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04-BS-9 · May 2014

Question 7 of 8: Electrostatic Energy of a Point Charge and Four Corner Charges

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

Notes on this paper

National Exams — May 2014 — 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 — electrostatics, Gauss's law, capacitance, magnetostatics, Faraday's law; Young & Freedman, University Physics with Modern Physics — induced EMF and AC quantities.

Question 7: Electrostatic Energy of a Point Charge and Four Corner Charges (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
Centre charge$+4e=6.4\times10^{-19}$ C
Corner charges (4)$-e=-1.6\times10^{-19}$ C each
Square side $a$$1\times10^{-10}$ m

Find. The total electrostatic (assembly) energy of the five-charge system.

−e −e −e −e +4e side a = 10⁻¹⁰ m
Square of side a, −e at each corner, +4e at the centre.

Approach. Sum the pairwise Coulomb energies $U_{ij}=kq_iq_j/r_{ij}$ over all $\binom{5}{2}=10$ pairs: 4 centre–corner pairs (distance $a/\sqrt2$), 4 adjacent corner–corner pairs (distance $a$), and 2 diagonal corner–corner pairs (distance $a\sqrt2$).

  1. Centre–corner pairs (4, attractive). $$U_{\text{cc}}=4\cdot\frac{k(4e)(-e)}{a/\sqrt2}=-16\sqrt2\,\frac{ke^2}{a}$$
  2. Corner–corner pairs (6, repulsive: 4 adjacent + 2 diagonal). $$U_{\text{adj}}=4\cdot\frac{k(-e)(-e)}{a}=4\,\frac{ke^2}{a},\qquad U_{\text{diag}}=2\cdot\frac{k(-e)(-e)}{a\sqrt2}=\sqrt2\,\frac{ke^2}{a}$$
  3. Total. With $k=1/4\pi\varepsilon_0=8.988\times10^9\ \text{N}\cdot\text{m}^2/\text{C}^2$ and $ke^2/a=2.301\times10^{-18}$ J: $$U=(-16\sqrt2+4+\sqrt2)\frac{ke^2}{a}=(4-15\sqrt2)\frac{ke^2}{a}$$ $$U=(-17.213)(2.301\times10^{-18}\ \text{J})$$ $$U=\boxed{-3.96\times10^{-17}\ \text{J}=-247.6\ \text{eV}}$$
QuantityResult
Centre–corner energy (4 pairs)$-16\sqrt2\,ke^2/a$
Corner–corner energy (6 pairs)$(4+\sqrt2)\,ke^2/a$
Total $U$$-3.96\times10^{-17}$ J ($-247.6$ eV)