Q4Q Figure 22-25 shows four situatio... [FREE SOLUTION]

Chapter 22: Q4Q (page 651)

Figure 22-25 shows four situations in which four charged particles are evenly spaced to the left and right of a central point. The charge values are indicated. Rank the situations according to the magnitude of the net electric field at the central point, greatest first.

Short Answer

Expert verified

The rank of the situations according to the magnitude of the net electric field at the central point is $| E_{2} | > | E_{4} | > | E_{3} | > | E_{1} |$ .

Step by step solution

Understanding the concept of electric field

Considering the concept that the electric field lines move toward the negative charge and lines radially move away from the positive charge, the net electric field can be calculated. If the right direction is said to be positive, the net field at point P can be calculated due to the four arranged charges in the given four situations.

The magnitude of the electric field,

$| E | = \frac{k | q |}{r^{2}}$ (i)

Calculation of the rank according to the magnitude of the electric field

According to the concept and using equation (i), the magnitude of the net electric field at the central point can be given as follows: (considering right direction as positive)

Situation-(1):

$\begin{matrix} E_{1} = \frac{k e}{4 d^{2}} 鈭� \frac{k e}{d^{2}} + \frac{k e}{d^{2}} 鈭� \frac{k e}{4 d^{2}} \\ | E_{1} | = 0 \end{matrix}$

Situation-(2):

$\begin{matrix} E_{2} = \frac{k e}{4 d^{2}} + \frac{k e}{d^{2}} + \frac{k e}{d^{2}} + \frac{k e}{4 d^{2}} \\ | E_{2} | = \frac{5 k e}{2 d^{2}} \end{matrix}$

Situation-(3):

$\begin{matrix} E_{3} = 鈭� \frac{k e}{4 d^{2}} + \frac{k e}{d^{2}} 鈭� \frac{k e}{d^{2}} 鈭� \frac{k e}{4 d^{2}} \\ = 鈭� \frac{k e}{2 d^{2}} \\ | E_{3} | = \frac{k e}{2 d^{2}} \end{matrix}$

Situation-(4):

$\begin{matrix} E_{4} = 鈭� \frac{k e}{4 d^{2}} 鈭� \frac{k e}{d^{2}} 鈭� \frac{k e}{d^{2}} + \frac{k e}{4 d^{2}} \\ = 鈭� \frac{2 k e}{d^{2}} \\ | E_{4} | = \frac{2 k e}{d^{2}} \end{matrix}$