[Giffiths Ch.-] 8//8, :am :am, Useful fomulas V ˆ ˆ V V V = + θ+ φ ˆ and v = ( v ) + (sin θvθ ) + v θ sinθ φ sinθ θ sinθ φ φ. (6%, 7%, 7%) Suppose the potential at the suface of a hollow hemisphee is specified, as shown in the figue, whee V ( a, θ ) =, V( b, θ ) = V(cosθ 5cosθsin θ), V (, π ) =. V is a constant. (a) Show the geneal solution in the egion b a. Detemine the potential in the egion b a, using the bounday conditions. (c) Calculate the electic field in the egion b a. [Hint: P( x) =, P( x) = x, P( x) = (x )/, and P ( x ) = (5 x x )/.]. (%, %) A point chage q is situated a lage distance fom a neutal atom of polaizability α. (a) Find the induced dipole moment of the atom p. Find the foce between them (attactive o epulsive).. (%, %) A point chage q is situated at distance a fom the cente of a conducting sphee of adius. The sphee is maintained at the constant potential V. (a) Find the position and the value of the image chage. Veify that the tangential component of the electic field is zeo thoughout on the suface of the metal sphee. [Hint:. use the notations shown below.. Assume q lays on the z-axis] - -
4. (8%, 6%, 6%) Conside two infinite paallel metal plates sepaated by a distance s ae at potential and V as shown in the figue.. (a) Use Poisson s equation to find the potential V in the egion between the plates whee the x space change density is ρ = ρ. d Find the electic field E in the egion between the plates. (c) Use the bounday condition to detemine the chage densities on the plates. [Note: The electic field inside the metal plate is zeo E inside =.] V= V=V x= x=d 5. (8%, 6%, 6%) Conside a hollowed chaged sphee with adius and unifom chage density ρ as shown in the figue. The inne adius of the spheical cavity is /. (a) If the obseve is vey fa fom the chaged sphee, find the multiple expansion of the potential V in powe of / Find the dipole moment p. (c) Find the electic field E up to the dipole tem. [Note: Specify a vecto with both magnitude and diection.] ρ - -
. (a) (i) V ( a, θ ) = Bounday condition (ii) V ( b, ) V (cos 5cos sin ) V (5cos cos ) V P (ii) V (, θ = π ) = θ = θ θ θ = θ θ = ( + ) Geneal solution V(, θ) = ( A + B ) P(cos θ) = ( + ) + B.C. (i) V( a, θ) = ( Aa + Ba ) P(cos θ) = B = Aa = ( + ) B.C. (ii) Vb (, θ) = ( Ab + Bb ) P(cos θ) = VP (cos θ) = 4 Ab + Bb = V A = B = = π 4 B.C. (iii) V(, θ = ) = ( A + B ) P() = A = B = except =, 4 4 7 Vb Vba A = and B 7 7 = 7 7 Compaing the coefficiency, fo,,, 4,5,... b a b a V 4 V 4 7 4 5cos θ cosθ V(, θ ) = b ba 7 7 7 7 b a b a (c) V 4 V 4 7 4 5cos θ cosθ V(, θ ) = b ba 7 7 7 7 b a b a V V E= V = ˆ θˆ θ 6V 4 8V 4 7 5 5cos θ cosθ = b ba ˆ 7 7 7 7 b a b a 6V 4 8V 4 7 5 5cos θsinθ sinθ + b ba 7 7 7 7 ˆθ b a b a. (a) q The electic field of a point chage E= ˆ αq The induced dipole moment p= αe= ˆ - -
αq The total electic static enegy U = ( ) pe = ( )( ) 4 Note : ( ) comes fom the fact that the dipole moment p is induced by E. q The foce bewteen is F= U = α( ) ˆ attactive. 5 4 πε The diection of the induced dipole p is in line with the electic field E geneated by the chage q..(a) Assume the image chage q is placed at a distance b fom the cente of the sphee. It is equipotential on the suface of a gounded sphee. Using two bounday conditions at P and P q q At P : ( + ) = b a two equations and two unknowns ( q and b) q q At P : ( + ) = + b a+ b= (position), q = q(value of the image chage) a a The potential outside the sphee when V = q q V( ) = +, whee b= and q = q ˆ ˆ bz az a a The potential outside the sphee when V= V V () whee 4 πε V q q ˆ ˆ bz az = + + bzˆ = ( θ + θ b = b θ + b sin ( cos ) ( cos ), a z ˆ = ( sin θ + ( cos θ a) = ( acos θ + a ) On the suface of the metal sphee, E V V = V () = ˆ ˆ θ E θ V q q = = + θ 4 πε θ ( bcos θ + b ) ( acos θ + a ) q ( bsin θ) q( asin θ) = + / / 4 πε ( b cos θ + b ) ( a cos θ + a ) sinθ qb qa = + / / 4 πε ( bcos θ + b ) ( acos θ + a ) - 4 -
q q qb = a a = a / ( + bcos θ b ) ( cos ) ( a acos ) a a a qa 4 / / θ + θ + = ( acos θ + a ) / sinθ qa qa Eθ (@ = ) = + / / = 4 πε ( acos θ + a ) ( acos θ + a ) 4.(a) ρ x dv = V( x) = + c x+ c dx εd 6εd Use the bounday conditions to detemine the coefficiencies. V() = c = ρd V ρd V( d) = V = + cd c = + 6ε d 6ε ρx V ρd V( x) = + ( + ) x 6ε d d 6ε ρx V ρd V( x) = + ( + ) x 6ε d d 6ε ρ x ρ ρ x V d E= V( x) = ( + )ˆ x ε d d 6ε (c) V ρd E ˆ x= = ( + ) x εv ρd σ ( 6 x ε Eoutside E d ε = = inside ) = ( + ) d 6 = ρd V Ex= d ( )ˆ ε x ρd εv σx= d = ε( Eoutside Einside ) = ( ) d d 5. (a) Conside this poblem as two chage sphees, one with chage density ρ the othe with opposite chage density ρ. 4π 4π Vbig = ( ρ ) and Vsmall = ( ρ ( ) ) = ( + ( ) cos θ + ) - 5 -
Using the pinciple of supeposition, we find, π 4π 4 V = ( ρ ) ( ρ ( ) )( + ( )cos θ + ) 7 4π 4π 4π = ( ρ ) ( ρ ( ) )( )cos θ +, let Q= ρ 8 7Q Q = ( )cosθ + 8 8 4π Q= ρ 7Q Q V = ( )cosθ + 8 8 The fist tem is the monopole tem and the second tem is the dipole tem. Q So the dipole moment p= zˆ. 6 (c) 7Q Q V = ( )cosθ + 8 8 V V ˆ V E= V = ˆ θ φˆ θ sinθ φ 7Q p p = cosθ ˆ sinθθˆ 8-6 -