To solve the problem step by step, we will calculate the original capacitance, the new capacitance after replacing air with glass, the new potential difference, and the energy stored in the capacitor.
### Step 1: Calculate the Original Capacitance (C₀)
The formula for the capacitance of a parallel plate capacitor is given by:
\[
C = \frac{\epsilon A}{d}
\]
Where:
- \(C\) = capacitance
- \(\epsilon\) = permittivity of the dielectric material (for air, \(\epsilon = \epsilon_0\))
- \(A\) = area of the plates
- \(d\) = separation between the plates
Given:
- Area \(A = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 0.01 \, \text{m}^2\)
- Plate separation \(d = 8.85 \, \text{mm} = 8.85 \times 10^{-3} \, \text{m}\)
- Permittivity of free space \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}\)
Substituting the values:
\[
C_0 = \frac{(8.85 \times 10^{-12} \, \text{F/m})(0.01 \, \text{m}^2)}{8.85 \times 10^{-3} \, \text{m}} = \frac{8.85 \times 10^{-14}}{8.85 \times 10^{-3}} = 10 \times 10^{-12} \, \text{F} = 10 \, \text{pF}
\]
### Step 2: Calculate the New Capacitance (C)
When the air is replaced by glass with a dielectric constant \(\epsilon_r = 5\), the new capacitance \(C\) is given by:
\[
C = \epsilon_r C_0
\]
Substituting the values:
\[
C = 5 \times 10 \, \text{pF} = 50 \, \text{pF}
\]
### Step 3: Calculate the Charge (Q)
The charge on the capacitor when it was charged to 100 V is given by:
\[
Q = C_0 V_0
\]
Where \(V_0 = 100 \, \text{V}\):
\[
Q = 10 \, \text{pF} \times 100 \, \text{V} = 1000 \, \text{pC} = 100 \, \text{pC}
\]
### Step 4: Calculate the New Potential Difference (V)
Since the capacitor is isolated, the charge remains constant. The new potential difference \(V\) can be calculated using:
\[
V = \frac{Q}{C}
\]
Substituting the values:
\[
V = \frac{100 \, \text{pC}}{50 \, \text{pF}} = \frac{100 \times 10^{-12} \, \text{C}}{50 \times 10^{-12} \, \text{F}} = 2 \, \text{V}
\]
### Step 5: Calculate the Energy Stored (U)
The energy stored in the capacitor is given by:
\[
U = \frac{1}{2} C V^2
\]
Substituting the values:
\[
U = \frac{1}{2} \times 50 \times 10^{-12} \, \text{F} \times (2 \, \text{V})^2 = \frac{1}{2} \times 50 \times 10^{-12} \times 4 = 100 \times 10^{-12} \, \text{J} = 10^{-10} \, \text{J}
\]
Converting to microjoules:
\[
U = 10^{-10} \, \text{J} = 0.1 \, \mu\text{J} = 100 \, \text{nJ}
\]
### Summary of Results
1. Original capacitance \(C_0 = 10 \, \text{pF}\)
2. New capacitance \(C = 50 \, \text{pF}\)
3. New potential difference \(V = 2 \, \text{V}\)
4. Energy stored \(U = 0.1 \, \mu\text{J}\)
