Introductory Physics Volume Two

1.2 Coulomb’s Law 9
Example
Consider three charges as shown. We want to know the net force on
charge a due to charges b and c.
0.4
y (m)
4.0 mC
r b
2.0 nC
r a
0.2
r c
-3.0 nC
0.0
0.0
We find from the diagram that
0.2 0.4 0.6 0.8 1.0
x (m)
⃗r a = (0.2î + 0.4ĵ)m and q a = 4.0 × 10 −3 C
⃗r b = (0.8î + 0.5ĵ)m and q b = 2.0 × 10 −9 C
⃗r c = (0.7î + 0.2ĵ)m and q c = −3.0 × 10 −9 C
so that we can compute
⃗r a − ⃗r b = (−0.6î − 0.1ĵ)m and q a q b = 8.0 × 10 −12 C 2
⃗r a − ⃗r c = (−0.5î + 0.2ĵ)m and q a q c = −12.0 × 10 −12 C 2
so that
⃗F a = F ⃗ ab + F ⃗ ac
= q aq b
4πɛ 0
⃗r a − ⃗r b
|⃗r a − ⃗r b | 3 + q aq c
4πɛ 0
−0.6î − 0.1ĵ
⃗r a − ⃗r c
|⃗r a − ⃗r c | 3
−0.5î + 0.2ĵ
= 0.072N − 0.108N
| − 0.6î − 0.1ĵ|
3
| − 0.5î + 0.2ĵ| 3
−0.6î − 0.1ĵ
−0.5î + 0.2ĵ
= 0.072N − 0.108N
(0.6 2 + 0.1 2 )
3/2
(0.5 2 + 0.2 2 ) 3/2
= 0.32N(−0.6î − 0.1ĵ) − 0.69N(−0.5î + 0.2ĵ)
= (0.15î − 0.17ĵ)N
The net force is to the right and down at about a 45 ◦ .
If we reworked the previous example but with the charge q a doubled,
then we would find that the force on the charge q a would also be
doubled, ⃗ Fa = 2(0.15î − 0.17ĵ)N. In general the force on a charge is
proportional to the charge. For example, in the previous example
⃗F a = q a (37.5 î − 42.5 ĵ) N C .
So we find that the ratio of the force and charge is a constant.
⃗F a
q a
= (37.5 î − 42.5 ĵ) N C .