---
title: "Half the energy within radius √(ab)"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:cylindrical-capacitor-half-energy-radius"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/cylindrical-capacitor-half-energy-radius
site: BCT Engineering
---

# Half the energy within radius √(ab)

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2074 Chaitra, 5 marks; 2071 Shrawan, 5 marks; 2069 Asar, 5 marks (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/cylindrical-capacitor-half-energy-radius
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

A cylindrical capacitor has radii 'a' and 'b'. Show that half the energy stored lies within the cylinder whose radius is r = √(ab).

## Short answer

The energy stored in a cylindrical capacitor from radius a to r is U(r) = (λ²L/4πε₀) ln(r/a). Total energy is U = (λ²L/4πε₀) ln(b/a). Setting U(r) = U/2 yields ln(r/a) = ½ ln(b/a), so r = √(ab).

## Answer

**Setup:** a cylindrical capacitor of length L has an inner conductor of radius *a* and an outer cylinder of radius *b*. The charge per unit length is λ. We show that half the stored energy lies within radius r = √(ab).

```text
   outer cylinder (radius b)
   +--------------------+
   |   +------------+   |
   |   |  inner a   |   |
   |   +------------+   |
   +--------------------+
   field region: a < r < b
```
#### Derivation
1. By Gauss' law with a coaxial cylinder of radius r (a < r < b): E (2πrL) = λL/ε₀, so E = λ / (2πε₀ r). The field is zero outside.
2. Energy density: u = ½ ε₀ E² = λ² / (8π² ε₀ r²).
3. Take a thin cylindrical shell of radius r, thickness dr, volume dV = 2πrL dr. The energy in it is
 dU = u dV = [λ² / (8π²ε₀r²)] · 2πrL dr = (λ² L / 4πε₀) dr / r.
4. Total energy stored (a to b):
 U = (λ²L / 4πε₀) ln(b/a).
5. Energy stored between a and some radius r:
 U(r) = (λ²L / 4πε₀) ln(r/a).
6. Require U(r) = U/2:
 ln(r/a) = ½ ln(b/a) = ln √(b/a)
 r/a = √(b/a)

 **r = √(ab)**

So half the energy is stored within the cylinder of radius √(ab) (the geometric mean of a and b), and half outside it. Check with the capacitor formulas: U = q²/2C with C = 2πε₀L/ln(b/a), which gives the same total.

## Check yourself

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

1. Using Gauss's law, what is the field at radius r between the cylinders?
   A. E = λ/(4πε0 r^2)
   B. E = λ r/(2ε0)
   C. E = λ/(2πε0 r)
   D. E = λ/(2πε0 r^2)
   Answer: C. E(2πrL) = λL/ε0 gives E ∝ 1/r. (Key point: Field between cylinders)

2. Why does the energy in a thin shell take the form dU ∝ dr/r?
   A. Energy density is constant and the volume ∝ dr
   B. Energy density ∝ 1/r^2 and shell volume = 2πrL dr, so the product ∝ dr/r
   C. Energy density ∝ r and the volume ∝ 1/r^2
   D. Energy density ∝ 1/r and volume ∝ dr
   Answer: B. u = ½ε0E^2 ∝ 1/r^2, multiplied by dV ∝ r dr gives dr/r, whose integral is ln. (Key point: Energy in a shell)

3. Setting the energy between a and r equal to half the total gives ln(r/a) = ½ ln(b/a). What is r?
   A. r = √(ab), the geometric mean
   B. r = (a + b)/2, the arithmetic mean
   C. r = b/√2
   D. r = 2ab/(a + b)
   Answer: A. ln(r/a) = ln√(b/a) gives r = a√(b/a) = √(ab). (Key point: Half-energy condition)

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