Two particles of mass `m` and `2m` with charges 2q and q are placed in a uniform electric field E and allowed to move for same time. Find the ratio of their kinetic energie

Options

A. `8:1`
B. `4:1`
C. `2:1`
D. `16:1`

Answer

To solve the problem of finding the ratio of the kinetic energies of two particles with different masses and charges placed in a uniform electric field, we can follow these steps:

### Step 1: Calculate the Force on Each Particle
1. For the first particle (mass = m, charge = 2q):
\[ F_1 = qE = 2qE \]

2. For the second particle (mass = 2m, charge = q):
\[ F_2 = qE = qE \]

### Step 2: Calculate the Acceleration of Each Particle
1. For the first particle:
\[ a_1 = \frac{F_1}{m} = \frac{2qE}{m} \]

2. For the second particle:
\[ a_2 = \frac{F_2}{2m} = \frac{qE}{2m} \]

### Step 3: Calculate the Velocity of Each Particle After Time t
1. For the first particle (initial velocity \(u = 0\)):
\[ v_1 = u + a_1 t = 0 + \left(\frac{2qE}{m}\right)t = \frac{2qEt}{m} \]

2. For the second particle (initial velocity \(u = 0\)):
\[ v_2 = u + a_2 t = 0 + \left(\frac{qE}{2m}\right)t = \frac{qEt}{2m} \]

### Step 4: Calculate the Kinetic Energy of Each Particle
1. For the first particle:
\[ KE_1 = \frac{1}{2} m v_1^2 = \frac{1}{2} m \left(\frac{2qEt}{m}\right)^2 = \frac{1}{2} m \cdot \frac{4q^2E^2t^2}{m^2} = \frac{2q^2E^2t^2}{m} \]

2. For the second particle:
\[ KE_2 = \frac{1}{2} (2m) v_2^2 = \frac{1}{2} (2m) \left(\frac{qEt}{2m}\right)^2 = \frac{1}{2} (2m) \cdot \frac{q^2E^2t^2}{4m^2} = \frac{q^2E^2t^2}{4m} \]

### Step 5: Find the Ratio of Kinetic Energies
Now, we can find the ratio \( \frac{KE_1}{KE_2} \):
\[ \frac{KE_1}{KE_2} = \frac{\frac{2q^2E^2t^2}{m}}{\frac{q^2E^2t^2}{4m}} = \frac{2}{\frac{1}{4}} = 2 \times 4 = 8 \]

### Final Answer
The ratio of the kinetic energies of the two particles is:
\[ \frac{KE_1}{KE_2} = 8:1 \]