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04-BS-9 · May 2016

Question 5 of 8: Charge Distribution Producing a Given Piecewise Radial Field

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2016 — 04-BS-9 Basic Electromagnetics. Three-hour, closed-book exam (approved Casio/Sharp calculator only). Aids given: $\varepsilon_0=8.85\times10^{-12}$ F/m, $\mu_0=4\pi\times10^{-7}$ H/m, $e=1.6\times10^{-19}$ C. Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness.

Reference texts: Sadiku, Elements of Electromagnetics / Hayt & Buck, Engineering Electromagnetics — Coulomb's law and superposition, Gauss's law and divergence, Biot–Coulomb law and magnetic field of current loops and coaxial conductors, Faraday's law, Lorentz force, capacitance and field energy, DC circuit power distribution; Young & Freedman, University Physics with Modern Physics — refraction geometry.

Question 5: Charge Distribution Producing a Given Piecewise Radial Field (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

QuantityValue
Coefficient $A$$1.44\times10^{21}$ V/m$^2$
Boundary radius $R$$10^{-10}$ m
Field, $r<R$$\vec E=A(x,y,z)=A\,\vec r$
Field, $r>R$$\vec E=A(R/r)^3\,\vec r=AR^3\,\hat r/r^2$

Find. The volume charge density $\rho(r)$ (and total charge) producing this field.

uniform ρ, radius R E∝r (inside) E∝1/r² (outside)
Field grows linearly with $r$ inside radius $R$ and falls as $1/r^2$ outside — the signature of a uniformly charged sphere, continuous at $r=R$.

Approach. Apply the divergence aid directly: $\rho=\varepsilon_0\,\text{div}\,\vec E$ in each region (Gauss's law in differential form). Both branches, evaluated at $r=R$, must agree in magnitude for the field to be physical (continuous) — use that as a consistency check, then integrate the inside density over the sphere's volume for the total charge.

  1. Divergence and charge density inside ($r<R$), $\vec E=A(x,y,z)$. $$\text{div}\,\vec E=\frac{\partial(Ax)}{\partial x}+\frac{\partial(Ay)}{\partial y}+\frac{\partial(Az)}{\partial z}=A+A+A=3A$$ $$\rho=\varepsilon_0(3A)=(8.85\times10^{-12})(3)(1.44\times10^{21})$$ $$\rho=\boxed{3.823\times10^{10}\ \text{C/m}^3\ \text{(uniform)}}$$
  2. Divergence outside ($r>R$), $\vec E=AR^3\,\vec r/r^3$. This is the standard $\hat r/r^2$ form (times the constant $AR^3$), whose divergence is zero everywhere except the origin (well outside this region): $$\text{div}\,\vec E=0\ \Rightarrow\ \rho=0\ \text{for}\ r>R$$ So all the charge is confined within radius $R$ — no charge (volume or surface) exists outside the sphere.
  3. Continuity check at $r=R$ (confirms a single uniform sphere, no surface layer). Inside at $r\to R$: $E=AR$. Outside at $r\to R$: $E=A(R/R)^3R=AR$. The two branches agree exactly, so the field has no jump at $r=R$ and therefore no surface charge sits there either — consistent with $\rho=0$ outside.
  4. Total charge, from the uniform volume density. $$Q=\rho\cdot\frac43\pi R^3=(3.823\times10^{10})\cdot\frac43\pi(10^{-10})^3$$ $$Q=\boxed{1.601\times10^{-19}\ \text{C}\ \approx\ +e}$$
QuantityResult
Volume charge density $\rho$ ($r<R$)$3.823\times10^{10}$ C/m$^3$, uniform
Charge density ($r>R$)$0$ (no charge outside $R$)
Total charge $Q$$1.601\times10^{-19}$ C $\approx+e$
that mark is the lower half of the curly brace grouping the two branches. The field as printed, $\vec E=A(R/r)^3(x,y,z)$ for $r>R$, is continuous at $r=R$ (Step 3), which is exactly what the solution above uses.