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(D) Key Idea: The capacitance of a parallel plate capacitor is given by the relation,$C = \varepsilon_r \frac{\varepsilon_0 A}{d}$,where $\varepsilon_

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$\frac{5}{6}$

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(D) Key Idea: The capacitance of a parallel plate capacitor is given by the relation,$C = \varepsilon_r \frac{\varepsilon_0 A}{d}$,where $\varepsilon_r$ is the dielectric constant,$A$ is the area of the plates,and $d$ is the distance between the plates.Given: $d_1 = x$,$\varepsilon_{r1} = 3k$,$d_2 = 2x$,and $\varepsilon_{r2} = 5k$.The two dielectric layers act as two capacitors $C_1$ and $C_2$ connected in series.$C_1 = \frac{(3k) \varepsilon_0 A}{x}$ and $C_2 = \frac{(5k) \varepsilon_0 A}{2x}$.In a series combination,the charge $Q$ stored on each capacitor is the same.The potential difference across each layer is given by $V = \frac{Q}{C}$.Thus,$V_1 = \frac{Q}{C_1} = \frac{Qx}{3k \varepsilon_0 A}$ and $V_2 = \frac{Q}{C_2} = \frac{Q(2x)}{5k \varepsilon_0 A}$.The ratio of the potential differences is $\frac{V_1}{V_2} = \frac{Qx / (3k \varepsilon_0 A)}{2Qx / (5k \varepsilon_0 A)} = \frac{1/3}{2/5} = \frac{1}{3} 	imes \frac{5}{2} = \frac{5}{6}$.Therefore,the correct option is $(d)$.

Similar Questions

Two dielectric slabs of dielectric constants $K_1$ and $K_2$ are filled between the plates of a parallel plate capacitor as shown in the figure. If the area of each plate is $A$ and the separation between the plates is $d$,what is the equivalent capacitance of the capacitor?

$A$ parallel plate capacitor,partially filled with a dielectric slab of dielectric constant $K$,is connected to a cell of emf $V \text{ volt}$,as shown in the figure. The separation between the plates is $D$. Then:

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