Chapter 16, Problem 8P

To determine

The number of excess electrons and increase in the mass.

Answer to Problem 8P

The number of excess electrons is $2.62\times {10}^{14}$ and the increase in mass is $2.38\times {10}^{-16}\text{kg}$ .

Explanation of Solution

Given:

Net charge, $Q=-42\mu \text{N}$

Formula used:

Write the expression to calculate the number of electrons.

  $Q=ne$.... (1)

Where,

The number of electrons is $n$ .

The charge on electron is $e$ .

Calculation:

Substitute the values in equation (1).

  $\begin{array}{c}-42\times {10}^{-6}=n\times \left(-1.6\times {10}^{-19}\right)\\ n=\frac{42\times {10}^{-6}}{1.6\times {10}^{-19}}\\ =2.62\times {10}^{14}\end{array}$

Calculate the increase in the mass.

  $\begin{array}{c}m=2.62\times {10}^{14}\times 9.1\times {10}^{-31}\\ =2.38\times {10}^{-16}\text{kg}\end{array}$

Conclusion:

Therefore, the number of excess electrons is $2.62\times {10}^{14}$ and the increase in mass is $2.38\times {10}^{-16}\text{kg}$ .