Chapter 20, Problem 36A

To determine

Value of each charges.

Answer to Problem 36A

  $q_1=q_2=3.2\times {10}^{-19}\text{C}$

Explanation of Solution

Given:

Two identical charges with repulsive force $F=6.4\times {10}^{-9}\text{N}$

Distance between the charges, $d=3.8\times {10}^{-10}\text{m}$

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,

  $6.4\times {10}^{-9}=\frac{9\times {10}^9\times \left(q_1\right)\times \left(q_2\right)}{{\left(3.8\times {10}^{-10}\right)}^2}$

As both charges are identical,

So, let $q_1=q_2=q$

Therefore,

  $\begin{array}{c}6.4\times {10}^{-9}=\frac{9\times {10}^9\times \left(q\right)\times \left(q\right)}{{\left(3.8\times {10}^{-10}\right)}^2}\\ q^2=\frac{\left(6.4\times {10}^{-9}\right){\left(3.8\times {10}^{-10}\right)}^2}{9\times {10}^9}\\ q=\sqrt{\frac{\left(6.4\times {10}^{-9}\right){\left(3.8\times {10}^{-10}\right)}^2}{9\times {10}^9}}\\ =3.2\times {10}^{-19}\text{C}\end{array}$

Conclusion:

Therefore, the charges are $q_1=q_2=3.2\times {10}^{-19}\text{C}$ .