---
title: "Potential V=100xz/(x²+4), charge on conductor"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:potential-v-100xz-conductor-charge"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/potential-v-100xz-conductor-charge
site: BCT Engineering
---

# Potential V=100xz/(x²+4), charge on conductor

Electromagnetism · Chapter 2: Electric field

- Asked in: 2069 Chaitra, 7 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/potential-v-100xz-conductor-charge
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Given the potential field V = 100xz/(x² + 4) volts in free space: (a) Find D at the surface z = 0. (b) Show that the z = 0 surface is an equipotential surface. (c) Assume that the z = 0 surface is a conductor and find the total charge on that portion of the conductor defined by 0 < x < 2, −3 < y < 0.

## Short answer

At z=0, E=−100x/(x²+4)a_z V/m, so D = −8.854×10⁻¹⁰ x/(x²+4) a_z C/m². Since V=0 constant on z=0, it is equipotential. Integrating surface charge ρ_s = D·a_z over 0<x<2, −3<y<0 gives total charge −0.92 nC.

## Answer

Given V = 100xz/(x² + 4) volts in free space. D = ε₀E = −ε₀∇V.

#### (a) D at z = 0
```text
∂V/∂x = 100z (4 − x²)/(x²+4)²        →  0 at z = 0
∂V/∂y = 0
∂V/∂z = 100x/(x²+4)

E = −∇V = −100z(4−x²)/(x²+4)² a_x − 100x/(x²+4) a_z
At z = 0:  E = −100x/(x²+4) a_z  V/m

D(z=0) = ε₀E = −100ε₀ x/(x²+4) a_z  C/m²
       = −8.854×10⁻¹⁰ x/(x²+4) a_z
```

#### (b) z = 0 is an equipotential surface
Put z = 0 in V: V = 100x(0)/(x²+4) = 0 for every x and y. The potential is the same (0 V) everywhere on the plane, so it is an equipotential surface. This agrees with (a): E has no component tangential to the plane (E_x = E_y = 0 at z = 0).

#### (c) Charge on the conductor, 0 < x < 2, −3 < y < 0
Take the conductor as the plane z = 0 with the field in the region z > 0, so the outward normal is a_n = +a_z. Then ρ_s = D·a_n:

```text
ρ_s = −100ε₀ x/(x²+4)  C/m²

Q = ∫(y=−3→0) ∫(x=0→2) ρ_s dx dy
  = −100ε₀ (3) ∫₀² x/(x²+4) dx
  = −300ε₀ · ½ ln(x²+4) |₀²
  = −150ε₀ ln(8/4)
  = −150 × 8.854×10⁻¹² × 0.6931
  = −9.21×10⁻¹⁰ C
```

**Answer: Q ≈ −0.92 nC.**

## Check yourself

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

1. How is D found from V?
   A. D = ε₀∇V
   B. D = ε₀E = −ε₀∇V
   C. D = −∇·V
   D. D = V/ε₀
   Answer: B. E = −∇V and D = ε₀E. (Key point: Finding D)

2. At z = 0, D is:
   A. −100ε₀ x/(x²+4) a_z
   B. 100ε₀ z/(x²+4) a_x
   C. 0
   D. −100ε₀ x a_x
   Answer: A. Only ∂V/∂z survives at z = 0. (Key point: D at z = 0)

3. Why is z = 0 an equipotential surface?
   A. Because E_z = 0
   B. Because D is constant
   C. Because V is constant everywhere
   D. V = 0 for all x and y when z = 0
   Answer: D. Setting z = 0 makes V zero everywhere on the plane. (Key point: Equipotential)

4. On the conductor, ρ_s equals:
   A. the divergence of V
   B. E²
   C. D · a_n, here the normal component of D
   D. D × a_n
   Answer: C. On a conductor surface ρ_s = D_n. (Key point: Surface charge)

## Related in Electromagnetism

- [Total current through surface, H field](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Atotal-current-through-surface-h-rho) — Electromagnetism, chapter 3
- [Potential due to sheet and line charge](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Asheet-and-line-charge-potential) — Electromagnetism, chapter 2
- [Charge density and total charge from D](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Avolume-charge-density-d-field-sphere) — Electromagnetism, chapter 2
- [Total current through conical surface](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Acurrent-conical-surface-h-field) — Electromagnetism, chapter 3
- [Total current through z = 4 surface](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Acurrent-through-z-4-surface) — Electromagnetism, chapter 3
- [Energy stored for given potential field](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aenergy-stored-potential-field) — Electromagnetism, chapter 2

## Related in other subjects

- [Well defined problem, state-space search](https://www.bctengineer.com/question.md?key=semester_6%2Fartificial_intelligence%3Awell-defined-problem-state-space) — Artificial Intelligence, chapter 2
- [Numerical: free-space received power (165 W, 325 MHz)](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Afree-space-received-power-165w) — Wireless Communication, chapter 3
- [Field and electron acceleration from potential](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Apotential-field-point-electron-acceleration) — Engineering Physics, chapter 7
- [Ring on axis: potential then field](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aring-axial-potential-then-field) — Engineering Physics, chapter 7, asked 2×