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. `40 ms^(-1)`
B. `44 ms^(-1)`
C. `56 ms^(-1)`
D. `80 ms^(-1)`

Answer

To solve the problem step by step, we will follow the given information and apply the relevant physics concepts.

### Step 1: Identify the given values
- Mass (m) = 20 g = 20 × 10^(-3) kg = 0.02 kg
- Charge (q) = 3.0 mC = 3 × 10^(-3) C = 0.003 C
- Initial velocity (u) = 20 m/s
- Electric field (E) = 80 N/C
- Time (t) = 3 s

### Step 2: Calculate the acceleration (a)
The acceleration of the mass due to the electric field can be calculated using the formula:

\[ a = \frac{qE}{m} \]

Substituting the values:

\[ a = \frac{(3 \times 10^{-3} \, \text{C})(80 \, \text{N/C})}{0.02 \, \text{kg}} \]

Calculating the numerator:

\[ 3 \times 10^{-3} \times 80 = 0.24 \, \text{N} \]

Now, substituting this into the acceleration formula:

\[ a = \frac{0.24 \, \text{N}}{0.02 \, \text{kg}} = 12 \, \text{m/s}^2 \]

### Step 3: Use the equation of motion to find the final velocity (v)
We will use the equation of motion:

\[ v = u + at \]

Substituting the known values:

\[ v = 20 \, \text{m/s} + (12 \, \text{m/s}^2)(3 \, \text{s}) \]

Calculating the second term:

\[ 12 \times 3 = 36 \, \text{m/s} \]

Now substituting back into the equation:

\[ v = 20 \, \text{m/s} + 36 \, \text{m/s} = 56 \, \text{m/s} \]

### Final Answer
The velocity of the mass after 3 seconds in the electric field is **56 m/s**.

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