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 \(4\,\mu\text{F}\).

Hint:

If two identical capacitors differ only by dielectric constant \(K\), then \[ C_{\text{dielectric}}=K\,C_{\text{air}}. \] For capacitors in series, \[ C_{\text{eq}}=\frac{C_1C_2}{C_1+C_2}. \]

Answer 1

Concept: The capacitance of a parallel plate capacitor is given by \[ C=\frac{\varepsilon_0 A}{d} \] where \(A\) is the plate area and \(d\) is the separation between the plates. When a dielectric medium of dielectric constant \(K\) completely fills the space between the plates, the capacitance becomes \[ C'=K\frac{\varepsilon_0 A}{d}=KC. \] Since the two capacitors have identical plate area and plate separation, \[ C_Y=4C_X. \] For capacitors connected in series, \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_X} + \frac{1}{C_Y}. \]

Step 1: Assume the capacitance of capacitor X. Let \[ C_X=C. \] Since capacitor Y contains a dielectric of dielectric constant \(4\), \[ C_Y=4C. \]

Step 2: Use the series combination formula. Given, \[ C_{\text{eq}}=4\,\mu\text{F}. \] For series combination, \[ \frac{1}{4} = \frac{1}{C} + \frac{1}{4C}. \] Taking LCM, \[ \frac{1}{4} = \frac{5}{4C}. \] Multiplying both sides by \(4C\), \[ C=5\,\mu\text{F}. \] Therefore, \[ \boxed{C_X=5\,\mu\text{F}}. \]

Step 3: Calculate the capacitance of capacitor Y. Since \[ C_Y=4C_X, \] \[ C_Y=4\times5. \] \[ C_Y=20\,\mu\text{F}. \] Therefore, \[ \boxed{C_Y=20\,\mu\text{F}}. \] Final Answer: \[ \boxed{C_X=5\,\mu\text{F}} \] \[ \boxed{C_Y=20\,\mu\text{F}} \]