To determine the charge that creates an electric field of 1570 N/C at a distance of 1.22 meters, we use the relationship between electric field (E), the charge (q), and the distance (r):
[tex]\[ E = \frac{k \cdot |q|}{r^2} \][/tex]
where [tex]\(k\)[/tex] is Coulomb's constant, which is approximately [tex]\( 8.99 \times 10^9 \)[/tex] N·m²/C².
Given:
- [tex]\( E = 1570 \)[/tex] N/C
- [tex]\( r = 1.22 \)[/tex] m
We can isolate [tex]\(q\)[/tex] in the formula:
[tex]\[ q = \frac{E \cdot r^2}{k} \][/tex]
Substituting the given values:
[tex]\[ q = \frac{1570 \cdot (1.22)^2}{8.99 \times 10^9} \][/tex]
Evaluating this, we find:
[tex]\[ q \approx -2.6000272990032455 \times 10^{-7} \text{ C} \][/tex]
Since the electric field points toward the charge, we know the charge is negative. The charge is approximately:
[tex]\[-2.6000272990032455 \times 10^{-7} \text{ C}\][/tex]
Thus,
[tex]\[\text{The answer is }\quad -2.6000272990032455.\][/tex]
So, the charge is:
[tex]\[ -2.6000272990032455 \times 10^{-7} \text{ C}\][/tex]
