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 find the time \( t \) when the energy stored in the capacitor equals the energy stored in the inductor after connecting them at \( t = 0 \).

### Step-by-Step Solution:

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

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

3. **Condition for Equal Energy**:
We want to find the time \( t \) when the energies in the capacitor and inductor are equal:
\[ U_C = U_L \]
Therefore,
\[ \frac{Q^2}{2C} = \frac{1}{2} L I^2 \]

4. **Relating Current and Charge**:
The current \( I \) in the circuit can be expressed as the rate of change of charge \( Q \) on the capacitor:
\[ I = -\frac{dQ}{dt} \]
Substituting this into the energy equation gives:
\[ \frac{Q^2}{2C} = \frac{1}{2} L \left(-\frac{dQ}{dt}\right)^2 \]

5. **Differential Equation**:
Rearranging the equation leads to:
\[ \frac{Q^2}{C} = L \left(\frac{dQ}{dt}\right)^2 \]
Taking the square root of both sides, we have:
\[ \frac{dQ}{dt} = \pm \frac{Q}{\sqrt{LC}} \]

6. **Separation of Variables**:
We can separate variables to solve for \( Q \):
\[ \frac{dQ}{Q} = \pm \frac{dt}{\sqrt{LC}} \]

7. **Integrating**:
Integrating both sides gives:
\[ \ln |Q| = \pm \frac{t}{\sqrt{LC}} + C_1 \]
Exponentiating both sides leads to:
\[ Q = Q_0 e^{\pm \frac{t}{\sqrt{LC}}} \]
where \( Q_0 \) is the initial charge.

8. **Finding Time When Energies are Equal**:
We know that the energy in the capacitor will be half of the initial energy when:
\[ Q = \frac{Q_0}{\sqrt{2}} \]
Setting \( Q = \frac{Q_0}{\sqrt{2}} \) in our equation:
\[ \frac{Q_0}{\sqrt{2}} = Q_0 e^{-\frac{t}{\sqrt{LC}}} \]
Dividing both sides by \( Q_0 \) (assuming \( Q_0 \neq 0 \)):
\[ \frac{1}{\sqrt{2}} = e^{-\frac{t}{\sqrt{LC}}} \]

9. **Taking Logarithm**:
Taking the natural logarithm of both sides:
\[ -\frac{t}{\sqrt{LC}} = \ln\left(\frac{1}{\sqrt{2}}\right) \]
Simplifying gives:
\[ t = -\sqrt{LC} \ln\left(\frac{1}{\sqrt{2}}\right) = \sqrt{LC} \cdot \frac{1}{2} \ln(2) \]

10. **Final Result**:
Thus, the time \( t \) when the energy stored in the capacitor and the inductor becomes equal is:
\[ t = \frac{\sqrt{LC}}{2} \ln(2) \]