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04-BS-9 · May 2015

Question 6 of 8: Electric Field in a Capacitor from Stored Energy

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

Notes on this paper

National Exams — May 2015 — 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, Biot–Savart law, magnetic field of current sheets, Faraday's law, capacitance and field energy, mutual inductance; Young & Freedman, University Physics with Modern Physics — light propagation and reflection geometry.

Question 6: Electric Field in a Capacitor from Stored Energy (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
Stored energy $U$$1$ J
Plate area $A$$10\ \text{cm}^2=10^{-3}\ \text{m}^2$
Plate separation $d$$1$ mm $=10^{-3}$ m
Dielectricair, $\varepsilon_r=1$

Find. $\vec E$ between the plates.

E d = 1 mm, A = 10 cm², U = 1 J
Parallel-plate air capacitor; uniform E field fills the gap volume between the plates.

Approach. Use the electrostatic energy density $u=\tfrac12\varepsilon_0E^2$, uniform throughout the gap volume $A\,d$, and solve for $E$ from the total stored energy $U=u\cdot(Ad)$.

  1. Gap volume. $$V=Ad=(10^{-3})(10^{-3})$$ $$V=\boxed{1.0\times10^{-6}\ \text{m}^3}$$
  2. Solve for $E$ from the stored energy. $$U=\frac12\varepsilon_0E^2V\ \Rightarrow\ E=\sqrt{\frac{2U}{\varepsilon_0V}}=\sqrt{\frac{2(1)}{(8.85\times10^{-12})(1.0\times10^{-6})}}$$ $$E=\boxed{4.754\times10^{8}\ \text{V/m}}$$, directed from the positive plate to the negative plate.
QuantityResult
Gap volume $V$$1.0\times10^{-6}$ m$^3$
$E$ between the plates$4.754\times10^{8}$ V/m
Check: this field magnitude (~475 MV/m) is far beyond the ~3 MV/m dielectric breakdown strength of air — physically the capacitor as stated could not actually hold 1 J without arcing. The question is a direct plug-into-the-energy-formula exercise; the arithmetic answer is reported as asked, with this physical caveat noted rather than silently ignored.