,) A Rcos = B R cos A R = B R V ( int,) = cos; V ( ext,) = R cos A = B = R Lectue Cal Bombeg - Pof. of Physics 7

9 5.8 (see poblem 5.7) X (x) = B n cos [(n + ) ] x a ( ) = C n sinh[ ( n + ) y a] Y y bounday conditions: X ( ± a ) = odd B. C. : Y ( ± a ) = ±V V (x, y) = g n sinh[ ( n + ) y a]cos (n + ) x a n= g n = B n C n V (x,+ a ) = V = g n sinh[ ( n + ) ]cos (n + ) x a n= B.C.: Y ( + a ) = V sets coef. g n a +a f ( x ) cos [(n + ) x a]dx = g n sinh ( n + ) a a V sin (n + ) x a (n + ) & sin (n + ) x a a a a a a + sin (n + ) x a Be caeful to beak Fouie integal into pieces ' ( ) = g n sinh[ ( n + ) ] g n = 4V ( ) n ( n + ) sinh ( n + ) V (x, y) = 4V n= () n sinh ( n + ) y a ( n + ) sinh ( n + ) cos [ (n + ) x a ] Lectue Cal Bombeg - Pof. of Physics 8

10 5.3 Dipole potential Geneal solution in spheical coodinates V = pcos ( 4 V (,) = A + B ) & ' P cos = + ( ) Dipole in gounded conducting sphee V (,) = A + B & ' P cos ( ) Apply bounday conditions Only one tem = > V (,) Bcos = pcos 4 B = p 4 = R V (R,) = AR + B R & ' = A = B R 3 = p 4 R 3 Full Solution: V (,) = p 4 + ( & ' R 3 ) * cos Lectue Cal Bombeg - Pof. of Physics 9

11 Caps C and C cay chages Q and Q. Close the switch - find new chages Q and Q Chage conseved: Q + Q = Q + Q V = Q Potentials become equal: C = Q C ; Q = C C Q + ( C C ) Q = Q + Q Q = Q + Q [ ( )] ; + C C ( ) Q Q = C Q + Q C + C C Compae stoed enegy befoe and afte switch is closed ( ( )) = Q + Q ( ( )) + C C Q = Q + Q; Q = Q Q U + U Potentials become equal: V V = U + U = Q + Q & C C = Q + C Q C ' + & (Q) ( ) C + Q C ' & Q +Q Q ' & C ( ) C ( ) U + U = U + U + Q C ; C = C C C + C Lectue Cal Bombeg - Pof. of Physics

12 3 a,b Chage q above a gounded conduction sphee Detemine a) image chage and position, b) potential outside Potential at points (a) and (b) must be zeo. ( ) q = q R z z R ; q = q R z z R ( ) q = q R + z z + R ; q = q R + z z + R ( ) ( ) R z z R = R + z z + R ; R = z z R z z a z b q q z R R z a) z = R z ; q = q R z b) V (,) = q ' + 4 &' + z z cos q ( * + z z cos )* Lectue Cal Bombeg - Pof. of Physics

13 3 c Two chages q outside gounded conduction sphee Image: Two chages q inside make V= on sphee. What adius sphee makes foce F = on q? a = z, b = R a; q = q R a F = q + q q 4a ( a + b) + q q ( a b) = q R 4a a = R 4a a 3 + R a = 4a R a 3 = R 4a a 3 q ( a + R a) R a ( ) + ( R a ) ' & q q R + 3 R 4 + a a 4 ' + + R & + 3 R 4 ( + + * a a 4 ' - ) &, + 3 R 4 + &; a 4 q ( a R a) ' R assuming R a << 4a a 3 Lectue Cal Bombeg - Pof. of Physics R q b ( ± x) = x + 3x + R a 8 q a

14 4 Use this function: f n () = A n sin n a) What is the best n to match this cylindical potential? +V V 4 f = A sin f n = = A sin zeos:,, 3,, V +V citical points: 4, 3 4, 5 4, 7 4, b) Next intege m whee f m includes these zeos and citical points? R +V V +V 4 f 6 = A 6 sin6 3 4 zeos:, 6, 3,, 3, 5 6,, citical points:, 4, 5, 7, 3 4, 3 4,, V c) Find V(,q) that satisfies Laplace s equation and compae fist two tems with the functions above. Lectue Cal Bombeg - Pof. of Physics 3

15 4 c ( ) = A n n sin n V, ; Bounday: V R, n= ( ) = A n R n sin n n= V (R,) = ±V V ( R, )sin md = & A n R n sin n sin md = A m R m n= { } = V sin n sin md = mn = V m cos m + cos m { cos m + + cos m } m {cos m } + + cos m = (m = ) V ( R, )sin md = 8V m = A m Rm A m = 8V Rm m ; m = 4n, n =,,3, V (, ) = 8V n= 4n R 4n = 4 (m = ) = (m = 3,4,5) = 4 (m = 6) sin (4n ) 4n Lectue Cal Bombeg - Pof. of Physics 4

16 5 Solve Laplace s equation fo a single line chage (adius a), chage density λ. You may need to use Gauss s Law to detemine one of the unknown constants. Gauss s Law LE = L ; E = V (,) = Aln + B; A = E = V = A ˆ Potential on the suface of line chage V (a,) = = ln a + B; B = ln a V (,) = ( )ln + ( )ln a = ( )ln a Find potential of Line Dipole: V d (,) = ln a + ln a = ( ln + ln + ) ( ) + + φ λ +λ ± = ( d cos + d ) ln + = ln + ln ( d cos + d ) ln = ln + ln ( + d cos + d ) V d (,) = ( ln ln + ) & d cos d d = p cos line, whee p line = d Lectue Cal Bombeg - Pof. of Physics 5