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2. Electric Potential and Capacitance
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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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