Answer:
A. 2.76 × 10-1 meters
Explanation:
The electrostatic force between two charges is given by:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the distance between the charges
In this problem, we know:
[tex]F=5.3 \cdot 10^{-6} N[/tex]
[tex]q_1 = 1_2 = 6.7 \cdot 10^{-9} C[/tex]
[tex]k=9.0 \cdot 10^9 N/m^2 C^2[/tex]
So we can re-arrange the equation to find the distance between the two charges:
[tex]r=\sqrt{\frac{k q_1 q_2}{F}}=\sqrt{\frac{9.0\cdot 10^9 (6.7 \cdot 10^{-9})^2}{5.3\cdot 10^{-6}}}=0.276 m[/tex]
Which corresponds to option A, 2.76 × 10-1 meters.
