The force experienced by a charge of `2mu C` in an electric field is `3 xx 10^(-3)N`. The intensity of the electric field.

Options

A. `1.5 xx 10^(3) N//C`
B. `150 N//C`
C. `15 N//C`
D. `10 N//C`

Answer

To find the intensity of the electric field \( E \) experienced by a charge, we can use the formula:

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

where:
- \( E \) is the electric field intensity,
- \( F \) is the force experienced by the charge,
- \( Q \) is the charge.

### Step 1: Identify the given values
- The force \( F \) experienced by the charge is \( 3 \times 10^{-3} \, \text{N} \).
- The charge \( Q \) is \( 2 \, \mu C = 2 \times 10^{-6} \, \text{C} \).

### Step 2: Substitute the values into the formula
We can now substitute the known values into the formula:

\[ E = \frac{3 \times 10^{-3} \, \text{N}}{2 \times 10^{-6} \, \text{C}} \]

### Step 3: Perform the calculation
Now, we will perform the division:

\[ E = \frac{3 \times 10^{-3}}{2 \times 10^{-6}} = \frac{3}{2} \times \frac{10^{-3}}{10^{-6}} = \frac{3}{2} \times 10^{3} \]

Calculating \( \frac{3}{2} \):

\[ \frac{3}{2} = 1.5 \]

Thus, we have:

\[ E = 1.5 \times 10^{3} \, \text{N/C} \]

### Step 4: Final answer
The intensity of the electric field is:

\[ E = 1.5 \times 10^{3} \, \text{N/C} \]

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