To find the energy [tex]\( U_2 \)[/tex] of the capacitor at the moment when the capacitor is half-filled with the dielectric, follow these steps:
### Step 1: Convert the given dimensions to standard units
- The plate area [tex]\( A \)[/tex] is given as [tex]\( 30.0 \, \text{cm}^2 \)[/tex]. Convert this to square meters:
[tex]\[
A = 30.0 \times 10^{-4} \, \text{m}^2
\][/tex]
- The plate separation [tex]\( d \)[/tex] is given as [tex]\( 5.00 \, \text{mm} \)[/tex]. Convert this to meters:
[tex]\[
d = 5.00 \times 10^{-3} \, \text{m}
\][/tex]
### Step 2: Calculate the capacitance [tex]\( C_1 \)[/tex] with the dielectric fully inserted
The capacitance of a parallel-plate capacitor filled with a dielectric material of dielectric constant [tex]\( k \)[/tex] is given by:
[tex]\[
C_1 = \frac{k \epsilon_0 A}{d}
\][/tex]
### Step 3: Calculate the capacitance [tex]\( C_2 \)[/tex] when half of the dielectric is pulled out
When the capacitor is half-filled with the dielectric, the capacitance becomes a combination of two capacitors in parallel: one with the dielectric and one with air (vacuum):
[tex]\[
C_2 = \frac{\epsilon_0 A}{d} + \frac{k \epsilon_0 A}{2d}
\][/tex]
### Step 4: Calculate the energy [tex]\( U_2 \)[/tex]
The energy stored in a capacitor is given by:
[tex]\[
U = \frac{1}{2} C V^2
\][/tex]
Using the capacitance when half of the dielectric is pulled out [tex]\( C_2 \)[/tex]:
[tex]\[
U_2 = \frac{1}{2} C_2 V^2
\][/tex]
### Step 5: Substitute the values and solve
Given values:
[tex]\[
\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2
\][/tex]
[tex]\[
A = 30.0 \times 10^{-4} \, \text{m}^2
\][/tex]
[tex]\[
d = 5.00 \times 10^{-3} \, \text{m}
\][/tex]
[tex]\[
k = 4.00
\][/tex]
[tex]\[
V = 12.5 \, \text{V}
\][/tex]
First, calculate [tex]\( C_2 \)[/tex]:
[tex]\[
C_2 = \frac{(\epsilon_0 \cdot A)}{d} + \frac{(k \cdot \epsilon_0 \cdot A)}{2d}
\][/tex]
[tex]\[
C_2 = \frac{(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2) \cdot (30.0 \times 10^{-4} \, \text{m}^2)}{5.00 \times 10^{-3} \, \text{m}} + \frac{4 \cdot (8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2) \cdot (30.0 \times 10^{-4} \, \text{m}^2)}{2 \cdot 5.00 \times 10^{-3} \, \text{m}}
\][/tex]
Calculate each term:
[tex]\[
C_2 = \frac{(8.85 \times 10^{-12}) \cdot (30.0 \times 10^{-4})}{5.00 \times 10^{-3}} + \frac{4 (8.85 \times 10^{-12}) \cdot (30.0 \times 10^{-4})}{2 \cdot 5.00 \times 10^{-3}}
\][/tex]
Combine and simplify:
[tex]\[
C_2 \approx 1.326 \times 10^{-11} \, \text{F}
\][/tex]
Next, calculate the energy:
[tex]\[
U_2 = \frac{1}{2} \times 1.326 \times 10^{-11} \times (12.5)^2
\][/tex]
[tex]\[
U_2 \approx 1.24453125 \times 10^{-9} \, \text{J}
\][/tex]
Therefore, the energy [tex]\( U_2 \)[/tex] of the capacitor when it is half-filled with the dielectric is [tex]\( \boxed{1.24453125 \times 10^{-9} \, \text{J}} \)[/tex].
