Answer :
The magnitude of the electrical force between two +2.5 C charges separated by 4000 m is approximately 3.5 N.
To calculate the magnitude of the electrical force between two point charges, we use Coulomb's law:
[tex]\[ F = k \frac{q_1 q_2}{r^2} \][/tex]
Where:
- F is the magnitude of the electrical force,
[tex]- \( k \) is Coulomb's constant (\(8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2\)),[/tex]
- [tex]\( q_1 \) and \( q_2 \)[/tex] are the magnitudes of the charges [tex](\(+2.5 \, \text{C}\)[/tex] each in this case),
- r is the distance between the charges (4000 m).
Plugging in the values:
[tex]\[ F = (8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2) \frac{(2.5 \, \text{C}) \times (2.5 \, \text{C})}{(4000 \, \text{m})^2} \][/tex]
[tex]\[ F = (8.99 \times 10^9) \frac{6.25}{16000000} \][/tex]
[tex]\[ F \approx (8.99 \times 10^9) \times 3.90625 \times 10^{-7} \][/tex]
[tex]\[ F \approx 3.515 \, \text{N} \][/tex]
Rounded to one decimal place, the magnitude of the electrical force between the two charges is approximately [tex]\(3.5 \, \text{N}\)[/tex].
Correct question is:
If two point charges, with a charge of +2.5 coulomb each are separated by 4000m, what is the magnitude of the electrical force between them in N? Round your answer to one decimal.
