A parallel plate capacitor of capacitance `3muF` has total charge `+15muC` on one plate and total charge `-15muC` on the other plate. The separation between the plates is 1mm. The electric field between the plates has magnitude: (in N/C)

Options

A. 50
B. 100
C. 5000
D. 10000

Explanation

To find the electric field between the plates of a parallel plate capacitor, we can follow these steps:

### Step 1: Understand the given values
- Capacitance (C) = 3 µF = \(3 \times 10^{-6}\) F
- Charge on one plate (Q) = +15 µC = \(15 \times 10^{-6}\) C
- Charge on the other plate = -15 µC = \(-15 \times 10^{-6}\) C
- Separation between the plates (d) = 1 mm = \(1 \times 10^{-3}\) m

### Step 2: Calculate the potential difference (ΔV)
The relationship between charge (Q), capacitance (C), and potential difference (ΔV) is given by the formula:
\[ Q = C \cdot \Delta V \]
Rearranging this gives:
\[ \Delta V = \frac{Q}{C} \]

Substituting the values:
\[ \Delta V = \frac{15 \times 10^{-6}}{3 \times 10^{-6}} \]
\[ \Delta V = 5 \, \text{V} \]

### Step 3: Calculate the electric field (E)
The electric field (E) between the plates of a parallel plate capacitor is given by the formula:
\[ E = \frac{\Delta V}{d} \]

Substituting the values we have:
\[ E = \frac{5}{1 \times 10^{-3}} \]
\[ E = 5000 \, \text{N/C} \]

### Final Answer
The magnitude of the electric field between the plates is **5000 N/C**.

---