Chapter 16, Problem 73GP

To determine

To Calculate: The charge at the center of the square.

Whether there is stable or unstable equilibrium.

Answer to Problem 73GP

  $-7.66\times {10}^{-6}\text{C}$

There is unstable equilibrium.

Explanation of Solution

Given:

The charge at the corner of the square is $8.0\mu \text{C}$ and the side is 9.2 cm.

Formula used:

The charge is calculated by using the force formula,

  $F=k\frac{q_1{q}_2}{r^2}$

Where, r is the separation distance, k is the proportionally constant and ${\text{q}}_{\text{1}}\text{and}{\text{q}}_{\text{2}}$ are the testing charges.

The net force in each direction due to the rest of the charges are given as,

  ${\left(F_{net}\right)}_{x-direction}=k\frac{Q^2}{d^2}+k\frac{Q^2}{d^2}+k\frac{Q^2}{d^2}-k\frac{Q^2}{d^2}$

Where, k is proportionally constant, Q is the charge placed at the corner and d is the separation distance.

Calculation:

Consider the charge at the center which is placed at the center of the square but having the same magnitude as that of positive charge. The separation distance between the charges is 9.2cm. According to the symmetry, the net charge on one of the corner charges of square be zero and net force on each other corner charge will also be zero. Therefore, the force is

  $F_{41}=k\frac{Q^2}{d^2}\Rightarrow {\left(F_{41}\right)}_x=k\frac{Q^2}{d^2}\text{and}{\left(F_{41}\right)}_y=0$

  $F_{41}=k\frac{Q^2}{2d^2}\Rightarrow {\left(F_{42}\right)}_x=k\frac{Q^2}{2d^2}\cos {45}^0=k\frac{\sqrt{2}{Q}^2}{4d^2}\text{and }{\left(F_{42}\right)}_y=k\frac{Q^2}{2d^2}\cos {45}^0$

  $\begin{array}{l}{F}_{43}=k\frac{Q^2}{d^2}\Rightarrow {\left(F_{43}\right)}_x=0\text{and}{\left(F_{41}\right)}_y=k\frac{Q^2}{d^2}\\ F_{4q}=k\frac{qQ}{d^2/2}\Rightarrow {\left(F_{4q}\right)}_x=k\frac{2qQ}{d^2}\cos {45}^0=k\frac{\sqrt{2}qQ}{d^2}={\left(F_{4q}\right)}_y\end{array}$

The net force is calculated by using the above formula,

  $\begin{array}{l}{\left(F_{net}\right)}_{\text{x-direction}}=k\frac{Q^2}{d^2}+k\frac{Q^2}{d^2}+k\frac{Q^2}{d^2}+k\frac{Q^2}{d^2}=0\\ =k\frac{Q^2}{d^2}+k\frac{\sqrt{2}{Q}^2}{4d^2}+0-k\frac{\sqrt{2}qQ}{d^2}=0\end{array}$

So, the charge at the center is,

  $q=Q\left(\frac{1}{\sqrt{2}}+\frac{1}{4}\right)=7.66\times {10}^{-6}C$

This is an unstable equilibrium because of the center charge at slightly displaced towards right then it would closer to the right charge than the left and would attract more towards the right side. Similar displacement is observed at positive charge.

Conclusion:

So, the charge which is placed at the center having same magnitude but different charge. So, the charge is $-7.66\times {10}^{-6}\text{C}$ .