A parallel plate capacitor with air between the plates has a capacitance of 2pF. The capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 is

12pF

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24pF

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6pF

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18pF

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Solution

The correct option is A 24pF
We know,

$C = \frac{A k ε_{0}}{d}$

Given,
For $k = 1,$
$C = \frac{A ε_{0}}{d} = 2$

For k=6, and $d \to \frac{d}{2}$

$C = \frac{A \times 6 ε_{0}}{\frac{d}{2}}$

$C = \frac{12 A ε_{0}}{d} = 12 C = 24 p F$

Option $B$ is the correct answer

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Q. A parallel plate capacitor with air between the plates has a capacitance of 9

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. The separation between its plates is

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and thickness

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A parallel plate capacitor with air between the plates has a capacitance of $9 p F$ . The separation between its plates is $d$ . The space between the plate is now partially filled with two dielectrics. One of the dielectric has dielectric constant $K_{1} = 3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $K_{2} = 6$ and thickness $\frac{d}{3}$ . The remaining $\frac{d}{3}$ space is filled with air. The new capacitance is :

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A parallel plate capacitor with air between the plates has a capacitance of 2pF. The capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 is

The correct option is A 24pFWe know,C=Akε0dGiven,For k=1,C=Aε0d=2For k=6, and d→d2C=A×6ε0d2C=12Aε0d=12C=24pFOption B is the correct answer

The correct option is A 24pF
We know,

$C = \frac{A k ε_{0}}{d}$

Given,
For $k = 1,$
$C = \frac{A ε_{0}}{d} = 2$

For k=6, and $d \to \frac{d}{2}$

$C = \frac{A \times 6 ε_{0}}{\frac{d}{2}}$

$C = \frac{12 A ε_{0}}{d} = 12 C = 24 p F$

Option $B$ is the correct answer