2. A negative charge of -6.0times 10^-6C exerts an attractive force of 65 Non a second charge that is 0.050 m away. What is the magnitude of the second charge?

To solve this problem, we can use Coulomb's law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.<br /><br />The formula for Coulomb's law is:<br /><br />\[ F = \frac{{k \cdot |q_1 \cdot q_2|}}{{r^2}} \]<br /><br />Where:<br />- \( F \) is the force between the charges (in Newtons)<br />- \( k \) is the Coulomb constant (\( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \))<br />- \( q_1 \) and \( q_2 \) are the charges (in Coulombs)<br />- \( r \) is the distance between the charges (in meters)<br /><br />Given:<br />- \( F = 65 \, \text{N} \)<br />- \( q_1 = -6.0 \times 10^{-6} \, \text{C} \)<br />- \( r = 0.050 \, \text{m} \)<br /><br />We need to find the magnitude of the second charge (\( q_2 \)).<br /><br />Rearranging the formula to solve for \( q_2 \):<br /><br />\[ q_2 = \frac{{F \cdot r^2}}{{k \cdot |q_1|}} \]<br /><br />Substituting the given values:<br /><br />\[ q_2 = \frac{{65 \, \text{N} \cdot (0.050 \, \text{m})^2}}{{(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \cdot |-6.0 \times 10^{-6} \, \text{C}|}} \]<br /><br />\[ q_2 = \frac{{65 \cdot 0.0025}}{{8.99 \times 10^9 \cdot 6.0 \times 10^{-6}}} \]<br /><br />\[ q_2 = \frac{{0.1625}}{{53.94 \times 10^3}} \]<br /><br />\[ q_2 = 3.01 \times 10^{-6} \, \text{C} \]<br /><br />Therefore, the magnitude of the second charge is \( 3.01 \times 10^{-6} \, \text{C} \).