A mass `m=20 g` has a charge `q= 3.0 mC`. It moves with a velocity of 20 m/s and enters a region of electric field of 80 N/C in the same direction as the velocity of the mass. The velocity of the mass after 3 s in this region is

Options

A. 80 m/s
B. 56 m/s
C. 44 m/s
D. 40 m/s

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the given values
- Mass \( m = 20 \, \text{g} = 0.02 \, \text{kg} \) (convert grams to kilograms)
- Charge \( q = 3.0 \, \text{mC} = 3.0 \times 10^{-3} \, \text{C} \) (convert milliCoulombs to Coulombs)
- Initial velocity \( u = 20 \, \text{m/s} \)
- Electric field \( E = 80 \, \text{N/C} \)
- Time \( t = 3 \, \text{s} \)

### Step 2: Calculate the electric force acting on the mass
The electric force \( F \) on the mass due to the electric field is given by:
\[ F = q \cdot E \]
Substituting the values:
\[ F = (3.0 \times 10^{-3} \, \text{C}) \cdot (80 \, \text{N/C}) = 0.24 \, \text{N} \]

### Step 3: Calculate the acceleration of the mass
Using Newton's second law, the acceleration \( a \) can be calculated as:
\[ a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{0.24 \, \text{N}}{0.02 \, \text{kg}} = 12 \, \text{m/s}^2 \]

### Step 4: Calculate the final velocity after 3 seconds
Using the first equation of motion:
\[ v = u + a \cdot t \]
Substituting the values:
\[ v = 20 \, \text{m/s} + (12 \, \text{m/s}^2 \cdot 3 \, \text{s}) = 20 \, \text{m/s} + 36 \, \text{m/s} = 56 \, \text{m/s} \]

### Final Answer
The velocity of the mass after 3 seconds in the electric field is \( 56 \, \text{m/s} \).

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