Question

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

Answer 1

Step 1. Given data

Area of the parallel plate, $A=100m^2$

Plate separation, $d=10m$

Thickness, $t=5m$

Dielectric constant, $k=10$

Step 2. Finding the capacitance of parallel plate $1$ ,$C_1$ and the capacitance of parallel plate $2$, $C_2$

By using the formula of capacitance, $C$

$C=\frac{kA{\epsilon }_0}{d}$

$C_1=\frac{10\times 100{\epsilon }_0}{5}$ [Where, for plate $1$,$d=5m,k=10$]

$C_1=200{\epsilon }_0$

$C_2=\frac{100\times {\epsilon }_0}{5}$ [Where, for plate $2$, $d=5m,k=1$]

$C_2=20{\epsilon }_0$

Step 3. Finding the equivalent capacitance, $C_{eq}$

$C_{eq}=\frac{C_1\times C_2}{C_1+C_2}$

$C_{eq}=\frac{200{\epsilon }_0\times 20{\epsilon }_0}{200{\epsilon }_0+20{\epsilon }_0}$

$C_{eq}=18.18{\epsilon }_0$

$C_{eq}=18.18\times 8.85\times {10}^{-19}$ [Where, ${\epsilon }_0=8.85\times {10}^{-19}F{m}^{-1}$]

$C_{eq}=160.9\times {10}^{-12}$

$C_{eq}=161pF$

Comparing the above value with $C_{eq}=xpF$, we get

Therefore, $x=161pF$