A fully charged capacitor C with initial charge `q_(0)` is connected to a coil of self inductance L at t=0. The time at which the energy is stored equally between the electric and the magnetic fields is

Options

A. `(pi)/(4)sqrt(LC)`
B. `2pi sqrt(LC)`
C. `sqrt(LC)`
D. `pi sqrt(LC)`

Explanation

To solve the problem of finding the time at which the energy is stored equally between the electric field in the capacitor and the magnetic field in the inductor, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the System**:
- We have a capacitor \( C \) charged with an initial charge \( q_0 \) connected to an inductor with self-inductance \( L \) at \( t = 0 \).
- The energy in the capacitor is given by \( U_C = \frac{Q^2}{2C} \) and the energy in the inductor is given by \( U_L = \frac{1}{2}LI^2 \).

2. **Charge and Current Relationships**:
- The charge \( Q \) on the capacitor at any time \( t \) can be expressed as:
\[ Q = q_0 \cos(\omega t) \]
- The current \( I \) flowing through the circuit is the rate of change of charge:
\[ I = \frac{dQ}{dt} = -\omega q_0 \sin(\omega t) \]

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

4. **Substituting for \( Q \) and \( I \)**:
- Substitute \( Q \) and \( I \) into the energy equation:
\[ \frac{(q_0 \cos(\omega t))^2}{2C} = \frac{1}{2}L(-\omega q_0 \sin(\omega t))^2 \]
- Simplifying both sides:
\[ \frac{q_0^2 \cos^2(\omega t)}{2C} = \frac{L \omega^2 q_0^2 \sin^2(\omega t)}{2} \]

5. **Cancelling Common Terms**:
- Cancel \( \frac{q_0^2}{2} \) from both sides (assuming \( q_0 \neq 0 \)):
\[ \frac{\cos^2(\omega t)}{C} = L \omega^2 \sin^2(\omega t) \]

6. **Using the Resonant Frequency**:
- Recall that \( \omega = \frac{1}{\sqrt{LC}} \):
\[ \frac{\cos^2(\omega t)}{C} = L \left(\frac{1}{LC}\right) \sin^2(\omega t) \]
- This simplifies to:
\[ \cos^2(\omega t) = \sin^2(\omega t) \]

7. **Finding the Angle**:
- From \( \cos^2(\omega t) = \sin^2(\omega t) \), we can deduce:
\[ \tan^2(\omega t) = 1 \implies \tan(\omega t) = 1 \]
- This occurs when:
\[ \omega t = \frac{\pi}{4} \]

8. **Solving for Time \( t \)**:
- Therefore, we can express \( t \) as:
\[ t = \frac{\pi}{4\omega} \]
- Substituting \( \omega = \frac{1}{\sqrt{LC}} \):
\[ t = \frac{\pi}{4} \sqrt{LC} \]

### Final Answer:
The time at which the energy is stored equally between the electric and magnetic fields is:
\[ t = \frac{\pi}{4} \sqrt{LC} \]