# Step 1: Find the Force and Acceleration The ion is doubly ionized, so its charge: \[ q = 2 \times (+e) = 2 \times 1.6 \times 10^{-19} \text{ C} = 3.2 \times 10^{-19} \text{ C} \] The force from the electric field (\( \vec{E} \), negative y-direction): \[ F = qE = 3.2 \times 10^{-19} \times 500 = 1.6 \times 10^{-15} \text{ N} \text{ (in } -y \text{ direction)} \] The mass of chlorine-35 ion (\( m \)): \[ m \approx 35 \times 1.67 \times 10^{-27} = 5.845 \times 10^{-26} \text{ kg} \] Now, acceleration: \[ a = \frac{F}{m} = \frac{1.6 \times 10^{-15}}{5.845 \times 10^{-26}} \approx 2.74 \times 10^{10} \text{ m/s}^2 \] **Explanation:** The electric field applies a constant force in the negative y-direction, causing a constant acceleration downward. --- # Step 2: Find the Displacement and Path **X-direction:** - No force acts in the x-direction. - Initial velocity: \( v_{x} = 10,000 \) m/s - Time: \( t = 2.5 \) s \[ x = v_{x} t = 10,000 \times 2.5 = 25,000 \text{ m} \] **Y-direction:** - Initial velocity: \( v_{y} = \) - Acceleration: \( a_y = -2.74 \times 10^{10} \) m/s\(^2\) \[ y = v_{y} t + \frac{1}{2} a t^2 = + \frac{1}{2} \left(-2.74 \times 10^{10}\right) (2.5)^2 \] \[ y = .5 \times -2.74 \times 10^{10} \times 6.25 = -8.56 \times 10^{10} \text{ m} \] **Explanation:** The ion moves in the x-direction at a constant velocity while being accelerated downward in the y-direction, making its path parabolic (like projectile motion). --- # Final Answer - **Path:** A parabola, moving right and curving downward. - **Displacement after 2.5 s:** \[ \Delta \vec{r} = (25,000\,\text{m},\ -8.56 \times 10^{10}\,\text{m}) \] from its entry point.