A system of two parallel plates, each of area A, are separated by distance `d_(1)` and `d_(2)`. The space between them is filled with dielectrices of permitivities `epsilon_(1)` and `epsilon)_(2)`. The permittivity of free space is `epsilon_(0)`. The equivalent capacitance of the system is

Options

A. `(epsilon_(1)epsilon_(2)A)/(epsilon_(2)d_(1)+epsilon_(1)d_(2))`
B. `(epsilon_(1)epsilon_(2)epsilon_(0)A)/(epsilon_(1)d_(1)+epsilon_(2)d_(2))`
C. `(epsilon_(0)A)/(epsilon_(1)d_(1)+epsilon_(2)d_(2))`
D. `(epsilon_(0)A)/(epsilon_(1)d_(2)+epsilon_(2)d_(1))`

Explanation

To find the equivalent capacitance of a system of two parallel plates filled with dielectrics, we can follow these steps:

### Step 1: Understand the Configuration
We have two parallel plates, each with an area \( A \), separated by two different distances \( d_1 \) and \( d_2 \). The space between the plates is filled with two different dielectrics with permittivities \( \epsilon_1 \) and \( \epsilon_2 \).

### Step 2: Identify the Capacitance of Each Section
The capacitance of a parallel plate capacitor filled with a dielectric is given by the formula:
\[ C = \frac{\epsilon A}{d} \]
For the first section with dielectric \( \epsilon_1 \) and distance \( d_1 \):
\[ C_1 = \frac{\epsilon_1 A}{d_1} \]
For the second section with dielectric \( \epsilon_2 \) and distance \( d_2 \):
\[ C_2 = \frac{\epsilon_2 A}{d_2} \]

### Step 3: Determine the Equivalent Capacitance
Since the two capacitors \( C_1 \) and \( C_2 \) are in series, the equivalent capacitance \( C \) can be calculated using the formula for capacitors in series:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the expressions for \( C_1 \) and \( C_2 \):
\[ \frac{1}{C} = \frac{d_1}{\epsilon_1 A} + \frac{d_2}{\epsilon_2 A} \]

### Step 4: Combine the Terms
To combine the terms, we can find a common denominator:
\[ \frac{1}{C} = \frac{d_1 \epsilon_2 + d_2 \epsilon_1}{\epsilon_1 \epsilon_2 A} \]

### Step 5: Invert to Find \( C \)
Now, we can invert both sides to find \( C \):
\[ C = \frac{\epsilon_1 \epsilon_2 A}{d_1 \epsilon_2 + d_2 \epsilon_1} \]

### Final Answer
Thus, the equivalent capacitance of the system is:
\[ C = \frac{\epsilon_1 \epsilon_2 A}{d_1 \epsilon_2 + d_2 \epsilon_1} \]

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