18. When 1.6 mathrm(~kg) point charge is 1.5 mathrm(~m) from a second point charge, the gravitational force each one experiences is 573 mathrm(~N) . What is the mass of the second charge? Given: & Work: Formula: & Answer: 2.83 times 10^10 mathrm(~kg) b. 1.2 times 10^11 mathrm(~kg) c. 1.2 times 10^12 mathrm(~kg) d. 8.5 times 10^12 mathrm(~kg) 19. Two electrons are 20 mathrm(~m) apart at closest approach. What is the magnitude of the maximum electric force that they exert on each other? (c=-1.60 times 10^10 mathrm(C)) Given: & Work: Formula: & Answer: a. 1.2 times 10^-37 mathrm(~N) b. 5.8 times 10^11 mathrm(~N) c. 2.5 times 10^-10 mathrm(~N) d. 1.6 times 10^11 mathrm(~N) 20. The force of attraction that a -40.0 mathrm(C) point charge exerts on a +108 mathrm(C) point charge has magnitude 4.00 mathrm(~N) . How far apart are these two charges? Given: & Work: Formula: & Answer: a. 1.38 times 10^9 mathrm(~m) b. 1.9 times 10^4 mathrm(~m) c. 3.12 times 10^6 mathrm(~m) d. 9.71 times 10^4 mathrm(~m)

18. To find the mass of the second charge, we can use the formula for gravitational force:<br /><br />\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]<br /><br />Given:<br />- Gravitational force (F) = 573 N<br />- Distance (r) = 15 m<br />- Mass of the first charge (m1) = 1.6 kg<br /><br />We need to find the mass of the second charge (m2).<br /><br />Rearranging the formula to solve for m2:<br /><br />\[ m_2 = \frac{F \cdot r^2}{G \cdot m_1} \]<br /><br />Substituting the given values:<br /><br />\[ m_2 = \frac{573 \cdot (15)^2}{6.674 \cdot 10^{-11} \cdot 1.6} \]<br /><br />\[ m_2 = \frac{10275}{1.06784 \cdot 10^{-10}} \]<br /><br />\[ m_2 = 9.57 \times 10^{10} \, \text{kg} \]<br /><br />Therefore, the mass of the second charge is \( 9.57 \times 10^{10} \, \text{kg} \).<br /><br />19. To find the magnitude of the maximum electric force between two electrons, we can use Coulomb's law:<br /><br />\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]<br /><br />Given:<br />- Charge of each electron (q1 and q2) = -1.60 x 10^-19 C<br />- Distance (r) = 20 m<br /><br />Substituting the given values:<br /><br />\[ F = \frac{8.99 \cdot 10^9 \cdot |-1.60 \cdot 10^{-19} \cdot -1.60 \cdot 10^{-19}|}{(20)^2} \]<br /><br />\[ F = \frac{8.99 \cdot 10^9 \cdot 2.56 \cdot 10^{-38}}{400} \]<br /><br />\[ F = 5.75 \times 10^{-19} \, \text{N} \]<br /><br />Therefore, the magnitude of the maximum electric force that the two electrons exert on each other is \( 5.75 \times 10^{-19} \, \text{N} \).<br /><br />20. To find the distance between the two charges, we can use Coulomb's law:<br /><br />\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]<br /><br />Given:<br />- Force (F) = 4.00 N<br />- Charge 1 (q1) = -40.0 C<br />- Charge 2 (q2) = +108 C<br /><br />Rearranging the formula to solve for r:<br /><br />\[ r = \sqrt{\frac{k \cdot |q_1 \cdot q_2|}{F}} \]<br /><br />Substituting the given values:<br /><br />\[ r = \sqrt{\frac{8.99 \cdot 10^9 \cdot |-40.0 \cdot 108|}{4.00}} \]<br /><br />\[ r = \sqrt{\frac{8.99 \cdot 10^9 \cdot 4320}{4.00}} \]<br /><br />\[ r = \sqrt{9.71 \times 10^{13}} \]<br /><br />\[ r = 3.12 \times 10^6 \, \text{m} \]<br /><br />Therefore, the distance between the two charges is \( 3.12 \times 10^6 \, \text{m} \).