1. **Problem Statement:** Find the ratio of the magnitudes of the electric fields $|\vec{E}_1| : |\vec{E}_2|$ experienced by charges $q_1$ and $q_2$ due to charge $Q$, and determine the correct directions of these fields. 2. **Formula for Electric Field:** The electric field due to a point charge $Q$ at a distance $r$ is given by: $$\vec{E} = k \frac{|Q|}{r^2} \hat{r}$$ where $k$ is Coulomb's constant, and $\hat{r}$ is the unit vector pointing away from $Q$ if $Q$ is positive, or towards $Q$ if $Q$ is negative. 3. **Given Distances:** - Distance from $Q$ to $q_1$ is $3a$ - Distance from $Q$ to $q_2$ is $a$ 4. **Calculate Magnitudes:** $$|\vec{E}_1| = k \frac{|Q|}{(3a)^2} = k \frac{|Q|}{9a^2}$$ $$|\vec{E}_2| = k \frac{|Q|}{a^2}$$ 5. **Ratio of Magnitudes:** $$|\vec{E}_1| : |\vec{E}_2| = \frac{k \frac{|Q|}{9a^2}}{k \frac{|Q|}{a^2}} = \frac{1/9}{1} = \frac{1}{9}$$ 6. **Answer for Magnitude Ratio:** The ratio is $1 : 9$, which corresponds to option B. 7. **Direction of Electric Fields:** - Charge $Q$ is negative. - Electric field direction is towards negative charges. - Therefore, $\vec{E}_1$ and $\vec{E}_2$ point towards $Q$. 8. **Answer for Directions:** - $\vec{E}_1$ is towards $Q$ (option B) - $\vec{E}_2$ is towards $Q$ (option C) **Final answers:** - Ratio $|\vec{E}_1| : |\vec{E}_2| = 1 : 9$ - Directions: $\vec{E}_1$ towards $Q$, $\vec{E}_2$ towards $Q$