a proton placed in an electric field would experience an electrical force equal to its weight. the magnitude of electric field intensity E would be

Options

A. `m^2g/e`
B. `e^2m^2/g`
C. mg/e
D. e/mg

Answer

To solve the problem, we need to find the magnitude of the electric field intensity \( E \) such that the electric force acting on a proton is equal to its weight.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The weight of the proton (\( W \)) is given by the formula:
\[ W = mg \]
where \( m \) is the mass of the proton and \( g \) is the acceleration due to gravity.

2. **Electric Force on the Proton**:
- The electric force (\( F_e \)) acting on the proton when placed in an electric field \( E \) is given by:
\[ F_e = qE \]
where \( q \) is the charge of the proton.

3. **Set the Forces Equal**:
- According to the problem, the electric force is equal to the weight of the proton:
\[ qE = mg \]

4. **Substitute the Charge of the Proton**:
- The charge of a proton (\( q \)) is approximately \( 1.6 \times 10^{-19} \) coulombs.

5. **Rearranging the Equation**:
- To find the electric field intensity \( E \), rearrange the equation:
\[ E = \frac{mg}{q} \]

6. **Substituting Values**:
- Substitute the known values for \( m \) (mass of the proton, approximately \( 1.67 \times 10^{-27} \) kg) and \( g \) (approximately \( 9.81 \, \text{m/s}^2 \)):
\[ E = \frac{(1.67 \times 10^{-27} \, \text{kg})(9.81 \, \text{m/s}^2)}{1.6 \times 10^{-19} \, \text{C}} \]

7. **Calculating \( E \)**:
- Calculate the numerator:
\[ mg \approx 1.67 \times 10^{-27} \times 9.81 \approx 1.64 \times 10^{-26} \, \text{N} \]
- Now, calculate \( E \):
\[ E \approx \frac{1.64 \times 10^{-26}}{1.6 \times 10^{-19}} \approx 1.025 \times 10^{-7} \, \text{N/C} \]

8. **Final Result**:
- The magnitude of the electric field intensity \( E \) is approximately:
\[ E \approx 1.025 \times 10^{-7} \, \text{N/C} \]