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:

In series capacitors:

Same charge on each
Voltage divides inversely with capacitance
Larger capacitance → smaller voltage drop

Answer 1

Concept: For parallel plate capacitor: \[ C = \frac{\varepsilon A}{d} \] If dielectric constant \( K \) is introduced: \[ C' = K C \] For capacitors in series: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Step 1: Relation between capacitances. Since same geometry:

Capacitor X (air): \( C_X = C \)
Capacitor Y (dielectric \( K = 4 \)): \( C_Y = 4C \)

Step 2: Equivalent capacitance in series. \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C} + \frac{1}{4C} = \frac{5}{4C} \] Given: \[ C_{\text{eq}} = 4 \, \mu\text{F} \] \[ \frac{1}{4} = \frac{5}{4C} \] \[ C = 5 \, \mu\text{F} \]
Step 3: Individual capacitances. \[ C_X = 5 \, \mu\text{F} \] \[ C_Y = 4C = 20 \, \mu\text{F} \]
Step 4: Voltage distribution in series. In series:

Charge on each capacitor is same
Voltage inversely proportional to capacitance
Total voltage: \[ V = 6 \, \text{V} \] Using: \[ V_X : V_Y = \frac{1}{C_X} : \frac{1}{C_Y} = \frac{1}{5} : \frac{1}{20} = 4 : 1 \]
Step 5: Calculate individual voltages. \[ V_X = \frac{4}{5} \times 6 = 4.8 \, \text{V} \] \[ V_Y = \frac{1}{5} \times 6 = 1.2 \, \text{V} \] Final Answers:

[(a)] \( C_X = 5 \, \mu\text{F}, \quad C_Y = 20 \, \mu\text{F} \)
[(b)] \( V_X = 4.8 \, \text{V}, \quad V_Y = 1.2 \, \text{V} \)

Answer 2

Concept: The individual capacitances and the voltage split can both be obtained using the product-over-sum form of the series formula and a direct voltage-divider fraction, without separately calculating the charge.

Step 1: Relate the two capacitances.
Since X and Y have the same plate area and separation, and Y additionally has a dielectric of constant \( K = 4 \) while X has air (\( K = 1 \)) between its plates, \[ C_Y = 4 \, C_X \]

Step 2: Use the product-over-sum form for series capacitors.\[ C_{\text{eq}} = \frac{C_X C_Y}{C_X + C_Y} = \frac{C_X (4C_X)}{C_X + 4C_X} = \frac{4C_X^{2}}{5C_X} = \frac{4C_X}{5} \]

Step 3: Solve for \( C_X \) and \( C_Y \).
Given \( C_{\text{eq}} = 4 \, \mu\text{F} \): \[ 4 = \frac{4C_X}{5} \implies C_X = 5 \, \mu\text{F} \] \[ C_Y = 4 C_X = 20 \, \mu\text{F} \]

Step 4: Split the battery voltage using the voltage-divider fraction.
For capacitors in series, since the charge on each is identical, each one's share of the total voltage is proportional to the OTHER capacitor's value: \[ V_X = V \cdot \frac{C_Y}{C_X + C_Y}, \qquad V_Y = V \cdot \frac{C_X}{C_X + C_Y} \]

Step 5: Substitute the numbers.\[ V_X = 6 \times \frac{20}{25} = 4.8 \, \text{V}, \qquad V_Y = 6 \times \frac{5}{25} = 1.2 \, \text{V} \]

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