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

Answer: (A)

$1.50$

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:
$k = 1.50.$