Electric field in a region is given as `hatE=(10-5x)hati`. A charge particle of mass `5kg` and charge `Q(=1C)` is situated at origin and free to move given electric field. Then choose the correct options (Neglect gravity):

Options

A. Motion of charge particle is Oscillatory
B. Maximum displacement of charge particle from origin is `4SI` units
C. Maximum velocity gain by charge particle is `2 SI` units
D. The position of charge particle, when velocity gained by particle is maximum, is `2 SI` units

Explanation

To solve the problem step by step, we will analyze the given electric field, calculate the force acting on the charged particle, and determine the motion characteristics of the particle.

### Step 1: Understand the Electric Field
The electric field is given as:
\[ \hat{E} = (10 - 5x) \hat{i} \]
This indicates that the electric field varies with position \( x \).

### Step 2: Calculate the Force on the Charged Particle
The force \( F \) acting on a charged particle in an electric field is given by:
\[ F = Q \cdot E \]
Where:
- \( Q = 1 \, \text{C} \) (charge of the particle)
- \( E = 10 - 5x \)

Substituting the values:
\[ F = 1 \cdot (10 - 5x) = 10 - 5x \]

### Step 3: Relate Force to Motion
According to Newton's second law, \( F = ma \), where \( m \) is the mass of the particle. Given \( m = 5 \, \text{kg} \):
\[ 10 - 5x = 5a \]
Thus, we can express acceleration \( a \):
\[ a = \frac{10 - 5x}{5} = 2 - x \]

### Step 4: Identify the Type of Motion
The equation \( a = 2 - x \) indicates that the acceleration is proportional to the negative of the displacement \( x \). This is characteristic of simple harmonic motion (SHM).

### Step 5: Determine the Mean Position
To find the mean position (where the force is zero):
\[ 10 - 5x = 0 \implies 5x = 10 \implies x = 2 \]
This means the mean position is at \( x = 2 \, \text{SI units} \).

### Step 6: Calculate Maximum Displacement
In SHM, the maximum displacement (amplitude) from the mean position is equal to the distance from the mean position to the extreme position. Since the mean position is at \( x = 2 \):
- The particle can move 2 units to the left (0) and 2 units to the right (4).
- Therefore, the total maximum displacement from the origin is:
\[ \text{Total Displacement} = 2 + 2 = 4 \, \text{SI units} \]

### Step 7: Calculate Maximum Velocity
In SHM, the maximum velocity \( v_{\text{max}} \) can be calculated using:
\[ v_{\text{max}} = A \omega \]
Where \( A \) is the amplitude and \( \omega \) is the angular frequency. The angular frequency \( \omega \) can be derived from the equation of motion:
\[ \omega^2 = 1 \implies \omega = 1 \]
The amplitude \( A \) is 2 (the distance from the mean position to the extreme position):
\[ v_{\text{max}} = 2 \cdot 1 = 2 \, \text{SI units} \]

### Step 8: Position When Velocity is Maximum
In SHM, the maximum velocity occurs at the mean position. Thus, the position when the velocity is maximum is:
\[ x = 2 \, \text{SI units} \]

### Conclusion
Based on the analysis, the correct options are:
1. The maximum displacement of the charged particle from the origin is 4 SI units.
2. The maximum velocity gained by the charged particle is 2 SI units.
3. The position of the charged particle when the velocity gained is maximum is 2 SI units.