The electric filed intensity at a point in vacuum is equal to

Options

A. Zero
B. Force a proton would experience there
C. Force an electron would experience there
D. Force a unit positive charge would experience there

Answer

To determine the electric field intensity at a point in vacuum due to a source charge, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Source Charge**:
Let's denote the source charge as \( Q \). For this problem, we will consider \( Q = 1 \, \text{C} \) (1 coulomb).

2. **Define the Electric Field**:
The electric field \( E \) at a point in space due to a charge \( Q \) is defined as the force \( F \) experienced by a unit positive test charge placed at that point, divided by the magnitude of the test charge \( q \):
\[ E = \frac{F}{q} \]
For a unit positive charge, \( q = 1 \, \text{C} \), so the formula simplifies to:
\[ E = F \]

3. **Calculate the Force on the Test Charge**:
The force \( F \) experienced by a test charge \( q \) due to the source charge \( Q \) can be calculated using Coulomb's law:
\[ F = k \frac{|Q \cdot q|}{r^2} \]
where \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)) and \( r \) is the distance from the source charge to the point where the electric field is being measured.

4. **Substituting the Values**:
For a unit positive charge \( q = 1 \, \text{C} \):
\[ F = k \frac{|Q \cdot 1|}{r^2} = k \frac{Q}{r^2} \]
Thus, the electric field \( E \) at that point becomes:
\[ E = \frac{F}{1} = k \frac{Q}{r^2} \]

5. **Final Expression**:
Therefore, the electric field intensity \( E \) at a point in vacuum due to a source charge \( Q \) is given by:
\[ E = k \frac{Q}{r^2} \]

### Conclusion:
In this case, if \( Q = 1 \, \text{C} \), the electric field intensity at a distance \( r \) from the charge is:
\[ E = k \frac{1}{r^2} \]