91Ó°ÊÓ

Figure 21-18 shows four situations in which particles of charge$+q$or $-q$are fixed in place. In each situation, the particles on the xaxis are equidistant from the yaxis.

As the magnitude of the force depends on the magnitudes of charges and the separation value of the charges from the middle particle, the force value will be equal for the case of equal charges and equal distance from the particle. Considering the concept of like charges repel and opposites attract, we can get the direction of the x and y-components of the force acting on the particle. Thus, using this concept the net force and its direction can be determined.

Formula:

The magnitude of the electrostatic force due to the two charges $\left|F\right|=\frac{k\left|q_1\right|\left|\right|q_2|}{r^2}$,. (i)

Here, the forces from the two charges +q and +q acting on the middle particle +q are equal in magnitude as the charges have equal magnitudes and they are equally displaced from the middle particle.

Hence, the magnitudes of forces are equal

Let the distance of each particle from the particle be.

Here, the direction of the two forces will have y-components contribution to the net force as being equal in magnitude and opposite in direction along the x-axis at the middle particle; their x-components cancel out.

Thus, the magnitude of the net force is given using equation (i) as follows:

$$\begin{gathered} \left|F_{net}\right|=\frac{k\left|+q\right|\left|+q\right|}{r^2}\left|\sin \mathrm{θ}\right|+\frac{k\left|+q\right|\left|+q\right|}{r^2}\left|\sin \mathrm{θ}\right| \\ =\frac{2k{q}^2}{r^2}\left|\sin \mathrm{θ}\right| \\ =2F\left|\sin \mathrm{θ}\right| \end{gathered}$$

Now, the sine value for the case of magnitude will always be $\left|\sin \mathrm{θ}\right|=\left\{0,1\right\}$.

Thus, the net force is less than $2F$.

From the discussion in part (b), it is clear that the x-components of the forces cancel out.

From the discussion in part (b), it is clear that the y-components of the forces add up to give the net force.

From part (b) calculations, it is made clear that the net force direction is due to the adding y-components of the force.

From part (b), it can be seen that the force due to the charges are only acting in the downward direction on the particle.

Hence, the direction of the net force is negative y-axis.

From the above diagram according to the concept, the direction of the net force is towards the positive y-axis.

From the above diagram according to the concept, the direction of the net force is towards the negative x-axis.

From the above diagram according to the concept, the direction of the net force is towards the positive x-axis.