IF an electron enters into a space between the plates of a parallel plate capacitor at an an angle `alpha` with the plates an leaves at an angle `beta` to the plates find the ratio of its kinetic energy while entering the capacitor of that while leaving.

Options

A. `((sin beta)/(sin alpha))^(2)`
B. `((cos beta)/(cos alpha))^(2)`
C. `((cos alpha)/(cos beta))^(2)`
D. `((sin alpha)/(sinbeta))^(2)`

Answer

To solve the problem of finding the ratio of the kinetic energy of an electron while entering a parallel plate capacitor to that while leaving, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion of the Electron**:
- The electron enters the capacitor at an angle \( \alpha \) with respect to the plates and exits at an angle \( \beta \).
- The electric field between the plates exerts a force on the electron, which affects its horizontal motion but not its vertical motion.

2. **Components of Velocity**:
- Let the initial velocity of the electron when it enters the capacitor be \( V_1 \).
- The components of the velocity can be expressed as:
- Horizontal component: \( V_{1x} = V_1 \cos \alpha \)
- Vertical component: \( V_{1y} = V_1 \sin \alpha \)
- When the electron exits the capacitor with a velocity \( V_2 \), the components are:
- Horizontal component: \( V_{2x} = V_2 \cos \beta \)
- Vertical component: \( V_{2y} = V_2 \sin \beta \)

3. **Conservation of Vertical Component**:
- Since there is no net force acting in the vertical direction, the vertical component of velocity remains constant:
\[ V_{1y} = V_{2y} \]
- Thus, we have:
\[ V_1 \sin \alpha = V_2 \sin \beta \]

4. **Finding the Ratio of Velocities**:
- Rearranging the above equation gives:
\[ \frac{V_1}{V_2} = \frac{\sin \beta}{\sin \alpha} \]

5. **Kinetic Energy Calculation**:
- The kinetic energy of the electron is given by:
\[ KE = \frac{1}{2} m V^2 \]
- Therefore, the ratio of the kinetic energies while entering and leaving the capacitor is:
\[ \frac{KE_1}{KE_2} = \frac{\frac{1}{2} m V_1^2}{\frac{1}{2} m V_2^2} = \frac{V_1^2}{V_2^2} \]

6. **Substituting the Velocity Ratio**:
- Substituting the expression for \( \frac{V_1}{V_2} \):
\[ \frac{KE_1}{KE_2} = \left( \frac{\sin \beta}{\sin \alpha} \right)^2 \]

7. **Final Ratio**:
- Therefore, the final ratio of the kinetic energy while entering the capacitor to that while leaving is:
\[ \frac{KE_1}{KE_2} = \frac{\sin^2 \beta}{\sin^2 \alpha} \]

### Conclusion:
The ratio of the kinetic energy of the electron while entering the capacitor to that while leaving is given by:
\[ \frac{KE_1}{KE_2} = \frac{\sin^2 \beta}{\sin^2 \alpha} \]