---
title: "Average J over surface and ∂ρv/∂t"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:average-j-charge-rate-change"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/average-j-charge-rate-change
site: BCT Engineering
---

# Average J over surface and ∂ρv/∂t

Electromagnetism · Chapter 2: Electric field

- Asked in: 2073 Shrawan, 5+3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/average-j-charge-rate-change
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

A current density in a certain region is given as J = 20 sinθ cosφ a_r + (1/r) a_θ A/m². Find: (i) the average value of J_r over the surface r = 1, 0 < θ < π/2, 0 < φ < π/2; (ii) ∂ρ_v/∂t.

## Short answer

Average J_r over the specified surface is (∫ J_r dS)/(∫ dS) = 10 A/m². From the continuity equation, ∂ρ_v/∂t = −∇·J = −(40 sinθ cosφ)/r − cosθ/(r² sinθ) A/m³.

## Answer

Given J = 20 sinθ cosφ a_r + (1/r) a_θ A/m² in spherical coordinates.

#### (i) Average value of J_r over r = 1, 0 < θ < π/2, 0 < φ < π/2
Average = (∫ J_r dS) / (∫ dS), with dS = r² sinθ dθ dφ (r = 1).

```text
∫ J_r dS = ∫₀^(π/2) ∫₀^(π/2) 20 sinθ cosφ · sinθ dθ dφ
         = 20 ∫₀^(π/2) sin²θ dθ · ∫₀^(π/2) cosφ dφ
         = 20 · (π/4) · 1
         = 5π = 15.71

∫ dS = ∫₀^(π/2) sinθ dθ ∫₀^(π/2) dφ = (1)(π/2) = π/2 = 1.571

J_r(avg) = 5π/(π/2) = 10 A/m²
```

#### (ii) ∂ρ_v/∂t
From the continuity equation, ∇·J = −∂ρ_v/∂t.

```text
∇·J = (1/r²) ∂(r² J_r)/∂r + (1/(r sinθ)) ∂(sinθ J_θ)/∂θ + (1/(r sinθ)) ∂J_φ/∂φ

term 1: (1/r²) ∂(20 r² sinθ cosφ)/∂r = 40 sinθ cosφ / r
term 2: (1/(r sinθ)) ∂(sinθ · 1/r)/∂θ = cosθ/(r² sinθ)
term 3: 0

∇·J = 40 sinθ cosφ / r + cosθ/(r² sinθ)

∂ρ_v/∂t = −∇·J = −[ 40 sinθ cosφ / r + cosθ/(r² sinθ) ]  A/m³ (C/m³ per second)
```

**Answers:** (i) average J_r = 10 A/m²; (ii) ∂ρ_v/∂t = −(40 sinθ cosφ)/r − cosθ/(r² sinθ) A/m³.

## Check yourself

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

1. How is the average of J_r over a surface calculated?
   A. ∫ J_r dS only
   B. J_r at the centre
   C. ∫ J_r dS divided by ∫ dS
   D. ∫ J_r dv
   Answer: C. Average = flux integral divided by area. (Key point: Average J_r)

2. The average J_r is:
   A. 5π A/m²
   B. 10 A/m²
   C. 20 A/m²
   D. π/2 A/m²
   Answer: B. ∫J_r dS = 5π and area = π/2 gives 10. (Key point: Average value)

3. ∂ρ_v/∂t is found from:
   A. The continuity equation, ∂ρ_v/∂t = −∇·J
   B. Ohm's law
   C. ∇×J
   D. Poisson's equation
   Answer: A. ∇·J = −∂ρ_v/∂t. (Key point: Relation for ∂ρ_v/∂t)

4. The a_θ component 1/r contributes which term to ∇·J?
   A. 40 sinθ cosφ/r
   B. 0
   C. 1/r²
   D. cosθ/(r² sinθ)
   Answer: D. (1/(r sinθ)) ∂(sinθ/r)/∂θ = cosθ/(r² sinθ). (Key point: Divergence term)

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