Chapter 16, Problem 22P

To determine

To find: Number of electron transfer from one ball to the other.

Answer to Problem 22P

The number of electrons transferred $1.03125\times {10}^{12}$ .

Explanation of Solution

Given:

A charge $Q$ is transferred from an initially uncharged plastic ball

Identical ball $12\text{cm}$ away

The force of attraction is then 17mN

Formula used:

The Coulomb’s force is given as,

  $F=\frac{k{Q}_1{Q}_2}{r^2}$

Calculation:

The force of attraction $(F)=17\text{mN}$

Distance between the 2 balls $(r)=12\text{cm}$

The initial uncharged ball acquires a charge $+Q$ and the other ball acquires a charge $-\text{Q}$

Hence $F=\frac{k\left|Q\right|\left|Q\right|}{r^2}$

Or, $17\times {10}^{-3}=\frac{\left(9\times {10}^9\right)|Q{|}^2}{{0.12}^2}$

Or, $\left|Q\right|=1.65\times {10}^{-7}\text{C}$

Charge on a single electron $e=-1.6\times {10}^{-19}\text{C}$

Calculate the number of electrons,

  $\begin{array}{l}n=\frac{\left|Q\right|}{\left|e\right|}\\ =\frac{1.65\times {10}^{-7}\text{C}}{1.6\times {10}^{-19}\text{C}}\\ =1.03125\times {10}^{12}\end{array}$

Conclusion:

Therefore, the number of electrons transferred $1.03125\times {10}^{12}$ .