To find the magnitude of the electric field (E) experienced by a charge (q), we can use the formula:
\[ E = \frac{F}{q} \]
where:
- \( E \) is the electric field,
- \( F \) is the force experienced by the charge,
- \( q \) is the magnitude of the charge.
Given:
- \( F = 0.2 , \text{N} \)
- \( q = 4.0 \times 10^{-4} , \text{C} \)
Substituting the values into the formula:
\[ E = \frac{0.2 , \text{N}}{4.0 \times 10^{-4} , \text{C}} \]
Calculating:
\[ E = \frac{0.2}{4.0 \times 10^{-4}} = 0.2 \div 0.0004 = 500 , \text{N/C} \]
In scientific notation, this is:
\[ E = 5.0 \times 10^{2} , \text{N/C} \]
Thus, the magnitude of the electric field is \( 5 \times 10^{2} , \text{N/C} \).
The correct answer from the choices provided is:
5 x 10² N/C
1 answer