As shown in the figure two capacitors \( C_{1} \) \& \( C_{2} \) each having same gap between the plates x filled with different media of dielectric constants 3K and 6K respectively. If these two capacitors are connected to a battery, the ratio of potential differences across dielectric layers of \( C_{1} \) and \( C_{2} \) is:
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In a series capacitor configuration, the capacitor with a higher dielectric constant develops a lower potential drop across its plates because its increased capacitance allows it to store charge much more easily.
Step 1: Recall capacitance with a dielectric. A parallel plate capacitor filled with a dielectric of constant $K$ has capacitance: \[ C = \frac{K\varepsilon_0 A}{d} \] So with the same plate gap, a bigger $K$ means a bigger capacitance. Step 2: Use the series rule for charge. When capacitors are in series with a battery, the same charge $Q$ sits on each. So the voltage on each is $V = \dfrac{Q}{C}$, meaning voltage is inversely proportional to capacitance: \[ \frac{V_1}{V_2} = \frac{C_2}{C_1} \] Step 3: Compare the two capacitances. Both have the same gap $x$. The first has $K_1 = 3K$ and the second has $K_2 = 6K$, so: \[ C_1 \propto 3K, \qquad C_2 \propto 6K \] Step 4: Form the voltage ratio. \[ \frac{V_1}{V_2} = \frac{C_2}{C_1} = \frac{6K}{3K} \] Step 5: Simplify. \[ \frac{V_1}{V_2} = 2 \] Step 6: State the answer. The capacitor with the smaller dielectric constant gets the larger voltage, and the ratio is: \[ \boxed{\frac{V_1}{V_2} = 2} \]
