Question

Two parallel plate capacitors X and Y are connected in series to a 6 V battery. They have the same plate area and same plate separation but capacitor X has air between its plates, whereas capacitor Y contains a material of dielectric constant 4. Calculate the capacitances of X and Y, if the equivalent capacitance of the combination of X and Y is \( 4 \, \mu\text{F} \). Calculate the potential difference across the plates of X and Y.

Hint:

Calculate the common series charge first from $Q=C_{eq}V$. Then set $C_Y=4C_X$ and use $Q/C_X+Q/C_Y=6$ to determine both capacitances.

Answer 1

Capacitance and Potential Difference of Series Capacitors
We are given:
- Two capacitors \( X \) and \( Y \) connected in series to a battery of \( V = 6 \, \text{V} \).
- Same plate area \( A \) and separation \( d \).
- Capacitor \( X \) has air between plates, \( Y \) has dielectric of constant \( K = 4 \).
- Equivalent capacitance: \( C_\text{eq} = 4 \, \mu\text{F} \).
Step 1: Capacitance of Individual Capacitors
For a parallel plate capacitor:
\[ C = \frac{\varepsilon_0 \, K \, A}{d} \]
- For capacitor \( X \) (air, \( K = 1 \)):
\[ C_X = \frac{\varepsilon_0 A}{d} \]
- For capacitor \( Y \) (dielectric \( K = 4 \)):
\[ C_Y = \frac{4 \varepsilon_0 A}{d} = 4 C_X \]
--- Step 2: Relation Between Series Capacitances
For capacitors in series:
\[ \frac{1}{C_\text{eq}} = \frac{1}{C_X} + \frac{1}{C_Y} \]
Substitute \( C_Y = 4 C_X \):
\[ \frac{1}{4 \, \mu\text{F}} = \frac{1}{C_X} + \frac{1}{4 C_X} = \frac{1 + 0.25}{C_X} = \frac{1.25}{C_X} \]
\[ C_X = 1.25 \times 4 \, \mu\text{F} = 5 \, \mu\text{F} \]
\[ C_Y = 4 C_X = 4 \times 5 \, \mu\text{F} = 20 \, \mu\text{F} \]
--- Step 3: Potential Difference Across Each Capacitor
In series, charge \( Q \) on both capacitors is the same:
\[ Q = C_\text{eq} \, V = 4 \times 10^{-6} \times 6 = 24 \times 10^{-6} \, \text{C} \]
Voltage across each capacitor:
\[ V_X = \frac{Q}{C_X} = \frac{24 \times 10^{-6}}{5 \times 10^{-6}} = 4.8 \, \text{V} \]
\[ V_Y = \frac{Q}{C_Y} = \frac{24 \times 10^{-6}}{20 \times 10^{-6}} = 1.2 \, \text{V} \]
Check: \( V_X + V_Y = 4.8 + 1.2 = 6 \, \text{V} \) ✔
--- Step 4: Summary
- Capacitances: \( C_X = 5 \, \mu\text{F}, \quad C_Y = 20 \, \mu\text{F} \)
- Potential differences: \( V_X = 4.8 \, \text{V}, \quad V_Y = 1.2 \, \text{V} \)
Conclusion:
The series combination of capacitors with one dielectric results in a higher capacitance for the dielectric capacitor and a smaller potential drop across it, while the air capacitor takes a larger voltage share. This maintains the total voltage equal to the battery voltage.

Answer 2

Step 1: Express \( C_Y \) in terms of \( C_X \).
With identical geometry and a dielectric constant of 4 in Y only, \[ C_Y = 4 C_X \]

Step 2: Apply the series capacitance condition.\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_X} + \frac{1}{4C_X} = \frac{4 + 1}{4C_X} = \frac{5}{4C_X} \] With \( C_{\text{eq}} = 4 \, \mu\text{F} \): \[ \frac{1}{4} = \frac{5}{4C_X} \implies C_X = 5 \, \mu\text{F}, \qquad C_Y = 20 \, \mu\text{F} \]

Step 3: Set up simultaneous equations for the voltages.
In series, both capacitors carry the same charge, so \( V \propto 1/C \), giving \[ \frac{V_X}{V_Y} = \frac{C_Y}{C_X} = \frac{20}{5} = 4 \implies V_X = 4V_Y \] Also, the two voltages must add up to the battery voltage: \[ V_X + V_Y = 6 \]

Step 4: Solve the pair of equations.
Substituting \( V_X = 4V_Y \) into the sum: \[ 4V_Y + V_Y = 6 \implies 5V_Y = 6 \implies V_Y = 1.2 \, \text{V} \] \[ V_X = 4 \times 1.2 = 4.8 \, \text{V} \]

Final Answers:\[ C_X = 5 \, \mu\text{F}, \quad C_Y = 20 \, \mu\text{F}, \quad V_X = 4.8 \, \text{V}, \quad V_Y = 1.2 \, \text{V} \]