Question

In Fig. 21-15, a central particle of charge-qis surrounded by two circular rings of charged particles. What are the magnitude and direction of the net electrostatic force on the central particle due to the other particles? (Hint:Consider symmetry.)

Short answer

Answer:

The magnitude and direction of the force of the net electrostatic force on the central particle due to the other particles is$\frac{2k{q}^2}{r^2}$that is being directed towards the central charge in downward direction.

.

Step-by-step solution

Understanding the concept of force due to symmetry 

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. the final charge 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 and acting along a straight line at equal distances from the charged particle will cancel out. Thus, only the charge acts on the charge.

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

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

As the charge +2q has positive polarity, it will get attracted by the charge -q at the center in the downward direction towards the charge.

Hence, the magnitude of the force is$\frac{2k{q}^2}{r^2}$ that it is being directed towards the central charge in downward direction.