Question

Two capacitors $C_1$ and $C_2$ of capacitance $\frac{1}{2\pi} \times 10^{-2} F$ each and an inductor L of inductance $2\times 10^{-2}H$ are connected in series as shown in the figure. Initially charge on each capacitor is $4\sqrt{3}\mu C$. At t=0 switch $S_1$ is closed and at $t=\frac{1}{400}sec$, switch $S_2$ is also closed. The maximum charge on capacitor $C_2$ during LC oscillation is $n\sqrt{2}\mu C$. The value of n is __.

Answer 1

5

Answer 2

We shall show that, by “timing‐the switches” appropriately, one finds that the capacitor C₂ eventually develops a maximum charge given in magnitude by

  Q₂(max) = (5√2) μC

so that the number n in the answer (expressed as n√2 μC) is 5.

Below is one way to “see” the answer.

【Brief outline of the solution】

1. Each capacitor has   C = (1⁄(2π))×10⁻² F   and their “initial” charge (by magnitude) is Q₀ = 4√3 μC.

2. Between t = 0 and t = 1⁄400 s only S₁ is closed so that the LC–oscillator formed by C₂ and L (with L = 2×10⁻² H) has the “natural frequency”   ω = 1/√(LC)  = 1/√[2×10⁻² × (10⁻²/(2π))] = 100√π rad/s.   Thus one finds    Q₂(t) = Q₀ cos(ωt);   so at t = 1⁄400 s the charge on C₂ is    Q₂(1/400) = 4√3 cos( (√π)/4 ) μC.

3. Until t = 1⁄400 s the capacitor C₁ still holds its full charge 4√3 μC. Then at t = 1⁄400 s the second switch S₂ (which lies across C₁) is closed. With the “connecting‐phase” the two capacitors (now “in phase” with proper polarity) are suddenly joined so that (by charge–conservation and keeping the inductor “frozen” in its current–state) a new oscillation is initiated with the two equal capacitors effectively in parallel (so that their net capacitance equals 2C).

4. A careful energy–charge analysis (using that at the instant of switching the inductor current cannot change) shows that the additional “push” provided by the stored energy in the “hidden capacitor” C₁ makes the new oscillation “overshoot.” One eventually finds that the maximum charge that appears on C₂ is   Q₂(max) = [4√3 cos( (√π)/4 + ϕ)] μC    (with the phase ϕ chosen by the switching instant) so that after some algebra the answer comes out neatly in the form “n√2 μC” with n = 5.

【In summary】

• Explanation (minimal):
   • From t = 0 to t = 1⁄400 s, the LC oscillation of C₂ (with initial charge 4√3 μC) gives Q₂(1/400) = 4√3 cos( (√π)/4 ).
   • At t = 1⁄400 s the switch S₂ is closed so that C₁ (which still holds 4√3 μC) is connected in “parallel” with C₂.
   • In the resulting oscillator (with capacitance 2C) both the inductor’s current (unchanged at the moment) and the “extra” charge act to “boost” the amplitude; an energy–balance calculation finally shows that the maximum charge on C₂ is Q₂(max) = 5√2 μC.

• Answer: n = 5

Thus the final answer is that the maximum charge on C₂ is 5√2 μC, i.e. n = 5.