Sure! Let's solve this problem step-by-step to find the electric field [tex]\( E \)[/tex] at a given distance from a charge.
### Given Values:
- Coulomb's constant, [tex]\( k = 9.0 \times 10^9 \)[/tex] N·m²/C²
- Charge, [tex]\( q = 6.4 \times 10^{-5} \)[/tex] C
- Distance, [tex]\( r = 2.5 \times 10^{-2} \)[/tex] m
### Formula:
The electric field [tex]\( E \)[/tex] due to a point charge [tex]\( q \)[/tex] at a distance [tex]\( r \)[/tex] is given by the formula:
[tex]\[ E = \frac{k \cdot q}{r^2} \][/tex]
### Step-by-Step Solution:
1. Calculate [tex]\( r^2 \)[/tex]:
[tex]\( r = 2.5 \times 10^{-2} \)[/tex] m
[tex]\[
r^2 = (2.5 \times 10^{-2})^2 = 6.25 \times 10^{-4} \, m^2
\][/tex]
2. Use the formula for Electric Field:
[tex]\[
E = \frac{9.0 \times 10^9 \, \text{N·m}^2/\text{C}^2 \times 6.4 \times 10^{-5} \, \text{C}}{6.25 \times 10^{-4} \, \text{m}^2}
\][/tex]
3. Calculate the numerator:
[tex]\[
9.0 \times 10^9 \times 6.4 \times 10^{-5} = 5.76 \times 10^5 \times 10^4 = 5.76 \times 10^9 \, \text{N·m}^2/\text{C}
\][/tex]
4. Divide by [tex]\( r^2 \)[/tex]:
[tex]\[
E = \frac{5.76 \times 10^9}{6.25 \times 10^{-4}} = 921600000 \, \text{N/C}
\][/tex]
### Conclusion:
After performing the calculations, we find that the electric field [tex]\( E \)[/tex] at the test charge is [tex]\( 9.2 \times 10^8 \text{ N/C} \)[/tex].
So, the correct answer is:
[tex]\[ \boxed{9.2 \times 10^8 \, \text{N/C}} \][/tex]
Hence, the correct option is:
E. [tex]\( 9.2 \times 10^8 \)[/tex] newtons/coulomb
