Question

Two conducting parallel plates have area $A$, are separated by distance $d$, and are filled with a dielectric material of constant $K_1$. The capacitance of this system is $C$. A student plans to replace the dielectric $K_1$ with a second dielectric, $K_2$, where $K_2=2K_1$. However, the student only has half the volume of dielectric material to fully replace $K_1$. So the student fills half the gap with $K_1$ and half with $K_2$. Both $K_1$ and $K_2$ have a cross sectional area of $A$, but a thickness of $d/2$, as shown in the diagram above.
In terms of $C$, what is the new capacitance of this modified system?

• A. $\frac{C}{2}$
• B. $\frac{2C}{3}$
• C. $\frac{3C}{4}$
• D. $\frac{4C}{3}$
• E. $2C$

Answer 1

Answer: (D)

$\frac{4C}{3}$

In first case, the capacitance $C=\frac{A{K}_1{ϵ}_0}{d}$
In second case, there are two capacitors $C_1.C_2$ and they are in series.
Here, $C_1=\frac{A{K}_1{ϵ}_0}{d/2}=2C$
and $C=\frac{A{K}_2{ϵ}_0}{d/2}=\frac{A\left(2K_1\right){ϵ}_0}{d/2}=4C$ as $(K_2=2K_1)$
The new capacitance is the equivalent capacitance for $C_1,C_2$.
Thus, $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{\left(2C\right)\left(4C\right)}{2C+4C}=\frac{4C}{3}$