4. Two spheres with charges of +6.00times 10^-6C and -5.0times 10^-6C attract each other with a force of 7.0times 10^-2 Newto r=sqrt (k(vert q_(1)q_(2)vert )/(F)) Determine the separation distance between the two objects. 5.00 m B. 1.96 m C 2.99 m 3.86 m

To determine the separation distance between the two charged spheres, we can use the formula for the force between two charged objects:<br /><br />$F = k \frac{\vert q_1 q_2 \vert}{r^2}$<br /><br />where:<br />- $F$ is the force between the objects (in Newtons)<br />- $k$ is the Coulomb constant ($8.99 \times 10^9 \, \text{N m}^2/\text{C}^2$)<br />- $q_1$ and $q_2$ are the charges of the objects (in Coulombs)<br />- $r$ is the separation distance between the objects (in meters)<br /><br />Given:<br />- $F = 7.0 \times 10^{-2} \, \text{N}$<br />- $q_1 = +6.00 \times 10^{-6} \, \text{C}$<br />- $q_2 = -5.0 \times 10^{-6} \, \text{C}$<br /><br />We need to solve for $r$:<br /><br />$7.0 \times 10^{-2} = 8.99 \times 10^9 \frac{\vert 6.00 \times 10^{-6} \times -5.0 \times 10^{-6} \vert}{r^2}$<br /><br />$r^2 = 8.99 \times 10^9 \frac{\vert 6.00 \times 10^{-6} \times -5.0 \times 10^{-6} \vert}{7.0 \times 10^{-2}}$<br /><br />$r^2 = 8.99 \times 10^9 \frac{30.0 \times 10^{-12}}{7.0 \times 10^{-2}}$<br /><br />$r^2 = 8.99 \times 10^9 \times 4.29 \times 10^{-10}$<br /><br />$r^2 = 3.86 \times 10^{-1}$<br /><br />$r = \sqrt{3.86 \times 10^{-1}}$<br /><br />$r \approx 0.62 \, \text{m}$<br /><br />Therefore, the separation distance between the two charged spheres is approximately 0.62 meters.