A parallel plate capacitor with air between the plates has a capacitance of  9pF. The separation between its plates is d . The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constan  k1 = 3  and thickness  d/3 while the other one has dielectric constant k2 = 6 and thickness 2d/3   Capacitance of the capacitor is now

  • Option 1)

    20.25pF

  • Option 2)

    1.8 pF

  • Option 3)

    45 pF

  • Option 4)

    40.5 pF

 

Answers (2)
N neha
D Divya Saini

As we learnt in

Parallel Plate Capacitor -

C=\frac{\epsilon _{0}A}{d}

- wherein

Area - A seperation between two plates.

 

 

If K filled between the plates -

\dpi{100} {C}'=K\frac{\epsilon _{0}A}{d}={C}'=kC

 

 

- wherein

C\propto A

C\propto\frac{1}{d}

 

 

C = \frac{\varepsilon _{0}A}{d}= 9\times 10^{-12}F

With dielectric, C = \frac{\varepsilon _{0}kA}{d}

C_{1} = \frac{\varepsilon _{0}A\cdot 3}{d/3}=9C

C_{2} = \frac{\varepsilon _{0}A\cdot 6}{2d/3}=9C

\therefore C_{total}= \frac{C_{1}C_{2}}{C_{1}+C_{2}}     (as they are in series)

= \frac{9C\times 9C}{18C}= \frac{9}{2}\times C\: or\: \frac{9}{2}\times 9\times10^{-12} F

\Rightarrow C_{total}= 40.5\: \: pF


Option 1)

20.25pF

Incorrect

Option 2)

1.8 pF

Incorrect

Option 3)

45 pF

Incorrect

Option 4)

40.5 pF

Correct

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