Distribution of induced charge

Transcription

1 E&M Lecture 10 Topics: (1)Distribution of induced charge on conducting plate (2)Total surface induced charge on plate (3)Point charge near grounded conducting sphere (4)Point charge near floating conducting sphere (5)Point charge near isolated conducting sphere (6)Point charge near LIH dielectric block +Q

2 Distribution of induced charge Induced charge is related to the outward field at the surface: Find E using image charge system: E + = E = E = 2 Q r 2 Q 2QD cosθ = 2 r r 3 σ ind = ε o E = 2QD 4πr 3 E = σ ind ε o E r σ ind = ε o E E + E - +Q θ -Q D (-ve and varying with r)

3 Total induced charge ds Introduce parameter s r s s s = r 2 D 2 D σ ind has azimuthal symmetry: consider elemental annulus, radius s, thickness ds Q ind = σ ind 2πsds= 0 0 2QD 4π( s 2 + D 2 ) 3 2 2πsds = QD 2 0 ds ( 2 + D 2 ) ( s 2 + D 2 ) = QD ( s 2 + D 2 ) = Q

4 Total induced charge: implications 2 conclusions from the result: Q ind = Q (1)Induced charge equals the negative of original point charge - trivially true in this case only! (2)Induced charge equals the image charge - generally true! Consider Gauss s Law, concept of enclosed charge? Important exclusion of general value: Must not try to determine E in the region of image charge! In this case (behind infinite conductor) it is zero, which is not the answer the image charge would yield! a more complex example

5 Expect image charge will be a point charge on centre line, left of centre of sphere, magnitude not equal to Q, call it Q Point charge near grounded conducting sphere +Q By comparison with previous example: D (1)Distance D to centre of symmetry, radius a (2)Image (charge) in convex mirror? (3)-ve induced charge predominantly on side facing +Q (4)Boundary condition, zero potential on sphere surface a

6 Point charge near grounded conducting sphere Q P 2 Q P 1 b D Q distance b from centre, φ=0 at symmetry points P 1 and P 2 φ = 1 Q D + a + Q = 0 ( at P ) 1 a + b φ = 1 Q D a + Q = 0 ( at P ) a b 2 solving b = a2 D and Q = a D Q = Q ind by Gauss

7 Point charge near floating conducting sphere Q Q Q V On its own, floating the sphere at V relative to ground results in uniform +ve charge density over the surface. In the presence of Q, induced -ve charge predominantly on left; this complex system easily solved by 2 image charges: Q = a D Q at b= a2 D and Q = av Q ind = Q + Q by Gauss at centre

8 Point charge near isolated conducting sphere Q Q Q With no connection to ground, the sphere is at an unknown non-zero potential φ; easily solved by same 2 image charges: the potential is still determined by Q but in this case, the sphere is overall neutral: Q +Q =0 φ = ( ) Q a = Q a = a D Q a = Q D Q ind = Q + Q = 0 by Gauss (same potential as if sphere was absent!)

9 Point charge near LIH dielectric block contrasting E field patterns note distorted radial outside but proper radial inside

10 Point charge near LIH dielectric block 2 differences: (1) the appropiate boundary condition is D n at block surface (2) Need to evaluate E outside and inside block, requiring two separate image charge arrangements (results only): (see Lorrain,Corson & Lorrain pp ) Outside: remove dielectric block and locate image charge Q a distance D behind Q = ε 1 r ε r +1 Q Inside: remove dielectric block and replace original point charge Q by Q Q = 2 ε r +1 Q (image method good for Force problems!)