Question 7 of 7: Field and Potential of a Charged Spherical Conductor
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Phys-A3, Electromagnetics — National Exam, December 2019. 3-hour closed-book exam (Casio or Sharp approved calculators only); any FIVE of the seven questions constitute a complete paper and only the first five as they appear in a candidate's answer book are marked, each of equal value, with full justification required for marks. All seven printed questions are solved below as a complete study resource.
Reference texts: Sadiku, Elements of Electromagnetics (7th ed.) — plane waves and boundaries, transmission lines, waveguides, Gauss's and Faraday's laws; Hayt & Buck, Engineering Electromagnetics (9th ed.) — transmission-line transients and bounce diagrams; Pozar, Microwave Engineering (4th ed.) — transmission-line theory, rectangular waveguide cutoff; Balanis, Antenna Theory (4th ed.) — wave polarization and radiation fundamentals.
Question 7: Field and Potential of a Charged Spherical Conductor (20 marks)
Given. Conducting sphere, radius $a=1$ cm, centred at the origin, uniform surface charge density $\rho_s=20\ \mu\text{C/m}^2$, surrounded by free space; a $q=1$ nC test charge moved from $(5,0,0)$ cm to $(3,0,0)$ cm.
Charged sphere of radius 1 cm at the origin, with the two field points at $r=3$ cm and $r=5$ cm along the $x$-axis, both outside the sphere.
Find. Total charge $Q$; $\mathbf E(r)$ for $r\ >\ a$; the work to move the test charge from 5 cm to 3 cm; the potential difference between those two points.
Approach. Total charge from $\rho_s$ times the sphere's surface area; apply Gauss's law with a spherical Gaussian surface for $\mathbf E$; use $V(r)=Q/(4\pi\varepsilon_0 r)$ (spherically symmetric charge behaves like a point charge outside itself) to get the work and potential difference.
Part (a) — Total charge.
$$Q=\rho_s\,(4\pi a^2)=(20\times10^{-6})\,4\pi(0.01)^2=\boxed{25.1\ \text{nC}}.$$
Part (b) — Electric field for $r\ >\ a$ (Gauss's law). By spherical symmetry, choose a concentric spherical Gaussian surface of radius $r\ >\ a$; $\mathbf E=E_r\hat{\mathbf a}_r$ is constant in magnitude over it:
$$\oint\mathbf E\cdot d\mathbf S = E_r(4\pi r^2)=\frac{Q_{enc}}{\varepsilon_0}=\frac{Q}{\varepsilon_0}\ \Rightarrow\ \boxed{\mathbf E(r)=\frac{Q}{4\pi\varepsilon_0 r^2}\,\hat{\mathbf a}_r = \frac{225.9}{r^2}\,\hat{\mathbf a}_r\ \text{V/m}\ (r\text{ in m})}.$$
(All the enclosed charge sits at $r=a$, exactly as for an equivalent point charge $Q$ at the origin, for any $r\ >\ a$.)
Part (c) — Potential at the two field points (needed for the work calculation). Using $V(r)=Q/(4\pi\varepsilon_0 r)$ (reference $V(\infty)=0$):
$$V(5\ \text{cm})=\frac{225.9}{0.05}=4518\ \text{V},\qquad V(3\ \text{cm})=\frac{225.9}{0.03}=7530\ \text{V}.$$
Work to move the test charge from 5 cm to 3 cm. The work done BY AN EXTERNAL AGENT moving charge $q$ from A to B equals $q[V(B)-V(A)]$:
$$W=q\big[V(3\text{ cm})-V(5\text{ cm})\big]=(1\times10^{-9})(7530-4518)=\boxed{3.01\ \mu\text{J}}.$$
This is positive: since both the sphere and the test charge are positive, moving the test charge CLOSER to the sphere means working against the repulsive field, exactly as the sign confirms.
Part (d) — Potential difference between the two points.
$$\boxed{V_{3\text{cm}}-V_{5\text{cm}} = 7530-4518 = 3012\ \text{V}}$$
(the point at 3 cm sits 3012 V higher than the point at 5 cm, consistent with being nearer the positive sphere) — this is exactly $W/q$ from the previous step, as it must be.
