# 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) Option 2) Option 3) Option 4)

Answers (2)
N neha
D Divya Saini

As we learnt in

Parallel Plate Capacitor -

$\dpi{100} C=\frac{\epsilon _{0}A}{d}$

- 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

$\dpi{100} C\propto A$

$\dpi{100} C\propto\frac{1}{d}$

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

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)

Incorrect

Option 2)

Incorrect

Option 3)

Incorrect

Option 4)

Correct

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