Question

Assume that a flat sheet of bakelite has a common boundary with a flat sheet of polystyrene. In the polystyrene there is a uniform electric field with magnitude E= 1 Volt per meter directed toward the boundary with inclination of 45 degrees relative to a normal at the boundary. What is the electric field in the bakelite (both magnitude and angle of inclination with respect to the normal vector)?

Answer 1

This question is based on boundary value problem in electric field.
| Solution : Perpendicular : Bakelite Ep2 Gay Boundary Tangential Let, tolystyrene €ry Fiq-2 " The relative permitivity of Polystyrene - Epl o the relative Bermitivity of Bakelite = Ep 2 It b based on boundary Value Poroblem of electric field. From this we know ANTE ON2-) DNIE Normal component of electric flex density in medium I a DN2 that in mediuma. Eti - Et2 —42) Ext= tangential component of electric field Component in medium 1 a Et2 that in medium 2.

Now, Forrom Fig1. Eti = E COS 45° = COS 45°41) ENI = Esin 45°- Sin 45° [Ezt V/m] We know, D=EE SO, DHI= Em EoENI ... => DNI: Emi to Sina 5 (2) . Now As fer boundary coom Condition, . Dni = Dn2 = Epi Eo Sin 45° So, Enrz = Erigo Sin 45º = Ery sin45° Now, Edi = £42 = COS 45° Erz Epz Eo so Ez EN20 - Ez = J ENE+E * = 5 ErP of sin 45+ Co45 Angle with Aromal, tano - Et? EN2

So, tano = cos 45o Eero sin45 Since cos 45= sin 45. So, tano. En solo-tan" (en) |