A point charge ` +q` & mass ` 100 gm` experiences a force of ` 100N` at a point of distance ` 20 cm` from a long infinite uniformly charged wire , If it is released find its speed when it is at a distance ` 40` cm from wire .

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Force on the Charge
We know that the force \( F \) acting on the charge \( +q \) at a distance \( r \) from an infinite charged wire is given by:
\[ F = \frac{\lambda q}{2 \pi \epsilon_0 r} \]
where \( \lambda \) is the linear charge density of the wire, \( \epsilon_0 \) is the permittivity of free space, and \( r \) is the distance from the wire.

### Step 2: Set Up the Initial Condition
Given that at \( r = 20 \, \text{cm} = 0.2 \, \text{m} \), the force \( F \) is \( 100 \, \text{N} \):
\[ 100 = \frac{\lambda q}{2 \pi \epsilon_0 (0.2)} \]

### Step 3: Solve for \( \lambda q \)
Rearranging the equation gives us:
\[ \lambda q = 100 \cdot 2 \pi \epsilon_0 \cdot 0.2 \]
Calculating \( 2 \pi \cdot 0.2 \):
\[ \lambda q = 40 \pi \epsilon_0 \]

### Step 4: Work Done by the Electric Field
The work done \( W \) when the charge moves from \( r_1 = 0.2 \, \text{m} \) to \( r_2 = 0.4 \, \text{m} \) is given by:
\[ W = \int_{r_1}^{r_2} F \, dr = \int_{0.2}^{0.4} \frac{\lambda q}{2 \pi \epsilon_0 r} \, dr \]
Substituting \( \lambda q \):
\[ W = \int_{0.2}^{0.4} \frac{40 \pi \epsilon_0}{2 \pi \epsilon_0 r} \, dr = 20 \int_{0.2}^{0.4} \frac{1}{r} \, dr \]

### Step 5: Evaluate the Integral
The integral \( \int \frac{1}{r} \, dr \) gives us:
\[ W = 20 \left[ \ln r \right]_{0.2}^{0.4} = 20 (\ln(0.4) - \ln(0.2)) = 20 \ln\left(\frac{0.4}{0.2}\right) = 20 \ln(2) \]

### Step 6: Relate Work Done to Kinetic Energy
According to the work-energy theorem, the work done is equal to the change in kinetic energy:
\[ W = \Delta KE = \frac{1}{2} m v^2 \]
Setting the two expressions for work equal gives:
\[ 20 \ln(2) = \frac{1}{2} m v^2 \]

### Step 7: Substitute Mass and Solve for Speed
The mass \( m = 100 \, \text{g} = 0.1 \, \text{kg} \):
\[ 20 \ln(2) = \frac{1}{2} (0.1) v^2 \]
\[ v^2 = \frac{20 \ln(2) \cdot 2}{0.1} = 400 \ln(2) \]
\[ v = \sqrt{400 \ln(2)} = 20 \sqrt{\ln(2)} \]

### Final Answer
Thus, the speed of the charge when it is at a distance of \( 40 \, \text{cm} \) from the wire is:
\[ v = 20 \sqrt{\ln(2)} \, \text{m/s} \]