A small object, which has a charge `q = 7.5 mu C` and mass `m = 9.0 xx 10^(-5) kg`, is placed in a constant electric field. Starting from rest, the object accelerates to a speed of `2.0 xx 10^(3)m//s` in a time of 0.96 s. Determine the magnitude of the electric field.

Options

A. `2.5 xx 10^(4) N//C`
B. `4.5 xx 10^(4) N//C`
C. `3.4 xx 10^(4) N//C`
D. `1.7 xx 10^(4) N//C`

Answer

To determine the magnitude of the electric field acting on the charged object, we can follow these steps:

### Step 1: Calculate the acceleration of the object
The object starts from rest and reaches a speed of \( v = 2.0 \times 10^3 \, \text{m/s} \) in a time \( t = 0.96 \, \text{s} \). The formula for acceleration \( a \) is given by:

\[ a = \frac{v - u}{t} \]

where \( u \) is the initial velocity (which is 0 since it starts from rest).

Substituting the values:

\[ a = \frac{2.0 \times 10^3 \, \text{m/s} - 0}{0.96 \, \text{s}} = \frac{2.0 \times 10^3}{0.96} \approx 2083.33 \, \text{m/s}^2 \]

### Step 2: Relate the acceleration to the electric field
The only force acting on the charged object is due to the electric field. According to Newton's second law, the force \( F \) acting on the object can be expressed as:

\[ F = m \cdot a \]

where \( m \) is the mass of the object. The force due to the electric field \( E \) is given by:

\[ F = q \cdot E \]

where \( q \) is the charge of the object. Setting these two expressions for force equal gives:

\[ m \cdot a = q \cdot E \]

### Step 3: Solve for the electric field \( E \)
Rearranging the equation to solve for \( E \):

\[ E = \frac{m \cdot a}{q} \]

### Step 4: Substitute the known values
The mass \( m = 9.0 \times 10^{-5} \, \text{kg} \) and the charge \( q = 7.5 \, \mu C = 7.5 \times 10^{-6} \, \text{C} \). Now substituting the values into the equation:

\[ E = \frac{(9.0 \times 10^{-5} \, \text{kg}) \cdot (2083.33 \, \text{m/s}^2)}{7.5 \times 10^{-6} \, \text{C}} \]

Calculating the numerator:

\[ 9.0 \times 10^{-5} \cdot 2083.33 \approx 0.1875 \, \text{N} \]

Now substituting this value into the equation for \( E \):

\[ E = \frac{0.1875 \, \text{N}}{7.5 \times 10^{-6} \, \text{C}} \approx 25000 \, \text{N/C} = 2.5 \times 10^4 \, \text{N/C} \]

### Final Result
The magnitude of the electric field is approximately:

\[ E \approx 2.5 \times 10^4 \, \text{N/C} \]

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