Chapter 20, Problem 55A

To determine

Electrostatic force between the spheres.

Answer to Problem 55A

  $F=14.4\text{N}$

Explanation of Solution

Given:

Charge on charged sphere, $q=1.2\times {10}^{-5}\text{C}$

Distance between the spheres, $d=0.15\text{m}$

Initially second sphere is neutral.

Formula Used:

From Coulomb’s law,

  $F=\frac{K{q}_1{q}_2}{d^2}$

Where,

  $F$ is electrostatic force

  $K$ is coulomb’s constant

  $q_1$ and $q_2$ are charges and

  $d$ is distance between the charges.

Calculation:

When the charged sphere will come in contact with neutral sphere, equal charge will be distributed on both sphere.

So, $q_1=q_2=\frac{q}{2}=\frac{1.2\times {10}^{-5}}{2}=0.6\times {10}^{-5}\text{C}$

As, $K=9\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2$

On putting the values, the force is

  $\begin{array}{c}F=\frac{9\times {10}^9\times \left(0.6\times {10}^{-5}\right)\times \left(0.6\times {10}^{-5}\right)}{{\left(0.15\right)}^2}\\ F=14.4\text{N}\end{array}$

Conclusion:

Therefore, the electrostatic force between the spheres is $F=14.4\text{N}$ .