Introductory Physics Volume Two

1.7 More Examples 29
-4q +2q -q
E -1
E -4
E +2
2a
a
a
In the diagram, the electric field due to the −4q charge has been labeled
E −4 , etc. The electric field at the point indicated is just the
superposition of the individual electric fields:
E = E +2 − E −1 − E −4 = 1 (2q)
4πɛ 0 (2a) 2 − 1 q
4πɛ 0 a 2 − 1 (4q)
4πɛ 0 (4a) 2
−→ E = − 3 q
16πɛ 0 a 2 .
Example
Using the configuration of charges in the previous example, determine
the electric field a vertical distance a above the +2q charge.
First, let’s add a coordinate system, since this problem will involve
vectors in two dimensions:
y
E +2
(0, a)
E -4
E -1
-4q θ 1
+2q θ 2 -q
x
(-2a, 0) (0, 0) (a, 0)
The angles θ 1 and θ 2 can be computed using some geometry:
cos θ 1 = √ 2 sin θ 1 = √ 1
5 5
cos θ 2 = √ 1 sin θ 2 = √ 1
2 2
To compute the electric field at the point (0, a), we add the individual
electric fields as vectors: