There is a parallel plate capacitor of capacitance C. If half of the spaceis filled withdielectric of dielectric constant k = 5 as in the figure, find the percentage increase in capacitance.

The correct answer is 66.67%.

Step 1: Understand the Initial State
Based on the provided diagram, the label "initial" shows a parallel plate capacitor where the plates are separated by a distance of $d$ in air/vacuum. If we let the area of the plates be $A$, the initial capacitance $C$ is given by the formula:

$C_{\text{initial}}=\frac{{\epsilon }_{0}A}{d}=C$

Step 2: Analyze the Final State
The label "Final" in the diagram shows that half of the space is filled with a dielectric material of thickness $\frac{d}{2}$ and dielectric constant $k=5$. This arrangement divides the capacitor into two separate regions connected in series:
1. A dielectric-filled region of thickness $\frac{d}{2}$ with capacitance $C_1$.
2. An air-filled region of thickness $\frac{d}{2}$ with capacitance $C_2$.

Step 3: Calculate the Individual Capacitances
For the dielectric-filled section with $k=5$:

$C_1=\frac{k{\epsilon }_{0}A}{d/2}=2k\left(\frac{{\epsilon }_{0}A}{d}\right)=2kC$

Substituting $k=5$:

$C_1=2\left(5\right)C=10C$

For the remaining air-filled section:

$C_2=\frac{{\epsilon }_{0}A}{d/2}=2\left(\frac{{\epsilon }_{0}A}{d}\right)=2C$

Step 4: Find the Equivalent Capacitance
Since the two regions are in series, the equivalent final capacitance $C_{\text{final}}$ is calculated as:

$\frac{1}{C_{\text{final}}}=\frac{1}{C_1}+\frac{1}{C_2}$

$\frac{1}{C_{\text{final}}}=\frac{1}{10C}+\frac{1}{2C}$

$\frac{1}{C_{\text{final}}}=\frac{1+5}{10C}=\frac{6}{10C}=\frac{3}{5C}$

Taking the reciprocal to find $C_{\text{final}}$:

$C_{\text{final}}=\frac{5}{3}C$

Step 5: Calculate the Percentage Increase in Capacitance
The percentage increase is determined by comparing the change in capacitance to the initial value:

$\text{Percentage Increase}=\frac{C_{\text{final}}-C_{\text{initial}}}{C_{\text{initial}}}\times 100\%$

$\text{Percentage Increase}=\frac{\frac{5}{3}C-C}{C}\times 100\%$

$\text{Percentage Increase}=(\frac{5}{3}-1)\times 100\%$

$\text{Percentage Increase}=\frac{2}{3}\times 100\%\approx 66.67\%$