Question

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:

• A. 2
• B. 6
• C. 4
• D. 8

Hint:

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.

Answer 1

The Correct Option is A

Concept: The capacitance of a parallel plate capacitor filled with a dielectric material is given by \( C = \frac{K \varepsilon_0 A}{d} \). When capacitors are connected in series to a battery, the electric charge \( Q \) accumulated on their plates is completely identical, which means the potential difference across each capacitor is inversely proportional to its capacitance: \[ V = \frac{Q}{C} \implies \frac{V_1}{V_2} = \frac{C_2}{C_1} \]

Step 1: Determining individual capacitance relationships.
Both capacitors share identical structural plate gaps \( d = x \).

• For capacitor \( C_1 \), the dielectric constant is \( K_1 = 3K \implies C_1 \propto 3K \)

• For capacitor \( C_2 \), the dielectric constant is \( K_2 = 6K \implies C_2 \propto 6K \)

Step 2: Calculating the potential difference ratio.
Since they are connected in series across the circuit network loops: \[ \frac{V_1}{V_2} = \frac{C_2}{C_1} = \frac{6K}{3K} = 2 \] Thus, the ratio of potential differences is exactly 2.