Q150 / 180 · COMEDK 2023 (Evening Shift) · Physics
In the figure, first the capacitors are fully charged by closing the key $\mathrm{K}$. Then after opening the Key a dielectric material with dielectric constant 2 is filled in the space between the plates of both the capacitor. At this state the ratio of the Charge on the capacitor $C_1$ to that of $C_2$ is:
In the figure, first the capacitors are fully charged by closing the key $\mathrm{K}$. Then after opening the Key a dielectric material with dielectric constant 2 is filled in the space between the plates of both the capacitor. At this state the ratio of the Charge on the capacitor $C_1$ to that of $C_2$ is:- Correct answerA2 : 1
- B1 : 1
- C1 : 2
- D3 : 2
Solution
When key K is closed, both $C_1$ and $C_2$ ($= 1\ \mu F$ each) are connected to the 12V battery. Initial charge on both:
$Q_1 = Q_2 = 1\ \mu F \times 12\ V = 12\ \mu C$
When key K is opened, $C_2$ gets isolated — its charge stays constant at $12\ \mu C$. $C_1$ remains connected to the battery.
Dielectric with $k = 2$ is inserted in both. New capacitances: $C_1' = C_2' = 2\ \mu F$
For $C_1$ (still connected to battery, $V = 12\ V$):
$Q_1' = 2\ \mu F \times 12\ V = 24\ \mu C$
For $C_2$ (isolated, charge conserved):
$Q_2' = 12\ \mu C$
$\dfrac{Q_1'}{Q_2'} = \dfrac{24}{12} = \dfrac{2}{1}$
$Q_1 = Q_2 = 1\ \mu F \times 12\ V = 12\ \mu C$
When key K is opened, $C_2$ gets isolated — its charge stays constant at $12\ \mu C$. $C_1$ remains connected to the battery.
Dielectric with $k = 2$ is inserted in both. New capacitances: $C_1' = C_2' = 2\ \mu F$
For $C_1$ (still connected to battery, $V = 12\ V$):
$Q_1' = 2\ \mu F \times 12\ V = 24\ \mu C$
For $C_2$ (isolated, charge conserved):
$Q_2' = 12\ \mu C$
$\dfrac{Q_1'}{Q_2'} = \dfrac{24}{12} = \dfrac{2}{1}$
Hence, the correct answer is option: 'A'.