Question

Two air core capacitors of equal capacitance $C$ are connected in series. If one of them is filled with a dielectric substance of dielectric constant $k$, the effective capacitance becomes

• A. $(k + 1)C$
• B. $\frac{kC}{1 + k}$
• C. $\frac{2kC}{1 + k}$
• D. $\frac{(k + 1)C}{k}$

Hint:

For two capacitors in series, the product-over-sum rule is the fastest way. If $n$ identical capacitors are in series, it's just $C/n$, but here they become different due to the dielectric.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When a dielectric material of constant $k$ is inserted into a capacitor, its capacitance increases by a factor of $k$.
When capacitors are connected in series, the reciprocal of the effective capacitance is the sum of the reciprocals of individual capacitances.
Key Formula or Approach:
1. New capacitance with dielectric: \( C' = kC \).
2. Series combination: \( \frac{1}{C_{eff}} = \frac{1}{C_1} + \frac{1}{C_2} \) or \( C_{eff} = \frac{C_1 C_2}{C_1 + C_2} \).

Step 2: Detailed Explanation:
Initially, we have two capacitors, both with capacitance $C$.
After filling one with dielectric, the two capacitances are \( C_1 = C \) and \( C_2 = kC \).
Since they are connected in series:
\[ C_{eff} = \frac{C \cdot kC}{C + kC} \]
Factor out $C$ from the denominator:
\[ C_{eff} = \frac{kC^2}{C(1 + k)} \]
Cancel the common factor $C$:
\[ C_{eff} = \frac{kC}{1 + k} \]

Step 3: Final Answer:
The effective capacitance is \( \frac{kC}{1 + k} \).