In an oscillating LC circuit, the maximum charge on the capacitor is qm. Determine the charge on the capacitor, q, and the current through the inductor, I, when energy is shared equally between the electric and magnetic fields. Express your answers in terms of qm, L, and C.

a. q = qm/√2 b. qm/√(2LC)

Explanation:

a. Charge on the capacitor

Let U₁ = energy in inductor and U₂ = energy in capacitor and U = total energy in circuit.

So, U₁ + U₂ = U.

Since the energy is shared equally between the capacitor and inductor, U₁= U₂,so

2U₂ = U

and U₂ = U/2

Now U₂ = q²/2C where q is the charge on the capacitor with capacitance C and

U = (qm)²/2C where qm is the maximum charge on the capacitor.

Since U₂ = U/2,

Substituting the values for U₂ and U, we have

q²/2C = [(qm)²/2C]/2

q² = (qm)²/2

taking square-root of both sides, we have

q = qm/√2

b. The current in the inductor

Since the energy in the capacitor equals the energy in the inductor,

1/2LI² = 1/2q²/C where L is the inductance of the inductor and I the current through it.

I² = q²/LC

taking square-root of both sides, we have

I = q/√LC

substituting the value of q from above, we have

I = qm/√2 ÷ √LC

I = qm/√(2LC)

multiple

the motion of a pendulum is a classic example of mechanical energy conservation. a pendulum consists of a mass (known as a bob ) attached by a string to a pivot point. as the pendulum moves it sweeps out a circular arc, moving back and forth in a periodic fashion.