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

1.8 pF

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45 pF

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40.5 pF

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20.25 pF

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Solution

The correct option is C 40.5 pF
$C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} C_{1} = \frac{k_{1} ϵ_{0} A}{\frac{d}{3}} = 3 k_{1} . \frac{ϵ_{0} A}{d} ∴ C_{e f f} = 40.50 p F C_{2} = \frac{k_{2} ϵ_{0} A}{\frac{2 d}{3}} = \frac{3}{2} k_{2} \frac{ϵ_{0} A}{d}$

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Similar questions

Q. 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 plates is now 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{2 d}{3}

. Capacitance of the capacitor is now :

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 :