Question

What are the strength and direction of the electric field at the position indicated by the dot in the figure below, in which q = 2 nC, d = 12 cm, and the positive x-axis points to the right?

Answer 1

The strength or magnitude of the field at a given point
is defined as the force that would be exerted on a
positive test charge of 1 coulomb placed at that point;
the direction of the field is given by the direction of
that force.
E = F/Q = kQ/r²
in Newtons/coulomb OR volts/meter
k = 8.99e9 Nm²/C²


E1 due to q1 = kq/0.05² = 400kq
q = 8e-9
E1x = 400kq
E1y = 0

distance = √(12² + 5²) = 13 cm = 0.13 m
E2 due to q2 = kq/0.0169 = 59kq
angle = arctan 12/5 = 67.38º
E2x = –80kq cos 67.4 = –30.74kq
E2y = –80kq sin 67.4 = –73.85kq

resultant
Rx = 400kq –30.74kq = 369.26kq
Ry = –73.85kq
R = √(Rx² + Ry²) = 376.57kq = 6778.26 v/m
θ = arctan (Ry/Rx) = arctan (73.85/369.26) = 11º 18’
from the diagram, that is 11º 18’ S of E
or from the +x axis, that is 360–11.18 = 348.82º