Question

In Fig. 21-19, a central particle of charge -2qis surrounded by a square array of charged particles, separated by either distance dor d/2 along the perimeter of the square. What are the magnitude and direction of the net electrostatic force on the central particle due to the other particles? (Hint:Consideration of symmetry can greatly reduce the amount of work required here.)

Short answer

Answer:

The magnitude and direction of the net electrostatic force on the central particle due to the other particles is $\frac{6k{q}^2}{d^2}$that is directed towards right along the positive x-axis

Step-by-step solution

The given data

The distance between the charges particles is either d or d/2 along the perimeter.

Understanding the concept of force

Due to symmetrical conditions of equal polarities, magnitudes, and distance from the charge on which the force is acting on, the forces for these symmetrical pairs cancel out. Thus, using this concept, we can cancel out the symmetrical pairs and check the final charge that has a contribution to the net force value.

Formula:

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

Calculation of the magnitude and direction of the net electrostatic force

As per the concept of symmetry, all the pairs with equal polarities and magnitude and at equal distances from the charge particle -2q will cancel out. Thus, only charge +3q acts on the charge.

So, the magnitude of the net electrostatic force acting on the charge can be given using equation (i) as follows:

$$\begin{gathered} \left|F_{net}\right|=\frac{k\left|+3q\right|\left|-2q\right|}{d^2} \\ =-\frac{6k{q}^2}{d^2} \end{gathered}$$

Now, as charge +3q has positive polarity, it will get attracted by the charge -2q at the centre.

Hence, the magnitude of the force is$\frac{6k{q}^2}{d^2}$ that is directed towards right in the positive x-axis direction.