7.) A charge of 1.5^ast 10^-7C is placed .128 m from a second charge If the force between the charges is 22.5 N,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 = k * (q1 * q2) / r^2<br /><br />Where:<br />- F is the force between the charges (in Newtons)<br />- k is the Coulomb constant (8.99 x 10^9 N⋅m^2/C^2)<br />- q1 and q2 are the charges (in Coulombs)<br />- r is the distance between the charges (in meters)<br /><br />Given information:<br />- Charge 1 (q1) = 1.5 x 10^-7 C<br />- Distance (r) = 0.128 m<br />- Force (F) = 22.5 N<br /><br />We need to find the magnitude of the second charge (q2).<br /><br />Rearranging the Coulomb's Law formula to solve for q2:<br /><br />q2 = (F * r^2) / (k * q1)<br /><br />Substituting the given values:<br /><br />q2 = (22.5 N * (0.128 m)^2) / (8.99 x 10^9 N⋅m^2/C^2 * 1.5 x 10^-7 C)<br />q2 = 1.0 x 10^-7 C<br /><br />Therefore, the magnitude of the second charge is 1.0 x 10^-7 C.