Question

A potential difference of 300 V is applied to a combination of 2.0µF and 8.0µF capacitors connected in series. The charge on the 2.0µF capacitor is:

• A. \( 2.4 \times 10^{-4} \) C
• B. \( 4.8 \times 10^{-4} \) C
• C. \( 7.2 \times 10^{-4} \) C
• D. \( 9.6 \times 10^{-4} \) C

Hint:

In a series combination of capacitors, the charge on each capacitor is the same, and the voltage divides based on the capacitance values.

Answer 1

The Correct Option is B

Step 1: Use the formula for charge in a series combination of capacitors:
In a series combination, the charge on each capacitor is the same, and the voltage divides according to the capacitors' values. The equivalent capacitance for two capacitors in series is: \[ C_{\text{eq}} = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2}} \] Given: \[ C_1 = 2.0 \, \mu F, \, C_2 = 8.0 \, \mu F, \, V = 300 \, V \]
Step 2: Calculate equivalent capacitance: \[ C_{\text{eq}} = \frac{1}{\frac{1}{2.0} + \frac{1}{8.0}} = \frac{1}{\frac{5}{8}} = 1.6 \, \mu F \]
Step 3: Calculate the charge: \[ Q = C_{\text{eq}} \times V = 1.6 \, \mu F \times 300 \, V = 4.8 \times 10^{-4} \, C \]