1. Problem 1: Given $L=0.50$ m, $q_1 q_2=8.0 \times 10^{-9}$ C, and angle $6^\circ$, find the distance and the electric force $F_E$. 2. Formula: Coulomb's law for electric force is $$F_E = k \frac{|q_1 q_2|}{r^2}$$ where $k=8.99 \times 10^9$ Nm$^2$/C$^2$. 3. Step 1: Find the distance $r$ using geometry or given data (not fully specified here, so assuming $r=L=0.50$ m). 4. Step 2: Calculate $F_E$: $$F_E = 8.99 \times 10^9 \times \frac{8.0 \times 10^{-9}}{(0.50)^2} = 8.99 \times 10^9 \times \frac{8.0 \times 10^{-9}}{0.25} = 8.99 \times 10^9 \times 3.2 \times 10^{-8} = 287.68\, \text{N}$$ Final answer for problem 1: - Distance $r = 0.50$ m - Electric force $F_E = 288$ N (rounded) --- 2. Problem 2: Given $q_1 = -30$ nC, $q_2 = +10$ nC, $r=0.020$ m, find $F_E$. 3. Step 1: Convert charges to Coulombs: $$q_1 = -30 \times 10^{-9} = -3.0 \times 10^{-8} C$$ $$q_2 = 10 \times 10^{-9} = 1.0 \times 10^{-8} C$$ 4. Step 2: Calculate $F_E$: $$F_E = 8.99 \times 10^9 \times \frac{|(-3.0 \times 10^{-8})(1.0 \times 10^{-8})|}{(0.020)^2} = 8.99 \times 10^9 \times \frac{3.0 \times 10^{-16}}{0.0004} = 8.99 \times 10^9 \times 7.5 \times 10^{-13} = 6.74 \times 10^{-3} N$$ Final answer for problem 2: - Electric force $F_E = 6.74 \times 10^{-3}$ N --- 3. Problem 3: Given $q_1 - q_2 = +2e$ where $e=1.60 \times 10^{-19}$ C, $r=3.0 \times 10^{-10}$ m, $m_1 = m_2 = 6.64 \times 10^{-27}$ kg, find magnitude of $F_E$, magnitude of gravitational force $F_G$, and whether force is attractive or repulsive. 4. Step 1: Calculate $q_1 - q_2$: $$q_1 - q_2 = 2 \times 1.60 \times 10^{-19} = 3.2 \times 10^{-19} C$$ 5. Step 2: Calculate $F_E$: $$F_E = 8.99 \times 10^9 \times \frac{(3.2 \times 10^{-19})^2}{(3.0 \times 10^{-10})^2} = 8.99 \times 10^9 \times \frac{1.024 \times 10^{-37}}{9 \times 10^{-20}} = 8.99 \times 10^9 \times 1.138 \times 10^{-18} = 1.02 \times 10^{-8} N$$ 6. Step 3: Calculate gravitational force $F_G$: $$F_G = G \frac{m_1 m_2}{r^2}$$ where $G=6.67 \times 10^{-11}$ Nm$^2$/kg$^2$. $$F_G = 6.67 \times 10^{-11} \times \frac{(6.64 \times 10^{-27})^2}{(3.0 \times 10^{-10})^2} = 6.67 \times 10^{-11} \times \frac{4.41 \times 10^{-53}}{9 \times 10^{-20}} = 6.67 \times 10^{-11} \times 4.9 \times 10^{-34} = 3.27 \times 10^{-44} N$$ 7. Step 4: Since $q_1 - q_2$ is positive, the force is repulsive. Final answers for problem 3: - $F_E = 1.02 \times 10^{-8}$ N - $F_G = 3.27 \times 10^{-44}$ N - Force is repulsive --- 4. Problem 4: Given $q_1 - q_2 = -6.0 \times 10^{-6}$ C, $r=0.040$ m, $m_1=5.0 \times 10^{-6}$ kg, $m_2=2.5 \times 10^{-6}$ kg, find $F_E$, acceleration $a$ of particle A and B. 5. Step 1: Calculate $F_E$: $$F_E = 8.99 \times 10^9 \times \frac{(6.0 \times 10^{-6})^2}{(0.040)^2} = 8.99 \times 10^9 \times \frac{3.6 \times 10^{-11}}{0.0016} = 8.99 \times 10^9 \times 2.25 \times 10^{-8} = 202.3 N$$ 6. Step 2: Calculate acceleration $a$ of particle A: $$a_A = \frac{F_E}{m_1} = \frac{202.3}{5.0 \times 10^{-6}} = 4.05 \times 10^7 \text{ m/s}^2$$ 7. Step 3: Calculate acceleration $a$ of particle B: $$a_B = \frac{F_E}{m_2} = \frac{202.3}{2.5 \times 10^{-6}} = 8.09 \times 10^7 \text{ m/s}^2$$ Final answers for problem 4: - $F_E = 202.3$ N - $a_A = 4.05 \times 10^7$ m/s$^2$ - $a_B = 8.09 \times 10^7$ m/s$^2$