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Step-by-Step Solution
Step 1: Identify the Original Capacitance
The capacitor initially has air as the dielectric, and its given capacitance is $9 \text{ pF}$. Denoting the plate area by $A$, separation by $d$, and the permittivity of free space by $\epsilon_0$, we use the relation for a parallel plate capacitor: Cair = ϵ0 Ad. Since $C_{\text{air}} = 9 \text{ pF}$, we have ϵ0 Ad = 9 pF.
Step 2: Write the Formula for Capacitance with Layered Dielectrics
When two dielectric layers of thickness $d_1$ and $d_2$, and dielectric constants $k_1$ and $k_2$ respectively, are placed one on top of the other along the separation between the plates, the combination acts like two capacitors in series. The effective capacitance is given by: Cnew = ϵ0 Ad1k1 + d2k2.
Step 3: Substitute the Given Data
According to the problem:
$d_1 = \frac{d}{3}$ with $k_1 = 3$
$d_2 = \frac{2d}{3}$ with $k_2 = 6$
Thus, d1k1=d33=d9, d2k2=2d36=2d18=d9. Therefore, d1k1 + d2k2=d9 + d9=2d9.
Step 4: Compute the New Capacitance
Using $ \epsilon_0 A = 9 \text{ pF} \, \times d $ from Step 1, we get Cnew = ϵ0 Ad1k1+d2k2 = 9 d pF2d9 = 9 pF × 92 = 40.5 pF.
Step 5: Conclude the Correct Answer
Hence, the capacitance of the capacitor with the given layered dielectric configuration becomes $40.5 \,\text{pF}$.
Step-by-Step Solution
Step 1: Identify the Original Capacitance
The capacitor initially has air as the dielectric, and its given capacitance is $9 \text{ pF}$. Denoting the plate area by $A$, separation by $d$, and the permittivity of free space by $\epsilon_0$, we use the relation for a parallel plate capacitor: Cair = ϵ0 Ad. C_{\text{air}} \;=\; \frac{\epsilon_0 \, A}{d}. Since $C_{\text{air}} = 9 \text{ pF}$, we have ϵ0 Ad = 9 pF. \frac{\epsilon_0 \, A}{d} \;=\; 9 \,\text{pF}.
Step 2: Write the Formula for Capacitance with Layered Dielectrics
When two dielectric layers of thickness $d_1$ and $d_2$, and dielectric constants $k_1$ and $k_2$ respectively, are placed one on top of the other along the separation between the plates, the combination acts like two capacitors in series. The effective capacitance is given by: Cnew = ϵ0 Ad1k1 + d2k2. C_{\text{new}} \;=\; \frac{\epsilon_0 \, A}{ \frac{d_1}{k_1} \;+\; \frac{d_2}{k_2} }.
Step 3: Substitute the Given Data
According to the problem:
$d_1 = \frac{d}{3}$ with $k_1 = 3$
$d_2 = \frac{2d}{3}$ with $k_2 = 6$
Thus, d1k1=d33=d9, d2k2=2d36=2d18=d9. \frac{d_1}{k_1} = \frac{\tfrac{d}{3}}{3} = \frac{d}{9}, \quad \frac{d_2}{k_2} = \frac{\tfrac{2d}{3}}{6} = \frac{2d}{18} = \frac{d}{9}. Therefore, d1k1 + d2k2=d9 + d9=2d9. \frac{d_1}{k_1} \;+\; \frac{d_2}{k_2} = \frac{d}{9} \;+\; \frac{d}{9} = \frac{2d}{9}.
Step 4: Compute the New Capacitance
Using $ \epsilon_0 A = 9 \text{ pF} \, \times d $ from Step 1, we get Cnew = ϵ0 Ad1k1+d2k2 = 9 d pF2d9 = 9 pF × 92 = 40.5 pF. C_{\text{new}} \;=\; \frac{\epsilon_0 \, A}{ \frac{d_1}{k_1} + \frac{d_2}{k_2} } \;=\; \frac{9\,d\,\text{pF}}{\tfrac{2d}{9}} \;=\; 9 \,\text{pF} \;\times\; \frac{9}{2} \;=\; 40.5 \,\text{pF}.
Step 5: Conclude the Correct Answer
Hence, the capacitance of the capacitor with the given layered dielectric configuration becomes $40.5 \,\text{pF}$.
