Question

THE CORRECT ANSWER IS A. CAN YOU EXPLAIN HOW TO GET THIS ANSWER

(13) An interface between two regions exists on the plane y 0. The permittvity for the region where y <0 is & = 4, while that for the region where y>0 is & =E,where &, is the free-space permittvity. If the electric field for y<0 is E, = (2i+3jy-52) V/m , what is the electric field E, for y>0 if there is no surface charge on the interface? (a) B, =(2i+12j-52) V/m (b) E f8i+3-20?) V/m (c) E -12 Vim (d) None of the others le E (2f 54)V/m

Answer 1

For y < 0 E_1 = 4 E_0 E_1 = 2x + 3y - 5z v/m y > 0 E_2 = E_0 e_2 = ? here y = 0 is the boundary hence y - components are tangential & x, z components are normal components E_tan 1 = 2x - 5z v/m E_N1 = 3y v/m rightarrow from boundary conditions we have E_tan 1 = E_tan 2 implies E_tan 2 = 2x - 5z v/m & D_N 1 = D_N 2 implies E_1 E_N 1 = E_2 E_N2 implies E_N2 = E_1/E_2 E_N 1 = 4 E_0/E_0 (3y) E_N 2 = 12y implies E_2 = E_tan 2 + E_N 2 E_2 = 2x + 12y - 5z v/m therefore option a, E_2 = 2x + 12y - 5z v/m.