What potential difference must be applied to produce an electric field that can acceleration an electric charge to one length the velocity of light ?

Options

A. 1352 V
B. 2511 V
C. 2531 V
D. 3521 V

Answer

To find the potential difference required to accelerate an electric charge (in this case, an electron) to a velocity equal to one length the speed of light, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the known values**:
- Mass of the electron, \( m = 9.1 \times 10^{-31} \) kg
- Charge of the electron, \( Q = -1.6 \times 10^{-19} \) C
- Speed of light, \( c = 3 \times 10^8 \) m/s

2. **Determine the kinetic energy**:
The kinetic energy (\( KE \)) gained by the electron when it is accelerated to a velocity \( v \) is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
Here, we want to find the kinetic energy when \( v = c \):
\[ KE = \frac{1}{2} m c^2 \]

3. **Substituting the values**:
\[ KE = \frac{1}{2} \times (9.1 \times 10^{-31} \text{ kg}) \times (3 \times 10^8 \text{ m/s})^2 \]
\[ KE = \frac{1}{2} \times (9.1 \times 10^{-31}) \times (9 \times 10^{16}) \]
\[ KE = \frac{1}{2} \times 8.19 \times 10^{-14} \text{ J} \]
\[ KE \approx 4.095 \times 10^{-14} \text{ J} \]

4. **Relate the kinetic energy to potential difference**:
The work done on the charge by the electric field is equal to the kinetic energy gained, which can be expressed as:
\[ KE = Q \cdot V \]
Rearranging gives:
\[ V = \frac{KE}{Q} \]

5. **Substituting the values**:
Since the charge of the electron is negative, we take the absolute value:
\[ V = \frac{4.095 \times 10^{-14} \text{ J}}{1.6 \times 10^{-19} \text{ C}} \]
\[ V \approx 2.56 \times 10^5 \text{ V} \]

6. **Final answer**:
The potential difference required to accelerate the electron to the speed of light is approximately \( 2.56 \times 10^5 \) volts, or \( 256,000 \) volts.