Question

1. In class, we derived an expression that allowed us to calculate the electric field at any point along the axis of a dipole. It is also useful to be able to calculate the electric field at any point on a plane that bisects the dipole. This plane will cross through the middle of the dipole and be at right angles to the line that connects the positive and negative charges of the dipole. Show that the electric field at any point in this bisecting plane is given by: 1 P Edipole 41€, 13 where r is the distance from the dipole axis and p is the dipole moment as defined in the textbook. We are assuming that r >> S. (Hint: It is helpful to think about symmetry in solving this problem. Due to symmetry, one of the components of the total electric field will be zero.)

Answer 1

Solution: ܡܰܢ ܘܗܳܢܳܟ݂ Easo Eyes JE, sino h sino لانا,ا و v2 JE, sino | El= |eal - RE . Ęsino = { sino Hence to component are cancel equal & opposite to each other out since they are E = E ..Ep = due to symmetry. & Caso (-2) + Ez Coso tê! The caso ER - 2k9 caso (22+52) i But Colors

î - 2kqxs (8²32) EP, = l But P = 8x25 ) direction - - q to at – K. (4 x 25 (827 32) 312 – kop 7724 523312 Ep = we can is neglect ons the term = se 82 >>7² hince in denominator -KP (8²+0) 312 - - * { where to e

Answer 2

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