Chapter 20.2, Problem 9PP

To determine

Force between the charges.

Answer to Problem 9PP

  $F=1.6\times {10}^4\text{N}$ attractive.

Explanation of Solution

Given:

Negative charge, $q_1=-2\times {10}^{-4}\text{C}$

Positive charge, $q_2=8\times {10}^{-4}\text{C}$

Distance between the charges, $d=0.3\text{m}$

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

On putting the values,

  $\begin{array}{c}F=\frac{9\times {10}^9\times \left(-2\times {10}^{-4}\right)\times \left(8\times {10}^{-4}\right)}{{0.3}^2}\\ =-1.6\times {10}^4\text{N}\end{array}$

As, $F=-1.6\times {10}^4\text{N}$

Negative means the force is attractive in nature.

Conclusion:

Therefore, the electrostatic force between the charges is $F=1.6\times {10}^4\text{N}$ and it is attractive in nature.