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 capacitors are connected in series, their equivalent capacitance is found using reciprocal addition. Introducing a dielectric into a capacitor multiplies its capacitance by the dielectric constant \(k\).
Step 2: Key Formula or Approach:
Capacitance with dielectric: \(C' = kC\).
Equivalent series capacitance: \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\), which simplifies to \(C_{eq} = \frac{C_1 C_2}{C_1 + C_2}\).
Step 3: Detailed Explanation:
Initially, we have two capacitors, both with capacitance \(C\).
One of them is filled with a dielectric, changing its capacitance to \(C_1 = kC\).
The other capacitor remains air-cored, so \(C_2 = C\).
Now, calculate the equivalent capacitance \(C_{eq}\) for the series connection:
\[ C_{eq} = \frac{C_1 \times C_2}{C_1 + C_2} \] Substitute the values:
\[ C_{eq} = \frac{(kC) \times C}{kC + C} \] Factor out \(C\) in the denominator:
\[ C_{eq} = \frac{kC^2}{C(k + 1)} \] Cancel one \(C\) from numerator and denominator:
\[ C_{eq} = \frac{kC}{k + 1} \] Step 4: Final Answer:
The effective capacitance becomes \(\frac{kC}{1 + k}\).