Stored Energy in a Circular Parallel-Plate Capacitor: 04-BS-9, May 2016, Q2

Question 2: Stored Energy in a Circular Parallel-Plate Capacitor (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.

Quantity	Value
Plate radius $r$	$5$ cm $=0.05$ m
Plate separation $d$	$0.5$ mm $=5\times10^{-4}$ m
Dielectric	air, $\varepsilon_r=1$
Charge magnitude $Q$ on each plate	$10^{-12}$ C

Find. The electric energy $U$ stored in the capacitor.

Circular parallel-plate air capacitor; charge $\pm10^{-12}$C on each plate.

Approach. Compute the capacitance from the plate geometry, then apply $U=Q^2/(2C)$ for a capacitor whose charge (rather than voltage) is the given quantity.

Capacitance of the plates. $$C=\frac{\varepsilon_0 A}{d}=\frac{(8.85\times10^{-12})\pi(0.05)^2}{5\times10^{-4}}$$ $$C=\boxed{1.390\times10^{-10}\ \text{F}}$$
Stored energy from the charge. $$U=\frac{Q^2}{2C}=\frac{(10^{-12})^2}{2(1.390\times10^{-10})}$$ $$U=\boxed{3.597\times10^{-15}\ \text{J}}$$

Quantity	Result
Capacitance $C$	$1.390\times10^{-10}$ F
Stored energy $U$	$3.597\times10^{-15}$ J

Check: the charge given ($10^{-12}$ C, i.e. about $6250\,e$) is extremely small for a 5 cm plate, so the resulting energy is correspondingly tiny — this follows directly from the given data, not from an arithmetic slip.

Concept & Theory

Topic. Capacitance of a parallel-plate capacitor and energy stored in terms of charge.

Key relations. $C=\varepsilon_0\varepsilon_r A/d$; energy in terms of charge, $U=Q^2/(2C)$ (equivalent to $U=\tfrac12 CV^2=\tfrac12 QV$).

Why this works. Neglecting fringing, the plates form a uniform field region whose capacitance depends only on geometry and dielectric; once $C$ is known, the energy for a specified charge follows directly from the energy-in-a-capacitor relation written in the charge (rather than voltage) form, which is the natural choice since $Q$, not $V$, is given.

Common pitfall. Using $U=\tfrac12 CV^2$ with an assumed or mis-derived voltage instead of the direct charge form $U=Q^2/(2C)$; or forgetting that "air dielectric" means $\varepsilon_r=1$, not $\varepsilon_0$ alone needing a further factor.

Source. Sadiku, Elements of Electromagnetics, Ch. 5 (capacitance, energy in an electrostatic field).

Question 2 of 8: Stored Energy in a Circular Parallel-Plate Capacitor

Question 2: Stored Energy in a Circular Parallel-Plate Capacitor (20 marks)