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: Use the formula for charge in a series combination of capacitors:
For two capacitors in series, the total charge \( Q \) is the same on both capacitors, and the equivalent capacitance is: \[ C_{\text{eq}} = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2}} \]
Step 2: Apply the formula for charge: The charge on the capacitors is calculated using: \[ Q = C_{\text{eq}} \times V \] Given the values of \( C_1 = 3 \, \mu F \), \( C_2 = 6 \, \mu F \), and \( V = 1000 \, V \), we find: \[ C_{\text{eq}} = \frac{1}{\frac{1}{3} + \frac{1}{6}} = 2 \, \mu F \] Thus, the charge is: \[ Q = 2 \times 10^{-6} \, F \times 1000 \, V = 2 \times 10^{-9} \, C \]