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)] \( V_X = 4.8 \, \text{V}, \quad V_Y = 1.2 \, \text{V} \)