Exercise 3: Three charges
Topic: Coulomb’s Law
Exercise: Three charged particles with q1 = - 50 nC, q2 = +50 nC и q3 = +30 nC are placed
on the corners of the 5.0 cm x 10.0 cm rectangle shown in Figure 1. What is the net force on
charge q3 due to the other two charges? Give your answer both in component form and as a
magnitude and direction.
Figure 1
Given:
q1 = - 5010-9 C; q2 = + 5010-9 C; q3 = + 3010-9C
r1 = 10 cm = 0.1 m; r = 5 cm = 0.05 m
Model: Model the charged particles as point charges.
Visualize: The pictorial representation of Figure 2 establishes a coordinate system. q1
and q3 are opposite charges, so force vector ⃗⃗⃗ is an attractive force toward q1. q2 and q3 are like
charges, so force vector ⃗⃗⃗ is a repulsive force away from q2. q1 and q2 have equal magnitudes,
but ⃗⃗⃗ has been drawn shorter than ⃗⃗⃗ because q2 is farther from q3. Vector addition has been
used to draw the net force vector ⃗⃗⃗ and to define the angle .
1
Figure 2
Solution:
The question asks for a force, so our answer will be the vector sum ⃗⃗⃗
write ⃗⃗⃗ and ⃗⃗⃗ in component form.
⃗⃗⃗
⃗⃗⃗ . We need to
The magnitude of force ⃗⃗⃗ can be found using Coulomb’s law:
| || |
Where we used r1 = 10 cm. The pictorial representation shows that ⃗⃗⃗ points downward, in the
negative y-direction, so
⃗⃗⃗
To calculate ⃗⃗⃗ , we first need the distance r2 between the charges:
r2 = √
= 11.2 cm = 0.112 m
2
The magnitude of ⃗⃗⃗ is thus:
| || |
This is only a magnitude. The vector ⃗⃗⃗ (Figure 3) is:
⃗⃗⃗
Figure 3
Where angle  is defined in the figure and the signs (negative x-component, positive ycomponent) were determined from the pictorial representation (Figure 2). From the geometry
of the rectangle:
 = tan-1 (
) = tan-1(2) = 63.40
Thus,
⃗⃗⃗
⃗
(
)
Now we can add ⃗⃗⃗ and ⃗⃗⃗ to find:
⃗⃗⃗
⃗⃗⃗
⃗⃗⃗
(
)
(
)
(
)
This would be an acceptable answer for many problems, but sometimes we need the net force
as a magnitude and direction. With angle  as defined in the figure, these are:
3
√
√
=
|
Thus ⃗⃗⃗
|
|
(
|
)
Assess: The forces are not large, but they are typical of electrostatic forces. Even so,
you’ll soon see that these forces can produce very large accelerations because the masses of the
charges objects are usually very small.
Solution with GeoGebra:
The solution with GeoGebra is a graphical and analytical solution of the given problem,
for different values of the distance r1. By moving the slider, the values for the distance r1 are
changed (from 0 to 0.1 m), and thus the values of the distance r2, as well as the values for the
forces.
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