A charge Q is given to the plates of an isolated capacitor of capacitance C. Now, the terminals of this capacitor are connected to the terminals of an inductor L at t = 0. After how much time will the energy stored in the capacitor and the Inductor become equal?

Options

A. `pisqrt((LC)//2)`
B. `pisqrt((LC)//4)`
C. `5pisqrt((LC)//2)`
D. `pisqrt((LC))`

Answer

To solve the problem, we need to determine the time at which the energy stored in the capacitor equals the energy stored in the inductor after they are connected in an LC circuit.

### Step-by-Step Solution:

1. **Initial Energy in the Capacitor:**
The initial energy stored in the capacitor when it has charge \( Q \) and capacitance \( C \) is given by the formula:
\[ U_C = \frac{Q^2}{2C} \]

2. **Energy in the Inductor:**
The energy stored in the inductor with current \( I \) is given by:
\[ U_L = \frac{1}{2} L I^2 \]

3. **Condition for Equal Energy:**
We need to find the time \( t \) when the energy in the capacitor equals the energy in the inductor:
\[ U_C = U_L \]
Substituting the expressions for \( U_C \) and \( U_L \):
\[ \frac{Q^2}{2C} = \frac{1}{2} L I^2 \]

4. **Expressing Current in Terms of Charge:**
In an LC circuit, the charge \( q(t) \) on the capacitor at time \( t \) can be expressed as:
\[ q(t) = Q \cos(\omega t) \]
where \( \omega = \frac{1}{\sqrt{LC}} \).

The current \( I \) can be expressed as the rate of change of charge:
\[ I = -\frac{dq}{dt} = -\frac{d}{dt}(Q \cos(\omega t)) = Q \omega \sin(\omega t) \]

5. **Substituting Current into Energy Equation:**
Substitute \( I \) into the energy equation:
\[ \frac{Q^2}{2C} = \frac{1}{2} L (Q \omega \sin(\omega t))^2 \]
Simplifying this gives:
\[ \frac{Q^2}{2C} = \frac{1}{2} L Q^2 \omega^2 \sin^2(\omega t) \]

6. **Cancelling \( \frac{Q^2}{2} \):**
Since \( Q^2 \) is common on both sides, we can cancel it (assuming \( Q \neq 0 \)):
\[ \frac{1}{C} = L \omega^2 \sin^2(\omega t) \]

7. **Substituting \( \omega \):**
Recall that \( \omega = \frac{1}{\sqrt{LC}} \):
\[ \frac{1}{C} = L \left(\frac{1}{\sqrt{LC}}\right)^2 \sin^2(\omega t) \]
This simplifies to:
\[ \frac{1}{C} = \frac{L}{LC} \sin^2(\omega t) \]
\[ 1 = \sin^2(\omega t) \]

8. **Finding the Time:**
The equation \( \sin^2(\omega t) = \frac{1}{2} \) implies:
\[ \sin(\omega t) = \frac{1}{\sqrt{2}} \]
This occurs at:
\[ \omega t = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
For the first positive solution:
\[ t = \frac{\pi}{4\omega} = \frac{\pi}{4} \sqrt{LC} \]

### Final Answer:
Thus, the time at which the energy stored in the capacitor and the inductor becomes equal is:
\[ t = \frac{\pi}{4} \sqrt{LC} \]