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, we need to find the charge on the capacitor at the instant when the electric energy and magnetic energy in the LC circuit are equal. Here's a step-by-step solution:

### Step 1: Understand the maximum charge and energy
The maximum charge on the capacitor is given as \( Q \). The maximum energy stored in the capacitor can be calculated using the formula:
\[ U_{\text{max}} = \frac{Q^2}{2C} \]
where \( C \) is the capacitance.

### Step 2: Set up the energy equations
At any instant, the electric energy (\( U_e \)) stored in the capacitor and the magnetic energy (\( U_m \)) stored in the inductor are equal. Let's denote the charge on the capacitor at this instant as \( Q' \). The electric energy at this instant is given by:
\[ U_e = \frac{(Q')^2}{2C} \]
The magnetic energy in the inductor can also be expressed in terms of the current \( I \). However, since we know that \( U_e = U_m \), we can express the total energy in the circuit as:
\[ U_{\text{total}} = U_e + U_m \]

### Step 3: Equate electric and magnetic energy
Since \( U_e = U_m \), we can write:
\[ U_e = U_m = \frac{(Q')^2}{2C} \]
Thus, we can express the total energy as:
\[ U_{\text{total}} = 2U_e = 2 \cdot \frac{(Q')^2}{2C} = \frac{(Q')^2}{C} \]

### Step 4: Relate total energy to maximum energy
The total energy in the LC circuit is also equal to the maximum energy, which we established as:
\[ U_{\text{total}} = U_{\text{max}} = \frac{Q^2}{2C} \]

### Step 5: Set the equations equal
Now we can set the two expressions for total energy equal to each other:
\[ \frac{(Q')^2}{C} = \frac{Q^2}{2C} \]

### Step 6: Simplify the equation
We can multiply both sides by \( C \) to eliminate the capacitance:
\[ (Q')^2 = \frac{Q^2}{2} \]

### Step 7: Solve for \( Q' \)
Taking the square root of both sides gives us:
\[ Q' = \frac{Q}{\sqrt{2}} \]

### Conclusion
Thus, the charge on the capacitor at the instant when the electric energy and magnetic energy are equal is:
\[ Q' = \frac{Q}{\sqrt{2}} \]

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