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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].
### 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].
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