A parallel plate capacitor consists of two plates of `2m xx 1m.` The space between the plates is of 1mm and filled with a dielectric of relative permittivity of 7. A potential difference of 300 volts is applied across the plates. Find the potential gradient

Options

A. `6 xx 10^(5) N//C`
B. `3 xx 10^(5) N//C`
C. `18 xx 10^(5) N//C`
D. `12 xx 10^(5) N//C`

Answer

To find the potential gradient in the parallel plate capacitor, we can follow these steps:

### Step 1: Identify the given values
- Area of the plates: \(2 \, \text{m} \times 1 \, \text{m}\)
- Separation between the plates: \(1 \, \text{mm} = 1 \times 10^{-3} \, \text{m}\)
- Potential difference (voltage) across the plates: \(300 \, \text{V}\)
- Relative permittivity (dielectric constant): \(7\) (not directly needed for potential gradient calculation)

### Step 2: Understand the formula for potential gradient
The potential gradient (also known as electric field strength, \(E\)) between the plates of a capacitor is given by the formula:
\[ E = \frac{\Delta V}{d} \]
where:
- \(E\) is the potential gradient (in volts per meter, V/m),
- \(\Delta V\) is the potential difference (in volts, V),
- \(d\) is the separation between the plates (in meters, m).

### Step 3: Substitute the known values into the formula
Substituting the known values into the formula:
\[ E = \frac{300 \, \text{V}}{1 \times 10^{-3} \, \text{m}} \]

### Step 4: Calculate the potential gradient
Calculating the above expression:
\[ E = \frac{300}{0.001} = 300000 \, \text{V/m} = 3 \times 10^5 \, \text{V/m} \]

### Step 5: State the final answer
The potential gradient in the parallel plate capacitor is:
\[ E = 3 \times 10^5 \, \text{V/m} \quad \text{or} \quad 3 \times 10^5 \, \text{N/C} \]