In Fig. 22-24, two particles of charge $\mathbf{-}\mathit{q}$are arranged symmetrically about the y axis; each produces an electric field at point Pon that axis. (a) Are the magnitudes of the fields at Pequal? (b) Is each electric field directed toward or away from the charge producing it? (c) Is the magnitude of the net electric field at Pequal to the sum of the magnitudes Eof the two field vectors (is it equal to 2E)? (d) Do the x components of those two field vectors add or cancel? (e) Do their y components add or cancel? (f) Is the direction of the net field at P that of the canceling components or the adding components? (g) What is the direction of the net field?

The whole concept of the problem is based on the direction of the individual fields due to the equally placed negative charges on the x-axis. As per the concept that the electric field lines move towards the negative charge, we can draw the x and y-components of the electric field. Thus, by canceling and adding the equal and opposite terms according to their polarity and magnitude, we can determine the direction and magnitude of the net electric field.

As both the charges have equal magnitudes and are at equal distances from point P, the contribution from both the charges is equal at the given point P according to equation (i).

Hence, the magnitudes of the fields at P are equal.

As the polarity of the two given charges are negative. Thus, according to the concept, the electric field due to the negative charge is always directed towards the negative charge.

Hence, the direction of the fields is towards the charges.

Now, the electric fields due to both the charges acting on the point P have two components. Thus, due to their similar polarities and as they are located on the x-axis, their x-components cancel each other and their y-components add on to give the net electric field at the point.

Thus, the net electric field is given using equation (i) as follows:

$E_{net}=-2E\sin \theta$

Hence, the magnitude of the net field is not equal to $2E$.

From the calculations and drawn figure in part (c), we can say that their x-components being equal in magnitude and opposite in direction cancel each other.

Hence, their x-components cancel out.

From the calculations and the figure was drawn in part (c), the net electric field is due to the contribution of their y-components only.

Hence, the field is due to the addition of the y-components.

From the calculations and the figure was drawn in part (c), the field direction is downwards at point P,which is toward the negative y-axis.