---
title: "Field of long charged rod, equatorial"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:charged-rod-field-potential-continuous"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/charged-rod-field-potential-continuous
site: BCT Engineering
---

# Field of long charged rod, equatorial

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2070 Chaitra, 5 marks
- Video: [Class 12 Physics | #30 Electric Field Strength due to a Uniformly Charged Rod at a General Point](https://www.youtube.com/watch?v=a4tazHiu3b0) — Physics Galaxy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/charged-rod-field-potential-continuous
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

Give a general method to calculate electric field and potential due to continuous charge distribution. Using your method, calculate electric field at an equatorial distance y due to a long charged rod having linear charge density λ.

## Short answer

General method: divide into dq = λ dl, treat as point charges, integrate E = ∫dE and V = ∫dV using symmetry. For a long rod, equatorial field: E = λ/(2πε₀ y) perpendicular to the rod, directed away for λ>0.

## Answer

#### General method for a continuous distribution
1. Divide the distribution into small elements dq (dq = λ dl for a line, σ dA for a surface, ρ dV for a volume).
2. Treat each dq as a point charge. Its field is dE = (1/4πε0) dq/r² along the line from dq to the point; its potential is dV = (1/4πε0) dq/r.
3. Resolve dE into components and use symmetry to cancel those that sum to zero.
4. Integrate: E = ∫dE (vector sum); V = ∫dV (scalar sum, easier). Check by E = −∇V.

#### Long charged rod, field at equatorial distance y
Rod of length 2L (very long), uniform λ, point P at perpendicular distance y from the middle.

```text
                P
               /|\
            r / |y \ r
             /  |    \
  ----------o---O-----o----------   rod
          -x    0    +x
```

1. Element dq = λ dx at x from the centre. r = √(x² + y²).
2. dE = (1/4πε0) λ dx / r². The x-components of elements at +x and −x cancel. Component along y: dE_y = dE cosθ, with cosθ = y/r.
3. Put x = y tanθ, dx = y sec²θ dθ, r = y secθ:

dE_y = (λ/4πε0) (y sec²θ dθ / y² sec²θ) cosθ = (λ/4πε0 y) cosθ dθ

4. Integrate from −θ₀ to +θ₀ (angles subtended by each end):

E = (λ/4πε0 y) · 2 sinθ₀ = λ L / (2πε0 y √(L² + y²))

5. For a long rod (L ≫ y), θ₀ → 90°, sinθ₀ → 1:

**E = λ / (2πε0 y)**, perpendicular to the rod, directed away for λ > 0.

## Check yourself

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

1. What is the first step in the general method for a continuous charge distribution?
   A. Integrate the total charge directly
   B. Divide it into small elements dq treated as point charges
   C. Apply Gauss's law first
   D. Assume all charge sits at the centre
   Answer: B. Divide the distribution into elements dq (λdl, σdA, ρdV), each acting as a point charge. (Key point: Method step)

2. In the rod problem, why do the x-components of dE vanish?
   A. The rod is uncharged on average
   B. dE has no x-component at all
   C. Elements at +x and −x give equal and opposite x-components
   D. They are cancelled by the potential
   Answer: C. Symmetric elements about the centre produce x-components that cancel, leaving only y. (Key point: Use of symmetry)

3. Which substitution simplifies the integral for the rod?
   A. x = y sinθ
   B. x = y tanθ
   C. x = y cosθ
   D. x = y/θ
   Answer: B. With x = y tanθ, dx = y sec^2θ dθ, r = y secθ, giving dE_y = (λ/4πε0 y) cosθ dθ. (Key point: Substitution)

4. For a very long rod, the field at distance y is
   A. λ/(2πε0 y^2)
   B. λ/(4πε0 y)
   C. λ/(2πε0 y)
   D. λ/(4πε0 y^2)
   Answer: C. As θ0 → 90°, sinθ0 → 1, so E = λ/(2πε0 y), falling as 1/y. (Key point: Long rod result)

## Related in Engineering Physics

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