The electric force experienced by a charge of `5xx10^(-6) C` is `25xx10^(-3) N`. Find the magnitude of the electric field at that position of the charge due to the source charges.

Answer

To find the magnitude of the electric field at the position of a charge due to source charges, we can use the relationship between electric force, electric field, and charge.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Charge \( q = 5 \times 10^{-6} \, \text{C} \)
- Electric Force \( F = 25 \times 10^{-3} \, \text{N} \)

2. **Use the Formula for Electric Field**:
The electric field \( E \) at a point due to a charge is given by the formula:
\[ E = \frac{F}{q} \]
where \( F \) is the electric force experienced by the charge and \( q \) is the charge itself.

3. **Substitute the Values into the Formula**:
\[ E = \frac{25 \times 10^{-3} \, \text{N}}{5 \times 10^{-6} \, \text{C}} \]

4. **Calculate the Electric Field**:
- First, simplify the fraction:
\[ E = \frac{25}{5} \times \frac{10^{-3}}{10^{-6}} \]
- This simplifies to:
\[ E = 5 \times 10^{3} \, \text{N/C} \]

5. **Final Result**:
The magnitude of the electric field at that position of the charge due to the source charges is:
\[ E = 5 \times 10^{3} \, \text{N/C} \]