Question

10$^{19}$ electrons are placed on an uncharged body. Calculate the charge produced on the body.

Hint:

Always remember: \[ 1 \, \text{electron} = -1.6 \times 10^{-19} \, \text{C} \] Adding electrons gives a \textbf{negative charge}, while removing electrons gives a \textbf{positive charge}.

Answer 1

Step 1: The charge on one electron is \( -1.6 \times 10^{-19} \, \text{C} \). This is a fundamental physical constant representing the elementary charge carried by a single electron, and the negative sign indicates that the electron is negatively charged.
Step 2: When \( 10^{19} \) electrons are added to an initially uncharged body, each carrying a charge of \( -1.6 \times 10^{-19} \, \text{C} \), the total charge acquired by the body is calculated as: \[ Q = n \times (-e) = 10^{19} \times (-1.6 \times 10^{-19}) = -1.6 \, \text{C}. \] Here, \( n = 10^{19} \) is the number of electrons added, and \( e = 1.6 \times 10^{-19} \, \text{C} \) is the magnitude of the charge on one electron. The product gives the total charge \( Q \), and the negative sign indicates that the charge is negative.
Step 3: Therefore, the body gains a negative charge of 1.6 C. This means that the body is now negatively charged due to the excess of electrons.