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

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:
\[k = 1.50.\]