Let's solve each part step-by-step as requested. --- ## (A) Calculate the electric field in the region between the parallel plates ### **Step 1:** Write the formula for the electric field between parallel plates The electric field \( E \) between two parallel plates separated by distance \( d \) with a voltage difference \( V \) is: \[ E = \frac{V}{d} \] ### **Step 2:** Substitute the given values - \( V = 1.5\, \text{kV} = 150\, \text{V} \) - \( d = 1\, \text{cm} = .01\, \text{m} \) \[ E = \frac{150\, \text{V}}{.01\, \text{m}} = 150,000\, \text{V/m} \] #### **Explanation** We used the standard formula for the electric field between plates, substituting the given voltage and separation. ### **Final Answer (A):** \[ \boxed{E = 1.5 \times 10^5~\text{V/m}} \] --- ## (B) What is the amount of charge \( Q \) that the molecule carries to be deflected by .3 mm at the end? ### **Step 1:** Find the acceleration due to the electric force - Electric force: \( F = QE \) - Acceleration: \( a = \frac{F}{m} = \frac{QE}{m} \) ### **Step 2:** Use kinematics to relate vertical deflection \( y \) to charge \( Q \) - Vertical displacement after time \( t \): \( y = \frac{1}{2} a t^2 \) - \( y = .3~\text{mm} = 3. \times 10^{-4}~\text{m} \) - \( a = \frac{QE}{m} \) So, \[ y = \frac{1}{2}\frac{QE}{m}t^2~\Rightarrow~ Q = \frac{2my}{Et^2} \] But we need \( t \): The time to traverse plates of length \( L = 5.~\text{cm} = .05~\text{m} \) at \( v = 55,000~\text{m/s} \): \[ t = \frac{L}{v} = \frac{.05}{55,000} = 9.09 \times 10^{-7}~\text{s} \] Now plug in all values: - \( m = 2.4 \times 10^{-15}~\text{kg} \) - \( y = 3. \times 10^{-4}~\text{m} \) - \( E = 1.5 \times 10^5~\text{V/m} \) - \( t = 9.09 \times 10^{-7}~\text{s} \) \[ Q = \frac{2 \times 2.4 \times 10^{-15} \times 3. \times 10^{-4}}{1.5 \times 10^5 \times (9.09 \times 10^{-7})^2} \] Calculate denominator: \[ 1.5 \times 10^5 \times (9.09 \times 10^{-7})^2 = 1.5 \times 10^5 \times 8.264 \times 10^{-13} \approx 1.2396 \times 10^{-7} \] Calculate numerator: \[ 2 \times 2.4 \times 10^{-15} \times 3. \times 10^{-4} = 14.4 \times 10^{-19} = 1.44 \times 10^{-18} \] So, \[ Q = \frac{1.44 \times 10^{-18}}{1.24 \times 10^{-7}} \approx 1.16 \times 10^{-11}~\text{C} \] #### **Explanation** We used the kinematics of constant acceleration under the electric field, and solved for the required charge. ### **Final Answer (B):** \[ \boxed{Q \approx 1.16 \times 10^{-11}~\text{C}} \] --- Let me know if you want the detailed steps for (C) and (D) as well!

Sure! Let's continue with parts (C) and (D). --- ## (C) How many seconds does it take for the molecule to pass through the plates? ### **Step 1:** Use the formula for time The time \( t \) taken to travel a distance \( L \) at constant velocity \( v \) is given by: \[ t = \frac{L}{v} \] ### **Step 2:** Substitute the values - Length of the plates \( L = 5.0\, \text{cm} = 0.05\, \text{m} \) - Velocity \( v = 55\, \text{km/s} = 55,000\, \text{m/s} \) Substituting the values: \[ t = \frac{0.05\, \text{m}}{55,000\, \text{m/s}} \approx 9.09 \times 10^{-7}\, \text{s} \] #### **Explanation** This calculation gives the time it takes for the molecule to travel the length of the plates, based on its initial speed. ### **Final Answer (C):** \[ \boxed{t \approx 9.09 \times 10^{-7}\, \text{s}} \] --- ## (D) What is the magnitude and direction of the magnetic field to have the molecule travel through undeflected? ### **Step 1:** Set up the magnetic force to balance the electric force The magnetic force \( F_B \) must equal the electric force \( F_E \) to keep the molecule undeflected: \[ F_B = F_E \] Where: - \( F_E = QE \) - \( F_B = QvB \) Thus, \[ QE = QvB \implies B = \frac{E}{v} \] ### **Step 2:** Substitute the values Using: - \( E = 1.5 \times 10^5\, \text{V/m} \) - \( v = 55,000\, \text{m/s} \) Substituting: \[ B = \frac{1.5 \times 10^5\, \text{V/m}}{55,000\, \text{m/s}} \approx 2.727\, \text{T} \] #### **Explanation** This calculation finds the required magnetic field strength to balance the electric force, ensuring the molecule travels straight through the plates. ### **Final Answer (D):** \[ \boxed{B \approx 2.73\, \text{T}} \] --- This concludes the calculations for all parts of the problem. If you have any further questions, feel free to ask!