A parallel plate capacitor with plate separation 5mm is charged up by a battery. It is found that on introducing a dielectric sheet of thickness 2 mm, while keeping the battery connections intact, the capacitor draws 25% more charge from the battery than before. The dielectric constant of the sheet is ____.
Correct Answer: 2
Approach Solution - 1
Problem Statement:
A parallel plate capacitor is composed of two conducting plates separated by a distance \( d = 5 \, \text{mm} = 0.005 \, \text{m} \). Initially, the capacitance \( C_0 \) without a dielectric and with air between the plates is given by: \[ C_0 = \frac{\varepsilon_0 \cdot A}{d}, \] where \( \varepsilon_0 \) is the permittivity of free space, and \( A \) is the area of the plates. With the introduction of a dielectric sheet of thickness \( t = 2 \, \text{mm} = 0.002 \, \text{m} \), the dielectric constant \( K \) is introduced. The effective separation becomes \( d - t = 3 \, \text{mm} = 0.003 \, \text{m} \). The capacitance with the dielectric, while keeping the battery connected, can be approximated using two series capacitors:
- One capacitor with thickness \( d - t \), with air between the plates and capacitance \( C_1 = \frac{\varepsilon_0 \cdot A}{d - t} \).
- A second capacitor with thickness \( t \), with the dielectric material, whose capacitance is \( C_2 = \frac{K \cdot \varepsilon_0 \cdot A}{t} \).
The total capacitance \( C \) is the equivalent of two capacitors in series. The formula for the total capacitance is given by:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d - t}{\varepsilon_0 \cdot A} + \frac{t}{K \cdot \varepsilon_0 \cdot A}. \]
Multiplying through by \( \varepsilon_0 \cdot A \) and simplifying, we obtain:
\[ \frac{1}{C} = \frac{d - t + \frac{t}{K}}{\varepsilon_0 \cdot A}. \]
Since it is stated that the charge increases by 25%, we have the relationship:
\[ \frac{C}{C_0} = 1.25. \] This implies: \[ C = 1.25 \cdot C_0. \]
Substituting \( C_0 = \frac{\varepsilon_0 \cdot A}{d} \) into this relationship, we get:
- \[ \frac{\varepsilon_0 \cdot A}{d - t + \frac{t}{K}} = 1.25 \cdot \frac{\varepsilon_0 \cdot A}{d}. \]
- Simplifying yields:
- Solving for \( K \), substitute \( d = 5 \), \( t = 2 \), we get:
- This simplifies to:
- Thus, we have:
- Solve for \( K \):
Conclusion: The dielectric constant is \( K = 2 \), which falls within the expected range, confirming the result is correct and matches the problem's constraints.
Approach Solution -2
Without the dielectric, the charge stored is given by:
\[ Q = \frac{A \epsilon_0 V}{d} \]
With the dielectric inserted:
\[ Q' = \frac{A \epsilon_0 V}{d - t + \frac{t}{K}} \]
Given:
\[ Q' = 1.25 Q \implies \frac{A \epsilon_0 V}{d - t + \frac{t}{K}} = 1.25 \cdot \frac{A \epsilon_0 V}{d} \]
Canceling common terms:
\[ \frac{d}{d - t + \frac{t}{K}} = 1.25 \]
Substituting \( d = 5 \, \text{mm} \) and \( t = 2 \, \text{mm} \):
\[ 1.25 \left( 5 - 2 + \frac{2}{K} \right) = 5 \]
Simplifying:
\[ 1.25 \left( 3 + \frac{2}{K} \right) = 5 \]
\[ 3 + \frac{2}{K} = 4 \]
\[ \frac{2}{K} = 1 \implies K = 2 \]
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