Show that this component is 1 0 0 n1 p e 0

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , returning to vector notation and collecting terms, we obtain the answer: (0) ∂ Ej 1 (x ) Qjk + ∇ F = q E (0) + ∇ p · E (0) 6 ∂xk 0 j,k 0 (b) Repeat the calculation of part a for the total torque. For simplicity, evaluate only one Cartesian component of the torque, say N1 . Show that this component is 1 ∂ ∂ (0) (0) N1 = p × E (0) + − Q 2 j Ej Q 3 j Ej 3 ∂x3 ∂x2 1 j j 0 2 As above N= d3 xρx × E (0) (0) N1 = d3 xρx × E (0) (0) 1 = p × E (0) (0) 3 1 + ∂E3 − ∂x′ j ∂ (0) Q 2 j Ej − ∂x2 j 1 ∂ 3 ∂x3 j ∂E2 ∂x′ j d3 xρx3 xj d3 xρx2 xj + j (0) Q3j Ej + ... x′ =0 x′ =0 Jackson 4.9 A point charge q is located in free space a distance d from the center of a dielectric sphere of radius a (a < d) and dielectric constant ǫ/ǫ0 . (a) Find the potential at all points in space as an expansion in spherical harmonics. Due to spherical symmetry, we can arrange our coordinate system such that the sphere is centered at the origin, the charge is located at z = d on the z axis. The problem is then azimuthally symmetric, and we can expand in Legendre polynomials. For r < a Φin = q 4πǫ Al l rl Pl (cos θ) al and for r > a q 1 + Φ0 4πǫ0 |x − dz | ˆ l r< q a = + Bl l+1 4πǫ0 r r> l Φout = l+1 Pl (cos θ) Macthing parrallel and perpendicular boundary conditions on electric field (D) at the surface r = a q 4πǫ l q Al ′ P (cos θ) sin θ = al 4πǫ0 Al = q 4π l l l Bl al − 1 Pl′ (cos θ) sin θ + dl+1 a ǫ a + Bl ...
View Full Document

Ask a homework question - tutors are online