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 \( x \) at which the kinetic energy of a charged particle becomes zero after being released in a non-uniform electric field given by \( E = E_0(1 - ax^2) \).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Particle**:
The electric force \( F \) acting on the charged particle is given by:
\[ F = qE = qE_0(1 - ax^2) \]

2. **Calculate the Work Done**:
The work done \( W \) by the electric force as the particle moves from the origin (0) to position \( x \) is given by:
\[ W = \int_0^x F \, dx = \int_0^x qE_0(1 - ax^2) \, dx \]

3. **Integrate the Work Done**:
We can break the integral into two parts:
\[ W = qE_0 \int_0^x (1 - ax^2) \, dx = qE_0 \left[ x - \frac{ax^3}{3} \right]_0^x \]
Evaluating this gives:
\[ W = qE_0 \left( x - \frac{ax^3}{3} \right) \]

4. **Set the Work Done Equal to Change in Kinetic Energy**:
The change in kinetic energy \( \Delta KE \) is given by:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 0 - 0 = 0 \]
Therefore, we have:
\[ W = 0 \]

5. **Set Up the Equation**:
From the work done, we can set up the equation:
\[ qE_0 \left( x - \frac{ax^3}{3} \right) = 0 \]
Since \( qE_0 \) is not zero, we can simplify this to:
\[ x - \frac{ax^3}{3} = 0 \]

6. **Factor the Equation**:
Rearranging gives:
\[ x\left(1 - \frac{ax^2}{3}\right) = 0 \]
This yields two solutions:
- \( x = 0 \) (the initial position)
- \( 1 - \frac{ax^2}{3} = 0 \)

7. **Solve for \( x \)**:
From the second equation:
\[ \frac{ax^2}{3} = 1 \implies ax^2 = 3 \implies x^2 = \frac{3}{a} \implies x = \sqrt{\frac{3}{a}} \]

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