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 find the electric field intensity at a point in vacuum, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Electric Field Intensity**:
The electric field intensity (E) at a point in space is defined as the force (F) experienced by a unit positive charge (q) placed at that point. Mathematically, it is expressed as:
\[ E = \frac{F}{q} \]

2. **Coulomb's Law**:
In vacuum, the electric force between two point charges \( Q_1 \) and \( Q_2 \) separated by a distance \( r \) is given by Coulomb's law:
\[ F = \frac{k \cdot Q_1 \cdot Q_2}{r^2} \]
where \( k \) is Coulomb's constant, \( k = \frac{1}{4 \pi \epsilon_0} \), and \( \epsilon_0 \) is the permittivity of free space.

3. **Substituting for Electric Field**:
If we want to find the electric field intensity due to charge \( Q_1 \) at a distance \( r \), we can set \( Q_2 \) as the test charge \( q \). Thus, the electric field intensity \( E \) can be expressed as:
\[ E = \frac{F}{q} = \frac{k \cdot Q_1 \cdot q}{r^2 \cdot q} = \frac{k \cdot Q_1}{r^2} \]

4. **Incorporating Permittivity**:
Since \( k = \frac{1}{4 \pi \epsilon_0} \), we can rewrite the electric field intensity as:
\[ E = \frac{Q_1}{4 \pi \epsilon_0 r^2} \]

5. **Conclusion**:
Therefore, the electric field intensity at a point in vacuum due to a point charge \( Q_1 \) is given by:
\[ E = \frac{Q_1}{4 \pi \epsilon_0 r^2} \]

### Final Answer:
The electric field intensity at a point in vacuum is given by:
\[ E = \frac{Q_1}{4 \pi \epsilon_0 r^2} \]