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 at a point where a force acts on a charge, we can use the formula:

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

where:
- \( E \) is the electric field intensity,
- \( F \) is the force acting on the charge, and
- \( q \) is the charge.

Given:
- \( F = 2.25 \, \text{N} \)
- \( q = 15 \times 10^{-4} \, \text{C} \)

### Step 1: Substitute the values into the formula

Substituting the given values into the formula for electric field intensity:

\[ E = \frac{2.25 \, \text{N}}{15 \times 10^{-4} \, \text{C}} \]

### Step 2: Simplify the expression

To simplify the expression, we can rewrite the denominator:

\[ E = \frac{2.25}{15 \times 10^{-4}} \]

### Step 3: Calculate the division

Now, let's calculate \( \frac{2.25}{15} \):

\[ \frac{2.25}{15} = 0.15 \]

Now, we can express this in terms of \( 10^{-4} \):

\[ E = \frac{0.15}{10^{-4}} \]

### Step 4: Convert to a more manageable form

To convert \( 0.15 \) into a fraction with a power of ten, we can multiply it by \( 10^2 \):

\[ 0.15 = \frac{15}{100} = \frac{15 \times 10^2}{100 \times 10^2} = \frac{15 \times 10^2}{100} \]

Thus, we can write:

\[ E = \frac{15 \times 10^2}{1} = 1500 \, \text{N/C} \]

### Final Answer

The intensity of the electric field at that point is:

\[ E = 1500 \, \text{N/C} \]

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