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

Question

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

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Answer:

C

Solution:

$C^{'} = \frac{A ε _{0}}{\frac{d _{1}}{3} + \frac{d _{2}}{6}} = \frac{A ε _{0}}{\frac{d}{9} + \frac{2 d}{18}} = \frac{18 A ε _{0}}{4 d} C^{'} = 40.5 PF$

TestHub · Accepted Answer

​ C ′ = 3 d 1 ​ ​ + 6 d 2 ​ ​ A ε 0 ​ ​ = 9 d ​ + 18 2 d ​ A ε 0 ​ ​ = 4 d 18 A ε 0 ​ ​ C ′ = 40.5 PF ​

Physics - Capacitor Question with Solution | TestHub

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