Chapter 20, Problem 7STP

To determine

To Choose: The correct option.

Answer to Problem 7STP

C. $7.7\times {10}^{-9}\text{C}$

Explanation of Solution

Given:

The excess number of electrons, $n=4.8\times {10}^{10}$

Formula used:

If the charge on an electron is $q_e$ and the number of electrons is $n$ , then the total charge of these electrons is,

  $Q_{tot}=q_{e}n$

Calculation:

Now, it is known that the charge on an electron is,

  $q_e=1.6\times {10}^{-19}\text{C}$

Then, the total charge due to given excess electrons will be,

  $\begin{array}{c}Q=n{q}_e\\ =4.8\times {10}^{10}\times 1.6\times {10}^{-19}\\ \approx 7.7\times {10}^{-9}\text{C}\end{array}$

Conclusion:

Thus, net charge on an electroscope that has an excess of electrons $4.8\times {10}^{10}$ is $7.7\times {10}^{-9}\text{C}$ .