Problem 71-9
Source: Problem 75 - page 817 - KNIGHT, Randall D. - Book:
Física - Uma Abordagem Estratégica - 2ª Edição - Ed. Bookman - 2009.
The identical small spheres shown in Figure 71-09.1 are loaded with +100 nC
and -100 nC. They are suspended, as shown, in the presence of a module electric
field equal to 100,000 N/C. What is the mass of each sphere?
Solution of the Problem 71-9
The Figure 71-09.2 shows the scheme of the forces acting on the negative charge and by
Figure 71-9.1 , we can easily see that θ = 10°. As the electric field is
directed from right to left in the figure, we conclude that there will be an electrical force,
FE, shifting the negative charge to the right. The wire tension can be decomposed
in two components: one on the x axis, like T senθ and the other on the y axis,
like T cosθ . And on the y axis, pointing down, we have the weight force,
PB.
In this problem it is important to understand that the electrical force, FE, is the force
resulting from two forces to which the charge B is subjected. One is the force of attraction
between charges, as they have opposite polarities; the other, is the force due
the presence of the electric field. Before we calculate these forces, let's calculate the value
the distance x between charges. Like this
x = 2 x 0.50 x sen 10° = 0.174 m
Now, to calculate the attraction force between the charges we are going to use Coulomb's law ,
according to eq. 71-04. Making the numerical substitution, we have
Fch = 9 x 109 x (100 x 10-9)2 / (0.174)2
Performing the calculation, we find
Fch = 2.973 x 10-3 N
On the other hand, the force to which the negative charge is subjected due to the
electric field, is given by eq. 71-02. So, making the numerical substitution, we have
Ffield = 100 x 10-9 x 105 = 0.01 N
With this data we can calculate the electrical force, FE, because
FE = Fch - Ffield = 7.027 x 10-3 N
The value of FE allows us to find the value of the tension in the wire. For this,
we will perform the algebraic sum on the x axis, as shown in Figure 71-09.2, because
we know that this sum must be equal to zero.
Working algebraically, we find
T = FE / sen θ = 7.027 x 10-3 / sen 10°
Performing the calculation, we find
T = 4.05 x 10-2 N
Recalling that PB = m g and making the algebraic sum of forces on the y axis,
which must also be equal to zero, we find