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 :

40.5

p F

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20.25

p F

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1.8

p F

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45

p F

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Solution

The correct option is C $40.5$ $p F$

The capacitance with air is $C_{0} = \frac{A ϵ_{0}}{d} = 9 p F$

After inserting dielectric, there are two capacitors $C_{1}, C_{2}$ and they are in series.

here, $C_{1} = \frac{A K_{1} ϵ_{0}}{d / 3} = \frac{9 A ϵ_{0}}{d} = 9 \times 9 = 81 p F$

and $C_{2} = \frac{A K_{2} ϵ_{0}}{2 d / 3} = \frac{9 A ϵ_{0}}{d} = 9 \times 9 = 81 p F$

Thus new capacitance $C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{81 \times 81}{81 + 81} = 40.5 p F$

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Similar questions

Q. 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

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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The correct option is C 40.5 pF The capacitance with air is C0=Aϵ0d=9pFAfter inserting dielectric, there are two capacitors C1,C2 and they are in series.here, C1=AK1ϵ0d/3=9Aϵ0d=9×9=81pFand C2=AK2ϵ02d/3=9Aϵ0d=9×9=81pFThus new capacitance Ceq=C1C2C1+C2=81×8181+81=40.5pF