Question

A positive point charge q is placed on the +Y-axis at y = a, and a negative point charge —q is placed
on the -y-axis at y = -a. A negative point charge - Q is located at some point on the + x-axis-
(a) In a free-bocy diagram. show the forces that act on the charge -Q.
(b) Find the x- and y-components of the net force that the two charges q and -q exert on -Q. (Your
answer should involve only k, q, Q, a and the coordinate x of the third charge.)
(c) What is the net force on the charge -Q when it is at the origin (x = O)?
(d) Graph the y-component of the net force on the charge - Q as a function of x for values ot x
between -4a and +4a.

Answer 1

The force acting between two charges is determined by Coulomb's law $\vec F_{12}=k\frac {q_1\cdot q_2}{r_{12}^2}\cdot \hat r_{12}$

In this record $\hat r_{12}$ - a unit vector in the direction from the first charge to the second. With the same sign of charges, the force is directed from the first charge to the second and is the repulsive force, for different signs of the charges, the force is directed in the opposite direction and is the force of attraction.

Answer a): In a free-body diagram (figure 1), we show the forces that act on the charge -Q. 

The force with which the first charge exerts on the third is indicated by $F_1=k\frac {qQ}{r^2}$ . It is an attractive force and is directed from the charge $-Q$ to the charge $q$. The distance between two charges is based on the Pythagorean formula $r^2=x^2+a^2$ . The second charge is located symmetrically with the first at the same distance from the third and with the same charge of the opposite sign. So the force with which the second charge exerts on the third is the repulsive force and is directed as shown in the figure 1. The magnitudes of these forces are the same $F_2=k\frac{qQ}{r^2}$ .

Answer b): To find the x- and y-components of the net force that the two charges q and -q exert on -Q we determine $sin\alpha=\frac {a}{r}=\frac {a}{\sqrt {x^2+a^2}}$ and $cos\alpha=\frac{x}{r}=\frac{x}{\sqrt{x^2+a^2}}$ . The horizontal components of the forces $\vec F_1$ and $\vec F_2$ are directed in opposite directions and are equal in magnitude. Therefore, the component of the total force along the direction X is $F_x=0$. The y- components of the both forces $\vec F_1$ and $\vec F_2$ are equals and they sum is

(1) $F_y=2\cdot k \frac{qQ}{x^2+a^2}\cdot sin\alpha=2\cdot k \frac{qQ\cdot a}{(x^2+a^2)^{3/2}}$

Answer c): The net force on the charge -Q when it is at the origin (x = O) from (1) is equals $F_y=2\cdot k\frac{qQ}{a^2}$

Answer d): The y-component of the net force on the charge - Q as a function of x for values of x between -4a and +4a is seen on figure 2. Here we put k=a=q=Q=1.