A charge of +8 micro-coulombs is placed at the bottom left corner of a rectangle, a charge of -8 micro-coulombs is placed at the top left corner of a rectangle, a charge of +8 micro-coulombs is placed at the bottom right corner of a rectangle, and a charge of +8 micro-coulombs is placed at the top right corner of a rectangle. The rectangle has a height of 20.6 cm and a width of 12.6 cm. (See sketch). What is the angle of the total electric field at the middle of the left side in degrees measured counter-clockwise from the +x axis?
In triangle PCD :
CD = 0.206/2 = 0.103 m , PC = 0.126 m
using Pythagorean theorem
PD = r = sqrt(PC2 + CD2) = sqrt(0.1262 + 0.1032) = 0.163 m
= tan-1(CD/PC) = tan-1(0.103/0.126) = 39.3 deg
PB = r = 0.163 m
r' = PE = PA = 0.206/2 = 0.103 m
E1 = E2 = k q1/r'2 = k q2/r'2 = (9 x 109) (8 x 10-6)/(0.103)2 = 6.8 x 106 N/C
E3 = E4 = k q3/r2 = k q3/r2 =(9 x 109) (8 x 10-6)/(0.163)2 = 2.7 x 106 N/C
Net electric field along X-direction is given as
Ex = - E3 Cos - E4 Cos = - (2.7 x 106) Cos39.3 - (2.7 x 106) Cos39.3 = - 4.2 x 106 N/C
Net electric field along Y-direction is given as
Ey = - E3 Sin + E4 Sin + E1 + E2 = - (2.7 x 106) Sin39.3 + (2.7 x 106) Sin39.3 + 6.8 x 106 + 6.8 x 106 = 13.6 x 106 N/C
angle from positive x-axis is given as
= tan-1(Ey /Ex) = tan-1((13.6 x 106)/(- 4.2 x 106)) = 107.2 deg
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