In L-C oscillation, maximum charge on capacitor can be Q. If at any instant, electric energy and magnetic enegy associated with circuit is equal, then charge on capacitor at the anstant is

Options

A. `(Q)/sqrt(2)`
B. `(Q)/(2)`
C. `sqrt(3)(Q)/(2)`
D. `(3Q)/(2)`

Answer

To solve the problem step by step, let's break down the concepts involved in LC oscillation and the relationship between electric and magnetic energy in the circuit.

### Step-by-Step Solution:

1. **Understanding the Maximum Charge (Q)**:
- The maximum charge that the capacitor can store is given as \( Q \).

2. **Energy Stored in the Capacitor**:
- The electric energy (\( U_E \)) stored in the capacitor when it has charge \( q \) is given by the formula:
\[ U_E = \frac{1}{2} \frac{q^2}{C} \]
- Here, \( C \) is the capacitance of the capacitor.

3. **Energy Stored in the Inductor**:
- The magnetic energy (\( U_B \)) stored in the inductor when the current is \( I \) is given by:
\[ U_B = \frac{1}{2} L I^2 \]
- Here, \( L \) is the inductance of the inductor.

4. **Condition Given in the Problem**:
- It is given that at a certain instant, the electric energy and magnetic energy are equal:
\[ U_E = U_B \]

5. **Expressing Magnetic Energy in Terms of Charge**:
- In LC oscillation, the current \( I \) can be related to the charge \( q \) on the capacitor as:
\[ I = -\frac{dq}{dt} \]
- At the instant when electric energy equals magnetic energy, we can express \( U_B \) in terms of the charge \( q \).

6. **Total Energy in the Circuit**:
- The total energy in the LC circuit is constant and equals the maximum energy stored in the capacitor when it is fully charged:
\[ U_{total} = \frac{1}{2} \frac{Q^2}{C} \]

7. **Setting Up the Equation**:
- Since \( U_E = U_B \), we can write:
\[ U_E + U_B = U_{total} \]
- Substituting \( U_E = U_B \) into the equation gives:
\[ 2U_E = U_{total} \]
- Therefore:
\[ 2 \left( \frac{1}{2} \frac{q^2}{C} \right) = \frac{1}{2} \frac{Q^2}{C} \]

8. **Simplifying the Equation**:
- This simplifies to:
\[ q^2 = \frac{Q^2}{2} \]

9. **Finding the Charge \( q \)**:
- Taking the square root of both sides gives:
\[ q = \frac{Q}{\sqrt{2}} \]

10. **Conclusion**:
- The charge on the capacitor at the instant when the electric energy equals the magnetic energy is:
\[ q = \frac{Q}{\sqrt{2}} \]

### Final Answer:
The charge on the capacitor at the instant when electric energy and magnetic energy are equal is \( \frac{Q}{\sqrt{2}} \).