Calculating Excess Electrons from Net Charge

This question asks us to determine the number of extra electrons on an object, given its total net charge and the charge of a single electron (elementary charge).

Understanding Electric Charge and Electrons

In physics, electric charge is a fundamental property of matter. Electrons are negatively charged particles. When an object has a net negative charge, it means it has more electrons than protons. The total negative charge is due to these extra electrons.

The charge of a single electron is denoted by '$e$', which is approximately $1.6 \times 10^{-19}$ Coulombs (C). The negative sign indicates its polarity.

Charge Calculation Steps

We are given:

• Net charge on the object, $Q = -5 \text{ nC}$.
• Elementary charge, $e = 1.6 \times 10^{-19} \text{ C}$.

First, let's convert the net charge from nanoCoulombs (nC) to Coulombs (C). Remember that $1 \text{ nC} = 10^{-9} \text{ C}$.

So, $Q = -5 \times 10^{-9} \text{ C}$.

The total charge ($Q$) on an object is related to the number of excess electrons ($n$) and the elementary charge ($e$) by the formula:

$ Q = n \times e $

Since the net charge is negative ($-5 \text{ nC}$), this indicates an excess of electrons. We need to find the number of these excess electrons, $n$. We can rearrange the formula to solve for $n$. We'll use the magnitude of the charge:

$ n = \frac{|Q|}{e} $

Now, substitute the given values into the formula:

$ n = \frac{|-5 \times 10^{-9} \text{ C}|}{1.6 \times 10^{-19} \text{ C}} $

$ n = \frac{5 \times 10^{-9} \text{ C}}{1.6 \times 10^{-19} \text{ C}} $

Let's perform the division:

1. Divide the numerical parts: $\frac{5}{1.6} = 3.125$.
2. Divide the powers of 10: $\frac{10^{-9}}{10^{-19}} = 10^{-9 - (-19)} = 10^{-9 + 19} = 10^{10}$.

Combining these results:

$ n = 3.125 \times 10^{10} $

Conclusion

The calculation shows that there are $3.125 \times 10^{10}$ excess electrons on the object to account for the net charge of $-5 \text{ nC}$.