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 dielectic 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 thickness dd, but a cross-sectional area of $A_2$, as shown in the diagram above.
In terms of $C$, what is the new capacitance of this modified system?

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

Answer 1

The correct option is C $\frac{3C}{2}$

Given : $K_1=1$ $K_2=2K_1$ $A_2=\frac{A}{2}$

Capacitance of old system $C=\frac{K_{1}A{ϵ}_o}{d}$

New system is made up of 2 capacitors connected in parallel with dielectrics $K_1$ and $K_2$ and plate area $A_2$.

Thus capacitance of new system $C^{\prime }=C_1+C_2=\frac{K_1{A}_2{ϵ}_o}{d}+\frac{K_2{A}_2{ϵ}_o}{d}$

$\therefore$ $C^{\prime }=(K_1+K_2)\frac{A_2{ϵ}_o}{d}=\left(3K_1\right)\frac{\frac{A}{2}{ϵ}_o}{d}=\frac{3C}{2}$