A circuits contains a capacitor and inductance each with negligible resistance. The capacitor is initially charged and the charging battery is disconnected. At subsequent time , the charge on the capacitor will

Options

A. increase exponentially
B. decrease exponentially
C. decrease linearly
D. remain constant

Answer

To solve the problem of how the charge on the capacitor varies over time in a circuit containing a capacitor and an inductor with negligible resistance, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have a capacitor (C) that is initially charged and an inductor (L) connected in parallel. The battery is disconnected, meaning no external voltage is applied.

2. **Energy Conservation**:
- The energy stored in the capacitor is given by \( U_C = \frac{1}{2} C V^2 \) or \( U_C = \frac{1}{2} \frac{Q^2}{C} \), where \( Q \) is the charge on the capacitor.
- The energy stored in the inductor is given by \( U_L = \frac{1}{2} L I^2 \), where \( I \) is the current through the inductor.

3. **Energy Transfer**:
- When the battery is disconnected, the energy will oscillate between the capacitor and the inductor. The total energy in the system remains constant, so:
\[ \frac{1}{2} C V^2 + \frac{1}{2} L I^2 = \text{constant} \]

4. **Using Kirchhoff's Law**:
- The relationship between the charge \( Q \) on the capacitor and the current \( I \) through the inductor can be expressed as:
\[ I = \frac{dQ}{dt} \]
- Thus, we can write the equation for the inductor:
\[ L \frac{dI}{dt} = \frac{Q}{C} \]

5. **Differential Equation**:
- Substituting \( I = \frac{dQ}{dt} \) into the inductor equation gives:
\[ L \frac{d^2Q}{dt^2} = -\frac{Q}{C} \]
- Rearranging this yields:
\[ \frac{d^2Q}{dt^2} + \frac{1}{LC} Q = 0 \]
- This is a second-order linear differential equation, which is characteristic of simple harmonic motion (SHM).

6. **Solution of the Differential Equation**:
- The general solution for this equation is:
\[ Q(t) = Q_0 \cos(\omega t + \phi) \]
- where \( \omega = \frac{1}{\sqrt{LC}} \), \( Q_0 \) is the maximum charge, and \( \phi \) is the phase constant.

7. **Conclusion**:
- The charge \( Q(t) \) on the capacitor oscillates sinusoidally over time with a fixed amplitude \( Q_0 \). Therefore, the charge does not decrease or increase; it simply oscillates between maximum and minimum values.

### Final Answer:
The charge on the capacitor will oscillate and remain constant in total energy, as there is no loss of charge due to negligible resistance.