Question

The series combination of two capacitors shown in the figure is connected across 1000V. The magnitude of the charges on the capacitors will be:

• A. \( 3 \times 10^{-9} \) C
• B. \( 2 \times 10^{-9} \) C
• C. \( 5 \times 10^{-9} \) C
• D. \( 3.5 \times 10^{-9} \) C

Hint:

In a series combination of capacitors, the charge is the same on all capacitors, and you can calculate the equivalent capacitance using the formula \( C_{\text{eq}} = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2}} \).

Answer 1

The Correct Option is B

Step 1: Equivalent capacitance in series
For capacitors in series:
\[ C_{eq} = \frac{C_1 C_2}{C_1 + C_2} \] Given: \( C_1 = 3\,\mu F,\ C_2 = 6\,\mu F \)

\[ C_{eq} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\,\mu F \] So, \[ C_{eq} = 2 \times 10^{-6} F \]

Step 2: Charge on capacitor
\[ Q = C_{eq} V \] \[ Q = (2 \times 10^{-6}) \times 1000 \] \[ Q = 2 \times 10^{-3} C \]

Final Answer:
\( Q = 2 \times 10^{-3} C \)