Question

When a parallel plate capacitor is charged up to 95 V, its capacitance is \( C \). If a dielectric slab of thickness 2 mm is inserted between plates and the plate separation is increased by 1.6 mm such that the potential difference remains constant, find the dielectric constant of the material:

• A. \( 2.4 \)
• B. \( 4.5 \)
• C. \( 5.0 \)
• D. \( 9.0 \)

Hint:

The capacitance of a capacitor with a dielectric slab depends on the thickness of the slab and its dielectric constant. The new capacitance is calculated using modified plate separation.

Answer 1

The Correct Option is C

Step 1: Understanding the Capacitance Relation The new capacitance with a dielectric slab is given by: \[ \frac{d - t + \frac{t}{K}}{d} = \frac{1}{K} \] where: \( d' = d + 1.6 \) mm, \( t = 2 \) mm (thickness of dielectric slab), \( K \) is the dielectric constant. Step 2: Solve for \( K \) \[ \frac{(d+1.6) - 2 + \frac{2}{K}}{d+1.6} = \frac{1}{K} \] Solving for \( K \), we get: \[ K = 5.0 \] Thus, the correct answer is \( 5.0 \).

Answer 2

Given:
- Initial voltage across capacitor, \( V = 95 \, V \)
- Initial capacitance, \( C \)
- Dielectric slab thickness, \( t = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Increase in plate separation, \( d_{inc} = 1.6 \, \text{mm} = 1.6 \times 10^{-3} \, \text{m} \)
- The potential difference \( V \) remains constant.

We need to find the dielectric constant \( K \) of the slab.

Step 1: Let the original plate separation be \( d \).
After inserting the dielectric slab and increasing plate separation, the new separation becomes:
\[ d' = d + d_{inc} = d + 1.6 \times 10^{-3} \]

Step 2: The capacitor now consists of two capacitors in series:
- Capacitor 1 with dielectric slab of thickness \( t = 2 \times 10^{-3} \) m and dielectric constant \( K \).
- Capacitor 2 with air gap of thickness \( d' - t = (d + 1.6 \times 10^{-3}) - 2 \times 10^{-3} = d - 0.4 \times 10^{-3} \) m.

Step 3: Capacitance of dielectric slab part:
\[ C_1 = \frac{K \varepsilon_0 A}{t} \] Capacitance of air gap part:
\[ C_2 = \frac{\varepsilon_0 A}{d' - t} \] where \( A \) is the plate area and \( \varepsilon_0 \) is permittivity of free space.

Step 4: Total capacitance \( C' \) after insertion:
For series capacitors,
\[ \frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{t}{K \varepsilon_0 A} + \frac{d' - t}{\varepsilon_0 A} = \frac{t}{K \varepsilon_0 A} + \frac{d' - t}{\varepsilon_0 A} \] \[ \Rightarrow C' = \frac{\varepsilon_0 A}{\frac{t}{K} + (d' - t)} = \frac{\varepsilon_0 A}{d' + t \left(\frac{1}{K} - 1 \right)} \]

Step 5: Initially, the capacitance was:
\[ C = \frac{\varepsilon_0 A}{d} \]

Step 6: The voltage remains constant, and charge \( Q = C V \), so charge is proportional to capacitance.
We assume the same applied voltage and need the ratio of new capacitance \( C' \) to initial capacitance \( C \). The problem implies \( C' = C \), meaning:
\[ C' = C \Rightarrow \frac{\varepsilon_0 A}{d' + t \left(\frac{1}{K} - 1\right)} = \frac{\varepsilon_0 A}{d} \] Simplify:
\[ d' + t \left(\frac{1}{K} - 1 \right) = d \]

Step 7: Substitute \( d' = d + 1.6 \times 10^{-3} \) m:
\[ d + 1.6 \times 10^{-3} + t \left( \frac{1}{K} - 1 \right) = d \] \[ 1.6 \times 10^{-3} + 2 \times 10^{-3} \left( \frac{1}{K} - 1 \right) = 0 \] \[ 2 \times 10^{-3} \left( \frac{1}{K} - 1 \right) = -1.6 \times 10^{-3} \] \[ \frac{1}{K} - 1 = -\frac{1.6 \times 10^{-3}}{2 \times 10^{-3}} = -0.8 \] \[ \frac{1}{K} = 1 - 0.8 = 0.2 \] \[ K = \frac{1}{0.2} = 5 \]

Therefore, the dielectric constant of the material is:
\[ \boxed{5.0} \]