- Asked 2/13
- 2075 Asoj · 5 marks
- 2067 Shrawan · 3 marks
Derive continuity equation
Short answer
Integral form: ∮_S J·dS = −dQ_in/dt = −d/dt ∫_v ρ_v dv, expressing conservation of charge. Point form: ∇·J = −∂ρ_v/∂t, obtained by applying divergence theorem. For steady currents, ∇·J = 0, implying KCL.
Derive the integral and point forms of the continuity equation.
The continuity equation expresses the law of conservation of charge: charge can neither be created nor destroyed. The current leaving a closed region equals the rate of decrease of charge inside it.
Integral form
Consider a closed surface S enclosing volume v and total charge Q_in. The net current flowing out through S is the total outward flux of the current density J:
I_out = ∮_S J · dS
By conservation of charge, this equals the rate at which the enclosed charge decreases:
∮_S J · dS = −dQ_in/dt = −d/dt ∫_v ρ_v dv ← integral form
Point form
Apply the divergence theorem to the left side: ∮ J · dS = ∫_v (∇ · J) dv. For a fixed volume, the time derivative can be taken inside the integral as a partial derivative:
∫_v (∇ · J) dv = −∫_v (∂ρ_v/∂t) dv
This must hold for any volume, however small, so the integrands must be equal:
∇ · J = −∂ρ_v/∂t ← point form
Remarks
- For steady (time-independent) currents ∂ρ_v/∂t = 0, so ∇ · J = 0 and ∮ J · dS = 0: current lines are continuous and closed, which is Kirchhoff's current law (net current into a node is zero).
- The point form means that if current diverges from a point, the charge density there must be decreasing with time.
- It combines with Ohm's law (J = σE) and Gauss's law to give the relaxation time T_r = ε/σ.
✓ Answer checked by Claude Opus 5.5 (high effort)
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