A parallel plate capacitor with air between the plates is charged to a potential difference of 500 V and then insulated. A plastic plate is inserted between the plates filling the whole gap. The potential difference between the plates now becomes 75V. The dielectric constant of plastic is

Options

A. 44107
B. 5
C. 20/3
D. 10

Explanation

To find the dielectric constant of the plastic plate inserted into the parallel plate capacitor, we can follow these steps:

### Step 1: Understand the initial conditions
The initial potential difference (V_initial) across the capacitor with air between the plates is given as:
- V_initial = 500 V
- The dielectric constant of air (k_air) = 1

### Step 2: Write the formula for capacitance
The capacitance (C) of a parallel plate capacitor is given by:
\[ C = C_0 \cdot k \]
where:
- \( C_0 \) is the capacitance without any dielectric (with air),
- \( k \) is the dielectric constant of the material inserted.

### Step 3: Determine the charge on the capacitor
Since the capacitor is insulated, the charge (Q) on the capacitor remains constant. The charge can be expressed as:
\[ Q = C \cdot V \]
For the initial condition (with air):
\[ Q_{initial} = C_0 \cdot V_{initial} = C_0 \cdot 500 \]

### Step 4: Analyze the final condition after inserting the plastic plate
After inserting the plastic plate, the potential difference (V_final) is given as:
- V_final = 75 V

The capacitance now becomes:
\[ C_{final} = C_0 \cdot k \]
Thus, the final charge can be expressed as:
\[ Q_{final} = C_{final} \cdot V_{final} = (C_0 \cdot k) \cdot 75 \]

### Step 5: Set the initial and final charge equal
Since the charge remains constant:
\[ Q_{initial} = Q_{final} \]
This gives us:
\[ C_0 \cdot 500 = (C_0 \cdot k) \cdot 75 \]

### Step 6: Simplify the equation
We can cancel \( C_0 \) from both sides (assuming \( C_0 \neq 0 \)):
\[ 500 = k \cdot 75 \]

### Step 7: Solve for the dielectric constant (k)
Now, we can solve for \( k \):
\[ k = \frac{500}{75} \]

### Step 8: Simplify the fraction
To simplify:
\[ k = \frac{500 \div 25}{75 \div 25} = \frac{20}{3} \]

### Conclusion
The dielectric constant of the plastic is:
\[ k = \frac{20}{3} \]

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