To solve the problem step by step, we need to determine the new capacitance of the capacitor after introducing two dielectric slabs.
### Step 1: Understand the Initial Conditions
- The initial capacitance \( C_0 \) of the capacitor is given as \( 2 \, \mu F \).
- The separation between the plates \( D \) is \( 4 \, mm \) (which is \( 0.004 \, m \)).
- Initially, there is air between the plates, which has a dielectric constant \( \epsilon_r = 1 \).
### Step 2: Determine the Capacitance with Dielectrics
When the two dielectric slabs are introduced, they are placed in series. Each slab has a thickness of \( 2 \, mm \) (or \( 0.002 \, m \)) and the dielectric constants are \( \epsilon_{r1} = 3 \) and \( \epsilon_{r2} = 5 \).
### Step 3: Calculate the Capacitance of Each Dielectric
The capacitance of a capacitor with a dielectric is given by:
\[
C = \frac{A \epsilon}{D}
\]
Where \( \epsilon = \epsilon_0 \epsilon_r \), \( \epsilon_0 \) is the permittivity of free space, and \( D \) is the separation.
For the first dielectric slab:
- Thickness \( D_1 = 2 \, mm = 0.002 \, m \)
- Dielectric constant \( \epsilon_{r1} = 3 \)
\[
C_1 = \frac{A \epsilon_0 \epsilon_{r1}}{D_1} = \frac{A \epsilon_0 \cdot 3}{0.002}
\]
For the second dielectric slab:
- Thickness \( D_2 = 2 \, mm = 0.002 \, m \)
- Dielectric constant \( \epsilon_{r2} = 5 \)
\[
C_2 = \frac{A \epsilon_0 \epsilon_{r2}}{D_2} = \frac{A \epsilon_0 \cdot 5}{0.002}
\]
### Step 4: Find the Equivalent Capacitance
Since the two capacitances \( C_1 \) and \( C_2 \) are in series, the equivalent capacitance \( C_{eq} \) is given by:
\[
\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}
\]
Substituting the expressions for \( C_1 \) and \( C_2 \):
\[
\frac{1}{C_{eq}} = \frac{0.002}{3 A \epsilon_0} + \frac{0.002}{5 A \epsilon_0}
\]
### Step 5: Simplify the Expression
Taking the common denominator:
\[
\frac{1}{C_{eq}} = \frac{0.002 \cdot 5 + 0.002 \cdot 3}{15 A \epsilon_0} = \frac{0.002 \cdot 8}{15 A \epsilon_0}
\]
Thus,
\[
C_{eq} = \frac{15 A \epsilon_0}{0.016}
\]
### Step 6: Relate to the Initial Capacitance
From the initial capacitance formula, we know:
\[
C_0 = \frac{A \epsilon_0}{D} = \frac{A \epsilon_0}{0.004}
\]
We can express \( A \epsilon_0 \) in terms of \( C_0 \):
\[
A \epsilon_0 = C_0 \cdot 0.004
\]
Substituting this into the equation for \( C_{eq} \):
\[
C_{eq} = \frac{15 \cdot C_0 \cdot 0.004}{0.016}
\]
### Step 7: Calculate the Equivalent Capacitance
Substituting \( C_0 = 2 \, \mu F \):
\[
C_{eq} = \frac{15 \cdot 2 \cdot 0.004}{0.016} = \frac{15 \cdot 2 \cdot 0.25}{1} = 7.5 \, \mu F
\]
### Final Answer
The new capacitance after inserting the dielectric slabs is \( 7.5 \, \mu F \).
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