A charge of [tex]$4.5 \times 10^{-5} C$[/tex] is placed in an electric field with a strength of [tex]$2.0 \times 10^4 \frac{ N }{ C }$[/tex]. What is the electric force acting on the charge?

[tex]\square[/tex] N



Answer :

To find the electric force acting on a charge placed in an electric field, you can use the formula:

[tex]\[ F = qE \][/tex]

where:
- [tex]\( F \)[/tex] is the electric force,
- [tex]\( q \)[/tex] is the charge,
- [tex]\( E \)[/tex] is the electric field strength.

Given:
- The charge, [tex]\( q = 4.5 \times 10^{-5} \, \text{C} \)[/tex]
- The electric field strength, [tex]\( E = 2.0 \times 10^4 \, \frac{\text{N}}{\text{C}} \)[/tex]

Step-by-step solution:
1. Identify the given values:
- [tex]\( q = 4.5 \times 10^{-5} \, \text{C} \)[/tex]
- [tex]\( E = 2.0 \times 10^4 \, \frac{\text{N}}{\text{C}} \)[/tex]

2. Substitute the given values into the formula:

[tex]\[ F = (4.5 \times 10^{-5} \, \text{C}) \times (2.0 \times 10^4 \, \frac{\text{N}}{\text{C}}) \][/tex]

3. Perform the multiplication:

[tex]\[ F = 0.9 \, \text{N} \][/tex]

Thus, the electric force acting on the charge is [tex]\( 0.9 \, \text{N} \)[/tex].