Insertion of dielectric in capacitor

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Question

In parallel plate capacitor set up, plate area is 2 $m^{2}$ and the plates are separated by 1 m. If the space between the plates are filled with a dielectric material of thickness 0.5 m and area 2 $m^{2}$ (see fig) the capacitance of the set up will be______ $ε_{0}$
(Take K=3.2)(Round off to the Nearest integer)

Moderate

Solution

$\begin{array}{l} C_{1} = \frac{K ε_{0} A}{d / 2}; C_{2} = \frac{ε_{0} A}{d / 2} \\ C_{1} a n d C_{2} a r e i n s e r i e s, \\ \Rightarrow \frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{d}{2 K ε_{0} A} + \frac{d}{2 ε_{0} A} \\ \Rightarrow \frac{1}{C} = \frac{d}{2 ε_{0} A} (\frac{K + 1}{K}) \\ \Rightarrow C = \frac{2 ε_{0} A K}{d (K + 1)} = \frac{2 \times 2 \times 3.2}{1 \times 4.2} ε_{0} = 3.04 ε_{0} \end{array}$

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Insertion of dielectric in capacitor

Remember concepts with our Masterclasses.

In parallel plate capacitor set up, plate area is 2m2 and the plates are separated by 1 m. If the space between the plates are filled with a dielectric material of thickness 0.5 m and area 2 m2 (see fig) the capacitance of the set up will be______ε0 (Take K=3.2)(Round off to the Nearest integer)

C1=Kε0Ad/2;C2=ε0Ad/2 C1 and C2 are in series,⇒1C=1C1+1C2=d2Kε0A+d2ε0A⇒1C=d2ε0AK+1K ⇒C=2ε0AKd(K+1)=2×2×3.21×4.2ε0=3.04ε0

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