What is electric field intensity when a force of 2.25 N acts on a charge of `15xx10^(4)C`?

Options

A. `180NC^(-1)`
B. `1.5NC^(-1)`
C. `1500NC^(-1)`
D. `15NC^(-1)`

Explanation

To find the electric field intensity (E) when a force (F) of 2.25 N acts on a charge (Q) of \( 15 \times 10^{4} \, C \), we can use the formula:

\[ E = \frac{F}{Q} \]

### Step-by-Step Solution:

1. **Identify the given values**:
- Force, \( F = 2.25 \, N \)
- Charge, \( Q = 15 \times 10^{4} \, C \)

2. **Substitute the values into the formula**:
\[ E = \frac{2.25 \, N}{15 \times 10^{4} \, C} \]

3. **Calculate the denominator**:
- First, calculate \( 15 \times 10^{4} \):
\[ 15 \times 10^{4} = 150000 \, C \]

4. **Perform the division**:
\[ E = \frac{2.25}{150000} \]

5. **Simplify the division**:
- To simplify, we can convert \( 150000 \) to scientific notation:
\[ 150000 = 1.5 \times 10^{5} \]
- Now, we can rewrite the equation:
\[ E = \frac{2.25}{1.5 \times 10^{5}} = \frac{2.25}{1.5} \times 10^{-5} \]

6. **Calculate \( \frac{2.25}{1.5} \)**:
\[ \frac{2.25}{1.5} = 1.5 \]

7. **Final calculation**:
\[ E = 1.5 \times 10^{-5} \, N/C \]

### Final Answer:
The electric field intensity is:
\[ E = 1.5 \times 10^{-5} \, N/C \]