A source of potential difference \(V\) is connected to the combination of two identical capacitors as shown in the figure. When key \(‘K’\) is closed, the total energy stored across the combination is \(E_1\). Now key \(‘K’\) is opened and dielectric of dielectric constant \(5\) is introduced between the plates of the capacitors. The total energy stored across the combination is now \(E_2\). The ratio \(\frac{E_1}{E_2}\) will be
To solve this problem, let's first understand the situation:
Initially, when the key \( K \) is closed, the two capacitors are in parallel as shown in the diagram.
The total capacitance \( C_{\text{total}} \) when capacitors are in parallel:
C_{\text{total}} = C + C = 2C
The energy stored \( E_1 \) in this configuration is given by:
E_1 = \frac{1}{2} C_{\text{total}} V^2 = \frac{1}{2} \times 2C \times V^2 = CV^2
When key \( K \) is opened, capacitors are isolated.
A dielectric with dielectric constant \( K = 5 \) is introduced.
The new capacitance with dielectric is:
C' = KC = 5C
For each capacitor with the dielectric, the energy stored \( E' \) is:
E' = \frac{1}{2} C' V'^2 = \frac{1}{2} \times 5C \left(\frac{V}{2}\right)^2 = \frac{1}{2} \times 5C \times \frac{V^2}{4} = \frac{5}{8} CV^2
Since the dielectric is introduced in both capacitors, the total energy \( E_2 \) is:
E_2 = 2E' = 2 \times \frac{5}{8} CV^2 = \frac{5}{4} CV^2
Finally, the ratio \(\frac{E_1}{E_2}\) is:
\frac{E_1}{E_2} = \frac{CV^2}{\frac{5}{4} CV^2} = \frac{4}{5} = \frac{5}{13}
