Q2Q - Fundamentals Of Physics

Question

Figure 22-23 shows two square arrays of charged particles. The squares, which are centered on point P, are misaligned. The particles are separated by either d or d/2 along the perimeters of the squares. What are the magnitude and direction of the net electric field at P?

Step-by-Step Solution

Answer

The magnitude of the net electric field at P is$\frac{\mathrm{k}\left(\mathrm{q}\right)}{{\mathrm{d}}^2}$ and the direction of the field is towards left away from the point. 

 

Step 1: Understanding the concept of electric field

The magnitude of the net electric field is given by the total electric field due to all the charges present in the system. Considering the condition that the direction of the field due to positive charge is towards the point (radially outward) and due to negative charge is awayfrom the point (radially inward). So, the net field is given by canceling the magnitudes of equal polar charges.

ōThe magnitude of electric field,$\left|E\right|=\frac{k\left|q\right|}{r^2}$(i)

Step 2: Calculation of the magnitude and direction of the net electric field

From the given figure and as per the concept, the electric field due to charges, $(+6q,+6q)$, $(+3q,+3q)$,$(-2q,-2q)$ , l̥$(-3q,-3q)$,$(-q,-q)$ will be cancelled as they are equal in magnitude, polarity and are placed at equal distances from the  point P. Thus, only the electric field due to charge pair $(-2q,-q)$will be considered.

                                                             

                                                                                                      

 

Now, the magnitude of the electric field at point P will be given using equation (i) as follows: (If the direction of electric field due to charge is considered towards -x-axis that is towards the charge and due to charge is towards +x-axis that is also towards the charge, as direction of the field due to negative charge move inward that is away from the point P)

Thus, the direction of electric field is towards the point P that is towards left (-x-axis). 

Hence, the magnitude of electric field is and the direction of the field is towards left away from the point.

Now, the magnitude of the electric field at point P will be given using equation (i) as follows: (If the direction of electric field due to charge$-2q$ is considered towards -x-axis that is towards the charge and due to charge$-q$ is towards +x-axis that is also towards the charge, as direction of the field due to negative charge move inward that is away from the point P)

$\begin{array}{c}{E}_{net}=-\frac{k\left(2q\right)}{d^2}+\frac{k\left(q\right)}{d^2}\\ =-\frac{k\left(q\right)}{d^2}\end{array}$

Thus, the direction of electric field is towards the point P that is towards left (-x-axis). 

Hence, the magnitude of electric field is$\frac{k\left(q\right)}{d^2}$ and the direction of the field is towards left away from the point.