The electric potential difference between two parallel plates is `2000 V`. If the plates are separated by 2 mm, then what is the magnitude of electrostatic force on a charge of `4 xx 10^(-6)C` located midway between the plates ?

Options

A. 4 N
B. 6 N
C. 8 N
D. `1.5 xx 10^(-6)N`

Answer

To solve the problem, we need to find the magnitude of the electrostatic force on a charge located midway between two parallel plates with a given potential difference and separation distance.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Electric potential difference (V) = 2000 V
- Separation between the plates (d) = 2 mm = 2 x 10^(-3) m
- Charge (q) = 4 x 10^(-6) C

2. **Calculate the Electric Field (E):**
The electric field (E) between two parallel plates can be calculated using the formula:
\[ E = \frac{V}{d} \]
Substituting the given values:
\[ E = \frac{2000 \, \text{V}}{2 \times 10^{-3} \, \text{m}} = \frac{2000}{0.002} = 10^6 \, \text{N/C} \]

3. **Calculate the Electrostatic Force (F):**
The electrostatic force (F) on a charge in an electric field is given by:
\[ F = q \cdot E \]
Substituting the values we have:
\[ F = 4 \times 10^{-6} \, \text{C} \times 10^6 \, \text{N/C} = 4 \, \text{N} \]

4. **Conclusion:**
The magnitude of the electrostatic force on the charge located midway between the plates is:
\[ F = 4 \, \text{N} \]

### Final Answer:
The magnitude of the electrostatic force on the charge is **4 N**.

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