A parallel plate capacitor has plate area
Question:

A parallel plate capacitor has plate area $100 \mathrm{~m}^{2}$ and plate separation of $10 \mathrm{~m}$. The space between the plates is filled up to a thickness $5 \mathrm{~m}$ with a material of dielectric constant of 10 . The resultant capacitance of the system is ‘x’ pF. The value of $\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \cdot \mathrm{m}^{-1}$

The value of ‘ $x$ ‘ to the nearest integer is

Solution:

$A=100 \mathrm{~m}^{2}$

Using $C=\frac{k \in_{0} A}{d}$

$\mathrm{C}_{1}=\frac{10 \epsilon_{0}(100)}{5}$

$=200 \epsilon_{0}$

$\mathrm{C}_{2}=\frac{\epsilon_{0}(100)}{5}=20 \epsilon_{0}$

$\mathrm{C}_{1} \& \mathrm{C}_{2}$ are in series so $\mathrm{C}_{\text {eqv. }}=\frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$

$=\frac{4000 \epsilon_{0}}{220}$

$=160.9 \times 10^{-12} \simeq 161 \mathrm{pF}$

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