Chapter 20, Problem 39A

To determine

Value of charges on two spheres.

Answer to Problem 39A

  $q=0.515\mu \text{C}\text{and 3}q=\text{1}\text{.545}\mu \text{C}$

Explanation of Solution

Given:

The given figure is,

Two chargesq and 3q with repulsive force $F=0.28\text{N}$

Distance between the charges, $d=16\text{cm}=0.16\text{m}$

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,

  $\begin{array}{l}0.28=\frac{9\times {10}^9\times \left(q\right)\times \left(3q\right)}{{\left(0.16\right)}^2}\\ 0.28=\frac{9\times {10}^9\times 3q^2}{{\left(0.16\right)}^2}\\ \Rightarrow q^2=\frac{\left(0.28\right){\left(0.16\right)}^2}{9\times {10}^9}\\ q=\sqrt{\frac{\left(0.28\right){\left(0.16\right)}^2}{3\times 9\times {10}^9}}\\ =0.515\times {10}^{-6}\text{C}\end{array}$

And

  $\begin{array}{l}3q=3\times 0.515\times {10}^{-6}\text{C}\\ \text{=1}\text{.545}\times {\text{10}}^{-6}\text{C}\end{array}$

Conclusion:

Therefore, the charges on spheres are, $q=0.515\mu \text{C}\text{and 3}q=\text{1}\text{.545}\mu \text{C}$ .