Question

Find the electric field at the location of q_a in the figure below, given that q_b = q_c = q_d = +1.85 nC, q = ?1.00 nC, and the square is 11.5 cm on a side. (The +x axis is directed to the right.)

magnitude	N/C
direction

Answer 1

let a be the side of the squre a = 0.115 m

distance of qd from qa = rd = sqrt(a^2+a^2) = sqrt(2)*a m

distance of q from qa = r = rd/2 = a/sqrt(2) m

Ec = Ecy = +k*qc/a^2 = (9*10^9*1.85*10^-9)/(0.115^2) = +1258.97 N/C

Eb = Eax = -k*qb/a^2 = -(9*10^9*1.85*10^-9)/(0.115^2) =

-1258.97 N/C

Edx = -k*qd*cos45/rd^2 = (9*10^9*1.85*10^-9*cos45)/
(2*0.115^2) = -445.12N/C

Edy = +k*qd*sin45/rd^2 = (9*10^9*1.85*10^-9*sin45)/

(2*0.115^2) = -445.12N/C

Eq due to q

Eqx = +k*q*cos45/r^2 = (9*10^9*1*10^-9*cos45*2)/

(0.115^2) = +962.42 N/C

Eqy = -k*q*sin45/r^2 = (9*10^9*1*10^-9*sin45*2)/

(0.115^2) = -962.42 N/C

net x component

Ex = Ebx + Ecx + Edx + Eqx

Ex = -1258.97 + 0 - 445.12 + 962.42

Ex = -741.67 N/C

net y component

Ey = Eby + Ecy + Edy + Eqy

Ey = 0 + 1258.97 + 445.12 - 962.42

Ey = 741.67 N/C

magnitude E = sqrt(Ex^2+Ey^2) = 1048.88 N/C <-------answer

direction = tan^-1(Ey/Ex) = tan^-1(1) = 45