Question

Two capacitors of capacitance \(2\mu F\) and \(6\mu F\) are connected in series. A potential difference of 800 V is applied to the outer plates of the two capacitors system. The charge on each capacitor will be :

• A. 1200 C
• B. 6000 C
• C. \(6000 \mu C\)
• D. \(1200 \mu C\)

Hint:

In series combination, charge remains same on all capacitors.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When capacitors are connected in series, the charge \( Q \) on each capacitor is the same. The equivalent capacitance \( C_{eq} \) is calculated using the reciprocal sum rule.
: Key Formula or Approach:
1. \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \) or \( C_{eq} = \frac{C_1 C_2}{C_1 + C_2} \).
2. \( Q = C_{eq} \cdot V \).
Step 2: Detailed Explanation:
Given \( C_1 = 2\mu \text{F} \), \( C_2 = 6\mu \text{F} \), and \( V = 800 \text{ V} \).
Calculate equivalent capacitance:
\[ C_{eq} = \frac{2 \times 6}{2 + 6} \mu \text{F} = \frac{12}{8} \mu \text{F} = 1.5 \mu \text{F} \]
Now, calculate the charge:
\[ Q = C_{eq} \times V \]
\[ Q = 1.5 \times 10^{-6} \text{ F} \times 800 \text{ V} \]
\[ Q = 1200 \times 10^{-6} \text{ C} = 1200\mu \text{C} \]
Since they are in series, both capacitors hold this same charge.
Step 3: Final Answer:
The charge on each capacitor is \( 1200\mu \text{C} \).