---
title: "Field inside uniformly charged sphere"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:sphere-uniform-charge-field-inside"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/sphere-uniform-charge-field-inside
site: BCT Engineering
---

# Field inside uniformly charged sphere

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2073 Chaitra, 5 marks
- Video: [Electric field of a uniformly charged solid sphere using Gauss' Law (how to use Gauss' Law)](https://www.youtube.com/watch?v=VgGLuTYa90M) — Zak's Lab
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/sphere-uniform-charge-field-inside
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

Charge of uniform volume density ρ = 3.2 μC/m³ fills a non conducting solid sphere of radius 5 cm. What is the magnitude of the electric field 3 cm from the sphere's center?

## Short answer

Using Gauss’s law inside the sphere, E = ρr/(3ε₀) = (3.2×10⁻⁶ × 0.03)/(3 × 8.854×10⁻¹²) ≈ 3.6×10³ N/C radially outward.

## Answer

Given: ρ = 3.2 μC/m³ = 3.2 × 10⁻⁶ C/m³, sphere radius R = 5 cm, field point r = 3 cm (inside, r < R).

#### Method (Gauss's law)
Take a concentric spherical Gaussian surface of radius r < R. Only the charge inside it counts:

q_enc = ρ (4/3)πr³

By symmetry E is radial and constant on the surface:

E (4πr²) = q_enc/ε0 = ρ (4/3)πr³ / ε0

**E = ρ r / (3ε0)**

#### Substitution
r = 3 cm = 0.03 m, ε0 = 8.854 × 10⁻¹² C²/N·m²

E = (3.2 × 10⁻⁶ × 0.03) / (3 × 8.854 × 10⁻¹²)
E = 9.6 × 10⁻⁸ / 2.656 × 10⁻¹¹
E = 3.61 × 10³ N/C

**Answer: E ≈ 3.6 × 10³ N/C (V/m), directed radially outward.**

## Check yourself

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

1. For a point inside a uniformly charged solid sphere (r < R), which charge counts in Gauss's law?
   A. The total sphere charge ρ(4/3)πR^3
   B. Only the charge inside the Gaussian radius: ρ(4/3)πr^3
   C. Zero, as in a conductor
   D. ρ 4πr^2
   Answer: B. Charge outside the Gaussian surface does not contribute for a spherical distribution. (Key point: Enclosed charge)

2. What is the field inside at radius r?
   A. E = ρr/(3ε0), growing linearly with r
   B. E = ρr/ε0
   C. E = ρR^3/(3ε0 r^2)
   D. E = ρ/(3ε0 r)
   Answer: A. E(4πr^2) = ρ(4/3)πr^3/ε0. (Key point: Formula)

3. What is E at r = 3 cm for ρ = 3.2 μC/m^3?
   A. About 6.0 x 10^3 N/C (using R = 5 cm)
   B. About 1.1 x 10^4 N/C (omitting the 3)
   C. About 3.6 x 10^-3 N/C
   D. About 3.6 x 10^3 N/C, radially outward
   Answer: D. E = (3.2 x 10^-6 x 0.03)/(3 x 8.854 x 10^-12) ≈ 3.6 x 10^3 N/C. (Key point: Value)

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