A particle of mass 1g and charge 2 C is released from a point A in a region of space containing electric field. Its speed is measured to be 2 `m//s` when it passes a point B some distance away from A. Potential difference the two points A and B is

Options

A. 1 mV
B. 2mV
C. 4 Mv
D. Both the points are a the same potentials

Answer

To find the potential difference between points A and B, we can follow these steps:

### Step 1: Identify the given values
- Mass of the particle, \( m = 1 \, \text{g} = 0.001 \, \text{kg} \)
- Charge of the particle, \( q = 2 \, \text{C} \)
- Initial speed at point A, \( u = 0 \, \text{m/s} \) (released from rest)
- Final speed at point B, \( v = 2 \, \text{m/s} \)

### Step 2: Use the kinematic equation
We will use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ (2)^2 = (0)^2 + 2as \]
This simplifies to:
\[ 4 = 2as \]
From this, we can express \( as \):
\[ as = 2 \]

### Step 3: Express acceleration in terms of force
The force acting on the particle due to the electric field is given by:
\[ F = qE \]
By Newton's second law, we also have:
\[ F = ma \]
Setting these equal gives:
\[ ma = qE \]
From this, we can express acceleration \( a \):
\[ a = \frac{qE}{m} \]

### Step 4: Substitute acceleration back into the equation
Substituting \( a \) back into the equation \( as = 2 \):
\[ \left(\frac{qE}{m}\right)s = 2 \]
Thus, we can express \( s \):
\[ s = \frac{2m}{qE} \]

### Step 5: Calculate the work done
The work done by the electric field is equal to the change in potential energy, which can be expressed as:
\[ \Delta U = F \cdot s = qEs \]
Substituting \( s \):
\[ \Delta U = qE \left(\frac{2m}{qE}\right) = 2m \]

### Step 6: Relate potential difference to change in potential energy
The potential difference \( \Delta V \) is given by:
\[ \Delta V = \frac{\Delta U}{q} \]
Substituting \( \Delta U \):
\[ \Delta V = \frac{2m}{q} \]

### Step 7: Substitute the known values
Now substituting the values of \( m \) and \( q \):
\[ \Delta V = \frac{2 \times 0.001 \, \text{kg}}{2 \, \text{C}} = \frac{0.002}{2} = 0.001 \, \text{V} \]
Thus,
\[ \Delta V = 1 \, \text{mV} \]

### Final Answer
The potential difference between points A and B is \( 1 \, \text{mV} \).

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