Question

An air capacitor and identical capacitor filled with dielectric medium of dielectric constant 5 are connected in series to a voltage source of 12V. The fall of potential across $C_1$ and $C_2$ are respectively

• A. 2V and l0V
• B. l0V and 2V
• C. 6V and 6V
• D. 4V and 8V
• E. 8V and 4V

Answer 1

The Correct Option is A

We are given two capacitors, \( C_1 \) (air capacitor) and \( C_2 \) (capacitor filled with a dielectric medium of dielectric constant 5), connected in series to a voltage source of 12V.

The voltage across each capacitor in a series combination is inversely proportional to their capacitances. The total capacitance \( C_{\text{total}} \) of capacitors in series is given by:

\[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} \] 

For \( C_1 \), the capacitance is that of an air capacitor, and for \( C_2 \), the capacitance increases due to the dielectric medium. The capacitance of \( C_2 \) is given by:

\[ C_2 = \kappa \cdot C_1 = 5 \cdot C_1 \]

Substitute \( C_2 = 5 \cdot C_1 \) into the total capacitance equation:

\[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{5C_1} = \frac{6}{5C_1} \]

Thus, the total capacitance is:

\[ C_{\text{total}} = \frac{5}{6} C_1 \]

Now, the voltage drop across each capacitor is proportional to the inverse of its capacitance. The voltage drop across \( C_1 \) is:

\[ V_1 = V \cdot \frac{C_2}{C_{\text{total}}} = 12 \cdot \frac{5C_1}{6C_1} = 10V \]

And the voltage drop across \( C_2 \) is:

\[ V_2 = V - V_1 = 12V - 10V = 2V \]

Thus, the voltage drops across the two capacitors are:

• Across \( C_1 \): \( 10V \)
• Across \( C_2 \): \( 2V \)

Correct Answer:

Correct Answer: (1) 2V and 10V

Answer 2

Let \( C_1 \) be the air capacitor with capacitance \( C \), and \( C_2 \) be the identical capacitor with dielectric, so its capacitance becomes \( 5C \). In series connection, the charge \( Q \) on both capacitors is the same. The voltage across a capacitor is given by: \[ V = \frac{Q}{C} \] Let the voltages across \( C_1 \) and \( C_2 \) be \( V_1 \) and \( V_2 \). Since they are in series, \[ V_1 + V_2 = 12\ V \] Also, \[ V_1 = \frac{Q}{C}, \quad V_2 = \frac{Q}{5C} \Rightarrow \frac{V_1}{V_2} = \frac{\frac{Q}{C}}{\frac{Q}{5C}} = 5 \Rightarrow V_1 = 5 V_2 \] Substitute in total voltage: \[ V_1 + V_2 = 12 \Rightarrow 5V_2 + V_2 = 12 \Rightarrow 6V_2 = 12 \Rightarrow V_2 = 2\ V,\quad V_1 = 10\ V \] So, the potential drops are 10 V across the air capacitor and 2 V across the dielectric-filled capacitor.