Given:
- The force of repulsion between two like-charged objects [tex]\( F = 8.2 \times 10^{-7} \)[/tex] newtons.
- The charge on each object [tex]\( q_1 = q_2 = 6.7 \times 10^{-9} \)[/tex] coulombs.
- The electrostatic constant [tex]\( k = 9.0 \times 10^9 \)[/tex] Nm²/C².
To find: The distance [tex]\( r \)[/tex] between the two charges.
We use Coulomb's Law, which relates the force between two point charges to the distance between them:
[tex]\[
F = k \frac{q_1 q_2}{r^2}
\][/tex]
We need to solve for [tex]\( r \)[/tex]. Rearranging the formula to isolate [tex]\( r \)[/tex]:
[tex]\[
r^2 = k \frac{q_1 q_2}{F}
\][/tex]
Plugging in the values:
[tex]\[
r^2 = \frac{(9.0 \times 10^9) \times (6.7 \times 10^{-9}) \times (6.7 \times 10^{-9})}{8.2 \times 10^{-7}}
\][/tex]
Evaluating the expression under the square root:
[tex]\[
r^2 \approx 0.49269512195121967
\][/tex]
Taking the square root of both sides to find [tex]\( r \)[/tex]:
[tex]\[
r \approx \sqrt{0.49269512195121967} \approx 0.7019224472484262 \, \text{meters}
\][/tex]
Therefore, the distance between the two charges is approximately [tex]\( 0.70 \)[/tex] meters. The correct answer is:
B. 0.70 meters
