Capacitance, Charge, Energy and Force of a Silicon-Filled Parallel-Plate Capacitor: 04-BS-9, December 2019, Q6

Question 6: Capacitance, Charge, Energy and Force of a Silicon-Filled 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 side (square)	$2$ m $\Rightarrow A=4$ m$^2$
Plate separation $d$	$1$ mm $=1\times10^{-3}$ m
Dielectric	silicon, $\varepsilon_r=12$
Potential difference $V$	$100$ V

Find. (a) capacitance $C$. (b) charge $Q$ on each plate. (c) stored energy $U$. (d) attractive force $F$ between the plates.

Side view: two 2m x 2m square plates separated by a 1mm silicon dielectric layer, potential difference 100V.

Approach. Use the parallel-plate capacitance formula with the dielectric's relative permittivity, then $Q=CV$, $U=\tfrac12CV^2$, and the constant-voltage force expression $F=\tfrac12\varepsilon_0\varepsilon_rA\,V^2/d^2$.

Part (a) — capacitance. $$C=\frac{\varepsilon_0\varepsilon_rA}{d}=\frac{(8.854\times10^{-12})(12)(4)}{1\times10^{-3}}$$ $$\boxed{C=4.250\times10^{-7}\ \text{F}=0.4250\ \mu\text{F}}$$
Part (b) — charge. $$Q=CV=(4.250\times10^{-7})(100)$$ $$\boxed{Q=4.250\times10^{-5}\ \text{C}=42.50\ \mu\text{C}}$$
Part (c) — stored energy. $$U=\tfrac12CV^2=\tfrac12(4.250\times10^{-7})(100)^2$$ $$\boxed{U=2.125\times10^{-3}\ \text{J}=2.125\ \text{mJ}}$$
Part (d) — force between the plates. Holding $V$ fixed, $F=\tfrac12\varepsilon_0\varepsilon_rA\,V^2/d^2$ (equivalently $Q^2/(2\varepsilon_0\varepsilon_rA)$, same numeric result): $$F=\tfrac12(8.854\times10^{-12})(12)(4)\frac{(100)^2}{(1\times10^{-3})^2}$$ $$\boxed{F=2.125\ \text{N, attractive (plates pulled together)}}$$

Quantity	Result
Capacitance $C$	$0.4250\ \mu$F
Charge $Q$	$42.50\ \mu$C
Stored energy $U$	$2.125$ mJ
Force between plates $F$	$2.125$ N, attractive

Concept & Theory

Topic. Parallel-plate capacitance, stored energy, and the electrostatic force between charged plates.

Key relations. $C=\varepsilon_0\varepsilon_rA/d$; $Q=CV$; $U=\tfrac12CV^2=\tfrac12QV=\tfrac12Q^2/C$; $F=Q^2/(2\varepsilon_0\varepsilon_rA)=\tfrac12\varepsilon_0\varepsilon_rAV^2/d^2$.

Why this works. With a uniform dielectric completely filling the gap, the parallel-plate formula's vacuum permittivity is simply scaled by $\varepsilon_r$; the plate force follows from the energy stored per unit gap length, and is always attractive regardless of the sign convention used to derive it (opposite charges on the two plates).

Common pitfall. Forgetting the factor of $\tfrac12$ in either the energy or the force formula, or using the vacuum $\varepsilon_0$ alone and forgetting to multiply by $\varepsilon_r=12$ for the silicon dielectric.

Source. Sadiku, Elements of Electromagnetics, Ch. 5–6 (capacitance, energy in the electrostatic field, force on dielectric/conducting plates).

Question 6 of 7: Capacitance, Charge, Energy and Force of a Silicon-Filled Parallel-Plate Capacitor

Question 6: Capacitance, Charge, Energy and Force of a Silicon-Filled Parallel-Plate Capacitor (20 marks)