An inductor of inductance 20 mH is connected across a charged capacitor of capacitance 5 `mu`F & resulting L-C circuit is set oscillating at its natural frequency . The maximum charge q is 200 `mu` C on the capacitor . Find the potential difference across the inductor when the charge is 100 `mu `C .

Answer

To solve the problem, we need to find the potential difference across the inductor when the charge on the capacitor is 100 µC. Here’s a step-by-step solution:

### Step 1: Understand the LC Circuit
An LC circuit consists of an inductor (L) and a capacitor (C) connected in parallel. The energy oscillates between the capacitor and the inductor.

### Step 2: Identify Given Values
- Inductance (L) = 20 mH = 20 x 10^-3 H
- Capacitance (C) = 5 µF = 5 x 10^-6 F
- Maximum charge (Q) = 200 µC = 200 x 10^-6 C
- Charge at the moment of interest (q) = 100 µC = 100 x 10^-6 C

### Step 3: Calculate the Potential Difference Across the Capacitor
The potential difference (V) across a capacitor is given by the formula:
\[ V = \frac{q}{C} \]
Substituting the values:
\[ V = \frac{100 \times 10^{-6}}{5 \times 10^{-6}} \]
\[ V = \frac{100}{5} = 20 \text{ volts} \]

### Step 4: Determine the Energy in the Circuit
The total energy in the LC circuit can be expressed in terms of the maximum charge:
\[ E = \frac{1}{2} C Q^2 \]
Substituting the values:
\[ E = \frac{1}{2} \times 5 \times 10^{-6} \times (200 \times 10^{-6})^2 \]
\[ E = \frac{1}{2} \times 5 \times 10^{-6} \times 40000 \times 10^{-12} \]
\[ E = \frac{1}{2} \times 5 \times 40 \times 10^{-18} \]
\[ E = 100 \times 10^{-18} = 10^{-16} \text{ joules} \]

### Step 5: Calculate the Potential Difference Across the Inductor
The potential difference across the inductor (V_L) can be calculated using the energy conservation principle. The total energy in the circuit is equal to the sum of the energy stored in the capacitor and the energy stored in the inductor:
\[ E = \frac{1}{2} C q^2 + \frac{1}{2} L I^2 \]
At the moment when the charge is 100 µC, we can find the current (I) using:
\[ I = \sqrt{\frac{2E}{L}} \]
Substituting the values:
\[ I = \sqrt{\frac{2 \times 10^{-16}}{20 \times 10^{-3}}} \]
\[ I = \sqrt{\frac{2 \times 10^{-16}}{20 \times 10^{-3}}} = \sqrt{10^{-14}} = 10^{-7} \text{ A} \]

Now, using the formula for potential difference across the inductor:
\[ V_L = L \frac{dI}{dt} \]
Since we are looking for the potential difference when the charge is 100 µC, we can use the relationship:
\[ V_L = V - V_C \]
Where \( V_C \) is the potential across the capacitor:
\[ V_C = \frac{q}{C} = 20 \text{ volts} \]

### Final Calculation
Since the total voltage across the LC circuit is constant, and we have calculated the voltage across the capacitor:
\[ V_L = V - V_C \]
\[ V_L = 20 - 20 = 0 \text{ volts} \]

### Conclusion
The potential difference across the inductor when the charge is 100 µC is 0 volts.