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An air filled parallel capacitor has cap...

An air filled parallel capacitor has capacity of 2pF. The separation of the plates is doubled and the interspaces between the plates is filled with wax. If the capacity is increased to 6pF, the dielectric constant of was is

A

2

B

3

C

4

D

6

Text Solution

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The correct Answer is:
To solve the problem step by step, we will analyze the given information and apply the relevant formulas for capacitors. ### Given: 1. Initial capacitance, \( C_1 = 2 \, \text{pF} \) 2. Final capacitance, \( C_2 = 6 \, \text{pF} \) 3. Initial separation of plates, \( d \) 4. Final separation of plates, \( D = 2d \) (the separation is doubled) 5. The interspace is filled with wax, and we need to find the dielectric constant \( k \) of the wax. ### Step 1: Write the formula for capacitance The capacitance \( C \) of a parallel plate capacitor is given by: \[ C = \frac{\epsilon A}{d} \] where: - \( C \) is the capacitance, - \( \epsilon \) is the permittivity of the material between the plates, - \( A \) is the area of the plates, - \( d \) is the separation between the plates. ### Step 2: Analyze the initial condition For the initial condition (air-filled capacitor): \[ C_1 = \frac{\epsilon_0 A}{d} \] Given \( C_1 = 2 \, \text{pF} \): \[ 2 \, \text{pF} = \frac{\epsilon_0 A}{d} \quad \text{(1)} \] ### Step 3: Analyze the final condition For the final condition (with wax): \[ C_2 = \frac{k \epsilon_0 A}{D} \] Since \( D = 2d \): \[ C_2 = \frac{k \epsilon_0 A}{2d} \] Given \( C_2 = 6 \, \text{pF} \): \[ 6 \, \text{pF} = \frac{k \epsilon_0 A}{2d} \quad \text{(2)} \] ### Step 4: Set up the ratio of the two capacitances From equations (1) and (2), we can set up the ratio: \[ \frac{C_2}{C_1} = \frac{\frac{k \epsilon_0 A}{2d}}{\frac{\epsilon_0 A}{d}} = \frac{k}{2} \] Substituting the known capacitances: \[ \frac{6 \, \text{pF}}{2 \, \text{pF}} = \frac{k}{2} \] This simplifies to: \[ 3 = \frac{k}{2} \] ### Step 5: Solve for the dielectric constant \( k \) To find \( k \): \[ k = 3 \times 2 = 6 \] ### Conclusion The dielectric constant of the wax is: \[ \boxed{6} \]

To solve the problem step by step, we will analyze the given information and apply the relevant formulas for capacitors. ### Given: 1. Initial capacitance, \( C_1 = 2 \, \text{pF} \) 2. Final capacitance, \( C_2 = 6 \, \text{pF} \) 3. Initial separation of plates, \( d \) 4. Final separation of plates, \( D = 2d \) (the separation is doubled) 5. The interspace is filled with wax, and we need to find the dielectric constant \( k \) of the wax. ...
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