A charged particle of chage `q` released in electric field `E= E_0(1-ax^2)` from origin. find position when its kinetic energy again becomes zero.

Options

A. `(sqrt(1/a))`
B. `(sqrt(2/a))`
C. `(sqrt(3/a))`
D. `2(sqrt(1/a))`

Answer

To solve the problem, we need to find the position \( x \) where the kinetic energy of a charged particle becomes zero after being released in the given electric field \( E = E_0(1 - ax^2) \). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Conditions
The charged particle is released from the origin (i.e., \( x = 0 \)) with an initial velocity of zero. Therefore, its initial kinetic energy \( KE_i \) is:
\[ KE_i = \frac{1}{2} mv^2 = 0 \]

### Step 2: Apply Work-Energy Theorem
According to the work-energy theorem, the work done on the particle by the electric field is equal to the change in kinetic energy:
\[ W = KE_f - KE_i \]
Since \( KE_i = 0 \), we have:
\[ W = KE_f \]
We need to find the position \( x \) where the final kinetic energy \( KE_f \) becomes zero again.

### Step 3: Calculate the Work Done by the Electric Field
The work done \( W \) when moving from \( x = 0 \) to \( x \) in an electric field \( E \) is given by:
\[ W = \int_0^x F \, dx = \int_0^x qE \, dx \]
Substituting \( E = E_0(1 - ax^2) \), we get:
\[ W = \int_0^x qE_0(1 - ax^2) \, dx \]

### Step 4: Perform the Integration
Now we perform the integration:
\[ W = qE_0 \int_0^x (1 - ax^2) \, dx \]
Calculating the integral:
\[ W = qE_0 \left[ x - \frac{ax^3}{3} \right]_0^x = qE_0 \left( x - \frac{ax^3}{3} \right) \]

### Step 5: Set Work Done to Zero
Since we want the final kinetic energy \( KE_f = 0 \), we set the work done equal to zero:
\[ qE_0 \left( x - \frac{ax^3}{3} \right) = 0 \]
This gives us two cases:
1. \( qE_0 = 0 \) (which is not possible since \( q \) and \( E_0 \) are non-zero)
2. \( x - \frac{ax^3}{3} = 0 \)

### Step 6: Solve for Position \( x \)
From the equation \( x - \frac{ax^3}{3} = 0 \), we can factor out \( x \):
\[ x \left( 1 - \frac{ax^2}{3} \right) = 0 \]
This gives us:
1. \( x = 0 \) (the initial position)
2. \( 1 - \frac{ax^2}{3} = 0 \) leading to:
\[ \frac{ax^2}{3} = 1 \implies x^2 = \frac{3}{a} \implies x = \sqrt{\frac{3}{a}} \]

### Final Answer
The position \( x \) where the kinetic energy of the charged particle again becomes zero is:
\[ x = \sqrt{\frac{3}{a}} \]