Chapter 20, Problem 35A

To determine

Force on each particle.

Answer to Problem 35A

  $F=90\text{N}$ towards the other charge.

Explanation of Solution

Given:

Negative charge, $q_1=-1.5\times {10}^{-5}\text{C}$

Positive charge, $q_2=1.5\times {10}^{-5}\text{C}$

Distance between the charges, $d=15\text{cm}=0.15\text{m}$

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:

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

On putting the values,

  $\begin{array}{c}F=\frac{9\times {10}^9\times \left(-1.5\times {10}^{-5}\right)\times \left(1.5\times {10}^{-5}\right)}{{0.15}^2}\\ =-90\text{N}\end{array}$

As, $F=-90\text{N}$

Negative means the force is attractive in nature.

So, force on each particle is $F=90\text{N}$ towards the other charge.

Conclusion:

Therefore, the electrostatic force on each particle is $F=90\text{N}$ towards the other charge.