Question 90 of 180

A composite parallel plate capacitor is made up of two different dielectric materials with different thickness (${t}_{1}$ and ${t}_{2}$) as shown in figure. The two different dielectric material are separated by a conducting foil $F$. The voltage of the conducting foil is _____ $V$.

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Answer: 60

Capacitance of each capacitor
${C}_{1}=\dfrac{A3{\epsilon }_{0}}{\dfrac{1}{2}}=6A{ϵ}_{0}$ and ${C}_{2}=\dfrac{A4{ϵ}_{0}}{1}=4A{\epsilon }_{0}$
Here, $A$ is area of plate.
Equivalent capacitance is
${C}_{eq}=\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}=\dfrac{6A{\epsilon }_{0}\times 4A{\epsilon }_{0}}{6A{\epsilon }_{0}+4A{\epsilon }_{0}}=\dfrac{24}{10}A{ϵ}_{0}$
Now, net charge stored is ${q}_{net}={C}_{eq}\left(\Delta V\right)=\dfrac{24}{10}A{\epsilon }_{0}\left(100\right)=240A{\epsilon }_{0}$
Potential drop across ${C}_{2}$ is $\Delta {V}_{2}=\dfrac{240A{\epsilon }_{0}}{4A{\epsilon }_{0}}=60V$
Thus, voltage of the conducting foil is ${V}_{foil}=60V$