Question

1. determine the number of excess electrons on a pith ball that has a net charge of -2.0 µc. 1.3 x 10¹³

Explanation:

Step1: Recall the charge of one electron

The charge of a single electron, \( e \), is \( -1.602 \times 10^{-19} \, \text{C} \). The magnitude of the charge of an electron is \( |e| = 1.602 \times 10^{-19} \, \text{C} \).

Step2: Identify the net charge of the pith ball

The net charge of the pith ball, \( Q \), is \( -2.0 \, \mu\text{C} \). First, convert microcoulombs to coulombs: \( 1 \, \mu\text{C} = 10^{-6} \, \text{C} \), so \( Q = -2.0 \times 10^{-6} \, \text{C} \). The magnitude of the net charge is \( |Q| = 2.0 \times 10^{-6} \, \text{C} \).

Step3: Calculate the number of excess electrons

The number of excess electrons, \( n \), is given by the total charge divided by the charge of one electron (in magnitude, since we're counting the number of electrons):
\( n = \frac{|Q|}{|e|} \)

Substitute the values:
\( n = \frac{2.0 \times 10^{-6} \, \text{C}}{1.602 \times 10^{-19} \, \text{C/electron}} \)

Calculate the result:
\( n \approx \frac{2.0 \times 10^{-6}}{1.602 \times 10^{-19}} \approx 1.25 \times 10^{13} \) (close to the given \( 1.3 \times 10^{13} \) when using approximate \( e \approx 1.6 \times 10^{-19} \, \text{C} \): \( \frac{2.0 \times 10^{-6}}{1.6 \times 10^{-19}} = 1.25 \times 10^{13} \approx 1.3 \times 10^{13} \))

Answer:

The number of excess electrons is approximately \( \boldsymbol{1.3 \times 10^{13}} \) (or more precisely \( \approx 1.25 \times 10^{13} \) using exact \( e \), but matching the given hint).