A charged particle enters at point `A` and comes out from point `B`. Its velocity vector makes angle `alpha` and `beta` with electric field at these two points, respectively. The ratio of kinetic energy of the charged particle at these two points `((K_(delta))/(K_(A)))` will be (Given `alpha=60^(circ)` and `beta=30^(circ)` ).
'(##CEN_KSR_PHY_JEE_C18_E01_014_Q05##)'

Answer

To find the ratio of kinetic energy of a charged particle at two points \( A \) and \( B \) when it enters and exits an electric field, we can follow these steps:

### Step 1: Understand the relationship between velocity and kinetic energy
The kinetic energy \( K \) of a charged particle is given by the formula:
\[ K = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

### Step 2: Identify the components of velocity
At point \( A \), the velocity \( V_A \) makes an angle \( \alpha \) with the electric field. The components of the velocity can be expressed as:
- Parallel component to the electric field: \( V_{A\parallel} = V_A \cos(\alpha) \)
- Perpendicular component to the electric field: \( V_{A\perp} = V_A \sin(\alpha) \)

At point \( B \), the velocity \( V_B \) makes an angle \( \beta \) with the electric field. The components of the velocity can be expressed as:
- Parallel component to the electric field: \( V_{B\parallel} = V_B \cos(\beta) \)
- Perpendicular component to the electric field: \( V_{B\perp} = V_B \sin(\beta) \)

### Step 3: Apply the work-energy theorem
The work done by the electric field on the charged particle results in a change in kinetic energy. The change in kinetic energy can be related to the parallel component of the velocity, as only this component is affected by the electric field.

### Step 4: Write the ratio of kinetic energies
The ratio of kinetic energies at points \( A \) and \( B \) can be expressed as:
\[ \frac{K_B}{K_A} = \frac{\frac{1}{2} m V_B^2}{\frac{1}{2} m V_A^2} = \frac{V_B^2}{V_A^2} \]

### Step 5: Relate the velocities using the angles
Using the relationship between the components of the velocities and the angles:
\[ \frac{V_B}{V_A} = \frac{\sin(\beta)}{\sin(\alpha)} \]
Thus,
\[ \frac{K_B}{K_A} = \left(\frac{\sin(\beta)}{\sin(\alpha)}\right)^2 \]

### Step 6: Substitute the given angles
Given \( \alpha = 60^\circ \) and \( \beta = 30^\circ \):
- \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

Now substituting these values:
\[ \frac{K_B}{K_A} = \left(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)^2 = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]

### Final Result
Thus, the ratio of kinetic energy at points \( B \) and \( A \) is:
\[ \frac{K_B}{K_A} = \frac{1}{3} \]

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