A force of 2.25 N acts on a charge of `15xx10^(-4)C`. The intensity of electric field at that point is

Options

A. `150NC^(-1)`
B. `15NC^(-1)`
C. `1500NC^(-1)`
D. `1.5NC^(-1)`

Explanation

To find the intensity of the electric field (E) at a point where a force (F) acts on a charge (q), 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:**
\[ E = \frac{F}{q} = \frac{2.25 \, \text{N}}{15 \times 10^{-4} \, \text{C}} \]

3. **Calculate the Denominator:**
- First, calculate \( 15 \times 10^{-4} \):
\[ 15 \times 10^{-4} = 0.0015 \, \text{C} \]

4. **Perform the Division:**
\[ E = \frac{2.25}{0.0015} \]

5. **Calculate the Result:**
- Performing the division:
\[ E = 1500 \, \text{N/C} \]

6. **Express the Result in Standard Units:**
- The intensity of the electric field can also be expressed as:
\[ E = 1500 \, \text{N/C} = 15 \times 10^{2} \, \text{N/C} \]

### Final Answer:
The intensity of the electric field at that point is \( 1500 \, \text{N/C} \) or \( 15 \times 10^{2} \, \text{N/C} \).

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