A force of 2.56 N acts on a charge of `16 xx 10^(-4) C`. The intensity of electric field at that point is :

Options

A. `1600 NC^(-1)`
B. `150 NC^(-1)`
C. `16 NC^(-1)`
D. `1.5 NC^(-1)`

Answer

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) = 2.56 N
- Charge (Q) = \( 16 \times 10^{-4} \) C

2. **Substitute the values into the formula**:
\[ E = \frac{2.56 \, \text{N}}{16 \times 10^{-4} \, \text{C}} \]

3. **Simplify the denominator**:
- Convert \( 16 \times 10^{-4} \) to a decimal:
\[ 16 \times 10^{-4} = 0.0016 \, \text{C} \]

4. **Calculate the electric field**:
\[ E = \frac{2.56}{0.0016} \]

5. **Perform the division**:
\[ E = 1600 \, \text{N/C} \quad \text{(or equivalently, 1600 V/m)} \]

### Final Answer:
The intensity of the electric field at that point is \( 1600 \, \text{N/C} \).

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