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

### Step 2: Calculate the potential difference (\( \Delta V \))
The potential difference across the capacitor can be calculated using the formula:
\[ \Delta V = \frac{Q}{C} \]
Substituting the values:
\[ \Delta V = \frac{15 \times 10^{-6}}{3 \times 10^{-6}} = 5 V \]

### Step 3: Calculate the electric field (\( E \))
The electric field \( E \) between the plates of a capacitor is given by the formula:
\[ E = \frac{\Delta V}{d} \]
Substituting the values:
\[ E = \frac{5}{1 \times 10^{-3}} = 5000 \, N/C \]

### Final Answer
The magnitude of the electric field between the plates is \( 5000 \, N/C \).

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