Two metal plates are separated by a distance `d` in a parallel plate condenser. A metal plate of thickness `t` and of the same area is inserted between the condenser plates. The value of capacitance increases by ….times.

Options

A. `(d-t)/(d)`
B. `(1-(t)/(d))`
C. `(t -(t)/(d))`
D. `(1)/((1-(t)/(d)))`

Explanation

To solve the problem, we need to analyze the effect of inserting a metal plate of thickness \( t \) between the plates of a parallel plate capacitor.

### Step 1: Understand the Initial Capacitance
The capacitance \( C \) of a parallel plate capacitor without any dielectric is given by the formula:
\[ C_0 = \frac{\varepsilon_0 A}{d} \]
where:
- \( C_0 \) is the initial capacitance,
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( d \) is the distance between the plates.

### Step 2: Determine the New Configuration
When a metal plate of thickness \( t \) is inserted between the plates, it effectively reduces the distance between the capacitor plates. The new effective distance \( d' \) between the plates can be calculated as:
\[ d' = d - t \]

### Step 3: Calculate the New Capacitance
The new capacitance \( C' \) of the capacitor with the inserted metal plate is given by:
\[ C' = \frac{\varepsilon_0 A}{d'} \]
Substituting \( d' \):
\[ C' = \frac{\varepsilon_0 A}{d - t} \]

### Step 4: Find the Ratio of the New Capacitance to the Old Capacitance
To find how many times the capacitance has increased, we take the ratio of the new capacitance \( C' \) to the initial capacitance \( C_0 \):
\[ \text{Increase Factor} = \frac{C'}{C_0} = \frac{\frac{\varepsilon_0 A}{d - t}}{\frac{\varepsilon_0 A}{d}} = \frac{d}{d - t} \]

### Step 5: Conclusion
Thus, the capacitance increases by a factor of:
\[ \frac{d}{d - t} \]