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 Newton. r=sqrt (k(vert q_(1)q_(2)vert )/(F)) Determine the separation distance between the two objects. 3.86 m 1.96 m c 5.00 m 2.99 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 />- $q_1 = +6.times 10^{-6} \, \text{C}$<br />- $q_2 = -5.0 \times 10^{-6} \, \text{C}$<br /><br />We need to solve for $r$:<br /><br />$r = \sqrt{k \frac{\vert q_1 q_2 \vert}{F}}$<br /><br />Substitute the given values:<br /><br />$r = \sqrt{(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \frac{\vert (+6.00 \times 10^{-6} \, \text{C}) \cdot (-5.0 \times 10^{-6} \, \text{C}) \vert}{7.0 \times 10^{-2} \, \text{N}}}$<br /><br />$rsqrt{(8.99 \times 10^9) \frac{(6.00 \times 10^{-6}) \cdot (5.0 \times 10^{-6})}{7.0 \times 10^{-2}}}$<br /><br />$r = \sqrt{(8.99 \times 10^9) \frac{30.0 \times 10^{-12}}{7.0 \times 10^{-2}}}$<br /><br />$r = \sqrt{(8.99 \times 10^9) \cdot (4.29 \times 10^{-12})}$<br /><br />$r = \sqrt{3.86 \times 10^{-2}}$<br /><br />$r \approx 0.62 \, \text{m}$<br /><br />Therefore, the separation distance between the two charged spheres is approximately 0.62 meters.