---
title: "Derive continuity equation"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:continuity-equation-derive-forms"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/continuity-equation-derive-forms
site: BCT Engineering
---

# Derive continuity equation

Electromagnetism · Chapter 2: Electric field

- Asked in: 2075 Asoj, 5 marks; 2067 Shrawan, 3 marks (asked 2 times)
- Video: [Continuity of Current: Theory and Derivation Explained](https://www.youtube.com/watch?v=lh9Kximd0Pg) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/continuity-equation-derive-forms
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Derive the integral and point forms of the 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.

## Answer

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:

```text
I_out = ∮_S J · dS
```

By conservation of charge, this equals the rate at which the enclosed charge decreases:

```text
∮_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:

```text
∫_v (∇ · J) dv = −∫_v (∂ρ_v/∂t) dv
```

This must hold for any volume, however small, so the integrands must be equal:

```text
∇ · 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The continuity equation expresses which physical law?
   A. Conservation of energy
   B. Conservation of momentum
   C. Conservation of charge
   D. Faraday's law of induction
   Answer: C. Charge can be neither created nor destroyed, so current leaving a region equals the rate of decrease of charge inside. (Key point: Physical principle)

2. Which is the integral form of the continuity equation?
   A. ∮ J · dS = + d/dt ∫ ρ_v dv
   B. ∮ J · dS = - d/dt ∫ ρ_v dv
   C. ∮ J · dl = - d/dt ∫ ρ_v dv
   D. ∮ J · dS = 0 always
   Answer: B. The net outward current equals the rate of decrease of enclosed charge, hence the minus sign. (Key point: Integral form)

3. How is the point form ∇·J = -∂ρ_v/∂t obtained from the integral form?
   A. Apply the divergence theorem and equate the integrands, since it holds for any volume
   B. Apply Stokes' theorem and set the line integral to zero
   C. Differentiate Gauss's law with respect to position
   D. Assume ρ_v is constant in time
   Answer: A. The divergence theorem converts the surface integral to ∫∇·J dv; since the equality holds for any volume the integrands are equal. (Key point: Getting the point form)

4. For steady (time-independent) currents the continuity equation gives:
   A. ∇·J = ρ_v
   B. ∇×J = 0 only
   C. J = 0
   D. ∇·J = 0, so current lines are closed (Kirchhoff's current law)
   Answer: D. With ∂ρ_v/∂t = 0, ∇·J = 0 and ∮J·dS = 0, which is Kirchhoff's current law. (Key point: Steady currents)

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