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Step-by-Step Solution
Step 1: Identify given parameters of the capacitor
• Width of the capacitor plates, W = 4\,\text{cm} = 4 \times 10^{-2}\,\text{m}
• Length of the capacitor plates, L = 8\,\text{cm} = 8 \times 10^{-2}\,\text{m}
• Separation between plates, d = 4\,\text{mm} = 4 \times 10^{-3}\,\text{m}
• Applied voltage, V = 20\,\text{V} .
Step 2: Note the dimensions of the dielectric slab
• Dielectric constant, k = 5
• Length of the dielectric slab, l = 1\,\text{cm} = 1 \times 10^{-2}\,\text{m}
• Width of the dielectric slab (same as plate width), W_{\text{dielectric}} = 4\,\text{cm} = 4 \times 10^{-2}\,\text{m}
• Thickness of dielectric slab (equal to plate separation), d_{\text{dielectric}} = 4 \times 10^{-3}\,\text{m} .
Step 3: Determine the areas involved
Since the total plate has an area of
A_{\text{total}} = L \times W = (8 \times 10^{-2}) \times (4 \times 10^{-2}) = 32 \times 10^{-4}~\text{m}^2 = 3.2 \times 10^{-3}~\text{m}^2.
The portion occupied by the dielectric slab has an area
A_{\text{dielectric}} = l \times W_{\text{dielectric}}
= (1 \times 10^{-2}) \times (4 \times 10^{-2})
= 4 \times 10^{-4}\,\text{m}^2.
Hence, the area of the plate left in air (not covered by the dielectric) is
A_{\text{air}} = A_{\text{total}} - A_{\text{dielectric}}
= 3.2 \times 10^{-3} - 4 \times 10^{-4}
= 2.8 \times 10^{-3}~\text{m}^2.
Step 4: Write the expression for the equivalent capacitance
When a dielectric slab covers only part of the plate, we can treat the capacitor as having two parallel sections – one with air and one with dielectric. The net (equivalent) capacitance is:
C_{\text{eq}}
= \frac{\epsilon_0 \, A_{\text{air}}}{d}
+ \frac{k\,\epsilon_0\, A_{\text{dielectric}}}{d}.
Substituting the values:
C_{\text{eq}}
= \frac{\epsilon_0 \times (2.8 \times 10^{-3})}{4 \times 10^{-3}}
+ \frac{5\,\epsilon_0 \times (4 \times 10^{-4})}{4 \times 10^{-3}}.
Step 5: Simplify the expression for C_{\text{eq}}
Combine terms carefully:
C_{\text{eq}}
= \epsilon_0 \left( \frac{2.8 \times 10^{-3}}{4 \times 10^{-3}} \right)
+ 5\,\epsilon_0 \left( \frac{4 \times 10^{-4}}{4 \times 10^{-3}} \right).
Compute each fraction:
1. For air portion:
\frac{2.8 \times 10^{-3}}{4 \times 10^{-3}} = 0.7.
2. For dielectric portion:
\frac{4 \times 10^{-4}}{4 \times 10^{-3}}
= \frac{4 \times 10^{-4}}{4 \times 10^{-3}}
= 0.1 \quad \text{and multiplying by } 5 \text{ gives } 0.5.
Hence,
C_{\text{eq}} = \epsilon_0 \times 0.7 + \epsilon_0 \times 0.5
= 1.2\,\epsilon_0.
Step 6: Calculate the total electrostatic energy
The energy stored in a capacitor connected to a voltage V is given by
U = \frac{1}{2}\,C_{\text{eq}}\,V^2.
Substitute C_{\text{eq}} = 1.2\,\epsilon_0 and V = 20\,\text{V} :
U = \frac{1}{2} \times (1.2\,\epsilon_0) \times (20)^2.
First, compute (20)^2 = 400 , so
U = 0.6\,\epsilon_0 \times 400
= 240\,\epsilon_0.
This is the energy in joules, with \epsilon_0 factored out.
Final Answer
The electrostatic energy of the system is
\boxed{240\,\epsilon_0 \,\text{J}}.
