---
title: "Energy stored for given potential field"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:energy-stored-potential-field"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/energy-stored-potential-field
site: BCT Engineering
---

# Energy stored for given potential field

Electromagnetism · Chapter 2: Electric field

- Asked in: 2076 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/energy-stored-potential-field
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Find the energy stored in free space for the region 2 mm < r < 3 mm, 0 < θ < 90°, 0 < φ < 90°, given the potential field V = (a) 200/r V and (b) 300/r² cosθ V.

## Short answer

(a) With V=200/r, E=(200/r²)a_r and W=½ε₀∫E²dv gives 46.4 μJ. (b) For V=300/r² cosθ, evaluating E² and integrating over the specified region yields W ≈ 36.7 J.

## Answer

**Given:** Region 2 mm < r < 3 mm, 0 < θ < 90°, 0 < φ < 90°, free space (ε₀ = 8.854 × 10⁻¹² F/m). Energy from the field: W_E = (1/2) ε₀ ∫ |E|² dv, with E = −∇V and dv = r² sinθ dr dθ dφ.

#### (a) V = 200/r

```text
E = −∇V = −(dV/dr) a_r = (200/r²) a_r,     E² = 4 × 10⁴ / r⁴

W_E = (1/2) ε₀ ∫₀^(π/2) ∫₀^(π/2) ∫_(0.002)^(0.003) (4×10⁴ / r⁴) r² sinθ dr dθ dφ
    = (1/2) ε₀ (4×10⁴) [ ∫ r⁻² dr ] [ ∫ sinθ dθ ] [ ∫ dφ ]
```

```text
∫ r⁻² dr = 1/0.002 − 1/0.003 = 500 − 333.33 = 166.67
∫₀^(π/2) sinθ dθ = 1,    ∫₀^(π/2) dφ = π/2 = 1.5708

W_E = 0.5 (8.854×10⁻¹²)(4×10⁴)(166.67)(1)(1.5708)
    = 4.636 × 10⁻⁵ J
```

**W_E = 46.4 μJ**

#### (b) V = (300/r²) cosθ

```text
E_r = −∂V/∂r = 600 cosθ / r³
E_θ = −(1/r) ∂V/∂θ = 300 sinθ / r³
E_φ = 0

E² = (3.6×10⁵ cos²θ + 9×10⁴ sin²θ) / r⁶
```

```text
W_E = (1/2) ε₀ ∫ ∫ ∫ (3.6×10⁵ cos²θ + 9×10⁴ sin²θ) r⁻⁶ r² sinθ dr dθ dφ
```

Radial part: ∫ r⁻⁴ dr = (1/3)(r₁⁻³ − r₂⁻³) = (1/3)(1.25×10⁸ − 3.704×10⁷) = 2.932 × 10⁷

Angular part (θ): ∫₀^(π/2) (3.6×10⁵ cos²θ + 9×10⁴ sin²θ) sinθ dθ = 3.6×10⁵ (1/3) + 9×10⁴ (2/3) = 1.2×10⁵ + 0.6×10⁵ = 1.8×10⁵

Angular part (φ): π/2.

```text
W_E = 0.5 (8.854×10⁻¹²)(1.8×10⁵)(1.5708)(2.932×10⁷)
    = 36.7 J
```

**Answer: (a) W_E = 46.4 μJ; (b) W_E ≈ 36.7 J.**

(The large value in (b) comes from the extremely strong field of this 1/r³ distribution at millimetre distances; such a potential would be unrealistic physically, but the working above follows the given V.)

## Check yourself

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

1. Which expression is used to find the energy stored from the potential field?
   A. W_E = ε₀ ∫ V dv
   B. W_E = (1/2) ∫ V² dv
   C. W_E = ∫ E dv
   D. W_E = (1/2) ε₀ ∫ |E|² dv with E = -∇V
   Answer: D. Energy is (1/2) ε₀ E² integrated over volume, with E found from -∇V. (Key point: Energy formula)

2. In the spherical region, the volume element is:
   A. dv = r dr dθ dφ
   B. dv = r sinθ dr dθ dφ
   C. dv = r² sinθ dr dθ dφ
   D. dv = sinθ dr dθ dφ
   Answer: C. dv = r² sinθ dr dθ dφ in spherical coordinates. (Key point: Volume element)

3. For V = 200/r, the field E is:
   A. (200/r) a_r
   B. (200/r²) a_r
   C. -(200/r²) a_r
   D. (200/r³) a_θ
   Answer: B. E = -dV/dr a_r = (200/r²) a_r, so E² = 4×10⁴/r⁴. (Key point: Part (a) field)

4. For V = (300/r²) cosθ, what are the field components?
   A. E_r = 600 cosθ/r³, E_θ = 300 sinθ/r³
   B. E_r = 300 cosθ/r³, E_θ = 600 sinθ/r³
   C. E_r = 600 cosθ/r³, E_θ = 0
   D. E_r = 300 cosθ/r², E_θ = 300 sinθ/r²
   Answer: A. E_r = -∂V/∂r = 600cosθ/r³ and E_θ = -(1/r)∂V/∂θ = 300 sinθ/r³. (Key point: Part (b) field components)

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