Question

A capacitor has air as dielectric medium and two conducting plates of area \(12 \, \text{cm}^2\) and they are \(0.6 \, \text{cm}\) apart. When a slab of dielectric having area \(12 \, \text{cm}^2\) and \(0.6 \, \text{cm}\) thickness is inserted between the plates, one of the conducting plates has to be moved by \(0.2 \, \text{cm}\) to keep the capacitance same as in previous case. The dielectric constant of the slab is:
\((\text{Given } \epsilon_0 = 8.834 \times 10^{-12} \, \text{F/m})\)

• A. \(1.50\)
• B. \(1.33\)
• C. \(0.66\)
• D. \(1\)

Answer 1

The Correct Option is A

We are given the relation:
\[ \frac{A \varepsilon_0}{d} = \frac{A \varepsilon_0}{(0.2 + \frac{d}{k})} \]

Step 1: Simplify the equation
By canceling out \( A \varepsilon_0 \) on both sides, we get:
\[ \frac{1}{d} = \frac{1}{0.2 + \frac{d}{k}} \] Multiplying both sides by \( d(0.2 + \frac{d}{k}) \), we have:
\[ 0.6 = 0.2 + \frac{0.6}{k} \]

Step 2: Solving for \( k \)
Rearranging the equation:
\[ 0.6 - 0.2 = \frac{0.6}{k} \] \[ 0.4 = \frac{0.6}{k} \] \[ k = \frac{0.6}{0.4} = \frac{3}{2} \]

Final Answer:
\[ k = \frac{3}{2} \]

Answer 2

The capacitance without the dielectric is:
\[C = \frac{A \epsilon_0}{d}.\]
With the dielectric inserted:
\[C = \frac{A \epsilon_0}{0.2 + \frac{d}{k}}.\]
Equating the capacitances:
\[\frac{A \epsilon_0}{0.6} = \frac{A \epsilon_0}{0.2 + \frac{0.6}{k}}.\]
Cancel \(A \epsilon_0\):
\[0.6 = 0.2 + \frac{0.6}{k}.\]
Rearranging:
\[0.6 - 0.2 = \frac{0.6}{k} \implies 0.4 = \frac{0.6}{k}.\]
Solving for \(k\):
\[k = \frac{0.6}{0.4} = \frac{3}{2} = 1.50.\]
Thus, the dielectric constant of the slab is:
\[k = 1.50.\]