Problem 71-6
Source: Entrance Exam FUVEST - SP - 2019.
Three small spheres charged with positive charge, Q, occupy the vertices of a triangle,
as shown in Figure 71-06.1. Inside the triangle, another small sphere is affixed,
with negative charge, -q. The distances of this charge to the other three can be obtained from
the figure.
Calculate the resulting electrical force (in module, direction and direction) on the charge q,
knowing that:
Q = 2 x 10-4 C,
q = - 2 x 10-5 C, x = 6 m, and
K = 9 x 109 N m2/C2
Solution of the Problema 71-6
We must first determine the value of m. Note that m is the hypotenuse of the right
triangle formed by the two charges Q, at the base of the equilateral triangle,
and the charge - q .
The sides of the right triangle measure x. We easily see that the angle formed between the
cateto x and the hypotenuse m is 45°. So we can write that
m = x / sen 45° = 6 / 0.707 = 8.5 meters
Now knowing the distance between the charges, we can calculate the forces between the charges. Note that
all forces in - q will be attractive, since the charges have opposite polarities.
The magnitude of the force of attraction between the charge that is located at the upper apex of the triangle,
which we call FQs, and the charge - q is
FQs = 9 x 109 2 x 10-4
2 x 10-5 / 62
Performing the calculation we find
FQs = 36 / 36 = 1 N
Let's calculate the magnitude of the force between the charges that are located at the
base of the equilateral triangle, which we call FQb, and the charge
- q .
Note that the two charges at the base of the triangle will exert, in modulus,
the same force on the charge - q . It will only change the direction and direction.
So, we have:
FQb = 9 x 109 2 x 10-4
2 x 10-5 / (6 √2)2
Performing the calculation we find
FQb = 36 / 72 = 0.5 N
To calculate the resulting force acting on the charge - q , we will base
in the diagram shown in Figure 71-06.2
Notice that the force FQs points upwards towards the upper vertex of the triangle.
The other two forces, FQb, make an angle of 45° with the negative vertical axis.
The sum of the two forces, represented by Fb, is their vectorial sum,
or Fb = 2 FQb cos 45° = 0.7 N. In this case, it points in the opposite direction
to the positive vertical axis. And as a result, the resulting force, FR, is the subtraction
between FQs and Fb, or
FR = FQs - Fb = 1 - 0.7 = 0.3 N
Like FQs > Fb, then the resulting force points in the direction
positive of the vertical axis, that is, it makes an angle of 90° with the horizontal.