2. Electric Potential and Capacitance
medium

A parallel plate capacitor has a capacity $C$. The separation between the plates is doubled and a dielectric medium is introduced between the plates. If the capacity now becomes $2C$, the dielectric constant of the medium is

A

$2$

B

$1$

C

$4$

D

$8$

Solution

(c) ${C_1} = {\varepsilon _0}\frac{A}{{{d_1}}}$ and ${C_2} = K{\varepsilon _0}\frac{A}{{{d_2}}}$
 $\frac{{{C_1}}}{{{C_2}}} = \frac{1}{K} \times \frac{{{d_2}}}{{{d_1}}}$ $ = \frac{C}{{2C}} = \frac{1}{K} \times \frac{{2d}}{d}$

$K = 4$

Standard 12
Physics

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