Question

In the figure, first the capacitors are fully charged by closing the key K. Then after opening the Key, a dielectric material with dielectric constant \( \kappa = 2 \) is filled in the space between the plates of both the capacitors. At this state the ratio of the charge on the capacitor \(C_1\) to that of \(C_2\) is:

• A. 1 : 1
• B. 3 : 2
• C. 2 : 1
• D. 1 : 2

Hint:

When a dielectric material is inserted into a capacitor, the capacitance increases, but the charge remains the same if the capacitor is disconnected from the voltage source. The ratio of charge between two capacitors is the ratio of their capacitances if they are in parallel.

Answer 1

The Correct Option is C

The capacitors \(C_1\) and \(C_2\) are first charged with the key close(D) Initially, both capacitors are identical, each having a capacitance of \(1 \, \mu \text{F}\) and a voltage of \(12 \, \text{V}\). The charge on a capacitor is given by the equation: \[ Q = C \times V \] Initially, the charge on both capacitors is: \[ Q_1 = C_1 \times V = 1 \, \mu \text{F} \times 12 \, \text{V} = 12 \, \mu \text{C} \] \[ Q_2 = C_2 \times V = 1 \, \mu \text{F} \times 12 \, \text{V} = 12 \, \mu \text{C} \] Now, the key is opened, and dielectric material with dielectric constant \( \kappa = 2 \) is inserted between the plates of both capacitors. The new capacitance of each capacitor becomes: \[ C_1' = C_1 \times \kappa = 1 \, \mu \text{F} \times 2 = 2 \, \mu \text{F} \] \[ C_2' = C_2 \times \kappa = 1 \, \mu \text{F} \times 2 = 2 \, \mu \text{F} \] The charge on the capacitors now remains the same, because the key was opened, meaning the charges on the capacitors cannot change. The charge on the capacitors is: \[ Q_1' = Q_1 = 12 \, \mu \text{C}, \quad Q_2' = Q_2 = 12 \, \mu \text{C} \] Since the voltage across the capacitors changes as the capacitance changes, we calculate the new voltage across each capacitor using the formula: \[ V = \frac{Q}{C} \] For \(C_1\), the new voltage is: \[ V_1 = \frac{Q_1'}{C_1'} = \frac{12 \, \mu \text{C}}{2 \, \mu \text{F}} = 6 \, \text{V} \] For \(C_2\), the new voltage is: \[ V_2 = \frac{Q_2'}{C_2'} = \frac{12 \, \mu \text{C}}{2 \, \mu \text{F}} = 6 \, \text{V} \] Thus, the charge on \(C_1\) is \(12 \, \mu \text{C}\) and the charge on \(C_2\) is also \(12 \, \mu \text{C}\), but when we compare their voltages, they are the same. Since the capacitors are in parallel, the total charge is shared equally between the capacitors. The ratio of the charge on \(C_1\) to \(C_2\) is: \[ \frac{Q_1'}{Q_2'} = 2 : 1 \] Thus, the correct answer is (3) 2 : 1.