Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area `A/3`and the separation between the plates is d, 2d and 3d respectively. The equivalent capacity of combination is `(epsilon_0` = absolute permittivity of free space)

Options

A. `( epsilon_(0)A)/(18d)`
B. `(11 epsilon_(0)A)/(18d)`
C. `(13 epsilon_(0)A)/(18d)`
D. `(17epsilon_(0)A)/(18d)`

Answer

To find the equivalent capacitance of three parallel plate capacitors connected in parallel, we can follow these steps:

### Step 1: Identify the parameters of each capacitor
- Each capacitor has a plate area of \( A/3 \).
- The separations between the plates are:
- For Capacitor 1 (C1): \( d \)
- For Capacitor 2 (C2): \( 2d \)
- For Capacitor 3 (C3): \( 3d \)

### Step 2: Calculate the capacitance of each capacitor
The formula for the capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\epsilon_0 A}{d} \]
where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( d \) is the separation between the plates.

Now, we can calculate the capacitance for each capacitor:

1. **Capacitance of C1**:
\[ C_1 = \frac{\epsilon_0 \left(\frac{A}{3}\right)}{d} = \frac{\epsilon_0 A}{3d} \]

2. **Capacitance of C2**:
\[ C_2 = \frac{\epsilon_0 \left(\frac{A}{3}\right)}{2d} = \frac{\epsilon_0 A}{6d} \]

3. **Capacitance of C3**:
\[ C_3 = \frac{\epsilon_0 \left(\frac{A}{3}\right)}{3d} = \frac{\epsilon_0 A}{9d} \]

### Step 3: Calculate the equivalent capacitance
Since the capacitors are connected in parallel, the equivalent capacitance \( C_{eq} \) is the sum of the individual capacitances:
\[ C_{eq} = C_1 + C_2 + C_3 \]
Substituting the values we calculated:
\[ C_{eq} = \frac{\epsilon_0 A}{3d} + \frac{\epsilon_0 A}{6d} + \frac{\epsilon_0 A}{9d} \]

### Step 4: Find a common denominator and add the fractions
The least common multiple of the denominators (3, 6, and 9) is 18. We can rewrite each term with a denominator of 18:
\[ C_{eq} = \frac{6\epsilon_0 A}{18d} + \frac{3\epsilon_0 A}{18d} + \frac{2\epsilon_0 A}{18d} \]
Now, combine the fractions:
\[ C_{eq} = \frac{(6 + 3 + 2)\epsilon_0 A}{18d} = \frac{11\epsilon_0 A}{18d} \]

### Final Answer
Thus, the equivalent capacitance of the combination is:
\[ C_{eq} = \frac{11\epsilon_0 A}{18d} \]