Question

A parallel plate capacitor has plate area $100{\text{m}}^2$ and plate separation of $10\text{m}$. The space between the plates is filled up to a thickness $5\text{m}$ with a material of dielectric constant of $10$. The resultant capacitance of the system is $x\text{pF}$.
The value of ${\epsilon }_0=8.85\times {10}^{-12}{\text{F m}}^{-1}$. The value of $x$ to the nearest integer is .

Answer 1

Given, Area, $A=100{\text{m}}^2$
Separation, $d=10\text{m}$
Thickness, $t=5\text{m}$
Dielectric constant, $K=10$
This capacitor can be thought as two capacitor in series, one with dilectric and another without dielectric.
$C_1=\frac{KA{\epsilon }_0}{t},C_2=\frac{A{\epsilon }_0}{d-t}$

$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{\frac{KA{\epsilon }_0}{t}\times \frac{A{\epsilon }_o}{d-t}}{\frac{KA{\epsilon }_0}{t}+\frac{A{\epsilon }_o}{d-t}}$

Here, $t=d-t=5\text{m}$

$C_{eq}=\frac{K{A}^2{\epsilon }_0^2}{t^2}\times \frac{t}{A{\epsilon }_0(1+K)}$

$C_{eq}=\frac{KA{\epsilon }_0}{t(1+K)}=\frac{10\times 100\times 8.85\times {10}^{-12}}{5(1+10)}$

$C_{eq}=\frac{8.85\times {10}^{-9}}{55}=0.1609090\times {10}^{-9}\text{F}$

$C_{eq}=160.90\times {10}^{-12}\text{F}$

$C_{eq}=161$ $\text{pF}$