Parallel-Plate Capacitor and a Levitated Charged Sphere: 04-BS-9, Undated paper, Q6

Question 6: Parallel-Plate Capacitor and a Levitated Charged Sphere (15 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, area	$2$ m, $A=4\ \text{m}^2$
Plate separation	$d=1$ cm $=0.01$ m (plates at $z=0,\,z=1$ cm)
Potential difference	$V=100$ V
Sphere mass, position	$m=1$ g, at $z=0.5$ cm (midway)

Find. (e) $C$. (f) $Q$ on each plate. (g) stored energy $U$. (h) equilibrium charge magnitude on the sphere.

The plates are horizontal sheets; the field between them is vertical (along z) and uniform. A negatively-charged sphere sits at the midplane, held up by the electric force against gravity.

Approach. Use the parallel-plate formulas $C=\varepsilon_0A/d$, $Q=CV$, $U=\tfrac12CV^2$; for the levitated sphere, balance the electric force $qE$ against gravity $mg$, using the same uniform field $E=V/d$ (independent of the sphere's exact height, since the field is uniform between ideal plates).

Part (e) — capacitance. $$C=\frac{\varepsilon_0A}{d}=\frac{(8.854\times10^{-12})(4)}{0.01}$$ $$\boxed{C=3.542\ \text{nF}}$$
Part (f) — charge on each plate. $$Q=CV=(3.542\times10^{-9})(100)$$ $$\boxed{Q=354.2\ \text{nC (one plate }+,\text{ the other }-)}$$
Part (g) — stored energy. $$U=\tfrac12CV^2=\tfrac12(3.542\times10^{-9})(100)^2$$ $$\boxed{U=17.71\ \mu\text{J}}$$
Part (h) — equilibrium charge on the sphere. The field is uniform, so its value is the same at $z=0.5$ cm as anywhere else between the plates: $$E=\frac{V}{d}=\frac{100}{0.01}=1.000\times10^4\ \text{V/m}$$ Equilibrium requires the (upward) electric force to balance gravity, $|Q_s|E=mg$: $$|Q_s|=\frac{mg}{E}=\frac{(0.001)(9.81)}{1.000\times10^4}$$ $$\boxed{|Q_s|=0.981\ \mu\text{C}}$$ Since the sphere carries negative charge, the electric force ($\vec F=Q_s\vec E$, opposite to $\vec E$ for $Q_s<0$) points upward only if $\vec E$ itself points downward — i.e. the top plate ($z=1$ cm) must be the positive one.

Quantity	Result
Capacitance $C$	$3.542$ nF
Charge $Q$	$354.2$ nC
Energy $U$	$17.71\ \mu$J
Sphere charge $\|Q_s\|$	$0.981\ \mu$C

Concept & Theory

Topic. Parallel-plate capacitance and electromechanical equilibrium in a uniform field (an electrostatic analogue of the Millikan oil-drop experiment).

Key relations. $C=\varepsilon_0A/d$; $Q=CV$; $U=\tfrac12CV^2=\tfrac12QV$; uniform-field magnitude $E=V/d$; equilibrium $|q|E=mg$.

Why this works. An ideal parallel-plate capacitor produces a field that is uniform in magnitude and direction everywhere between the plates (edge effects aside), so a charged object suspended anywhere in that gap sees exactly $E=V/d$ regardless of its precise height — the "at $z=0.5$ cm" detail is really just confirming the sphere is between the plates, not changing the field value used.

Common pitfall. Re-deriving $E$ from Coulomb's law at the sphere's specific height instead of recognising the uniform-field shortcut; and mismatching the sign of the required field direction against the stated sign of the sphere's charge.

Source. Sadiku, Elements of Electromagnetics, Ch. 5 (capacitance); Ch. 4 (force in a uniform field).

Question 6 of 7: Parallel-Plate Capacitor and a Levitated Charged Sphere

Question 6: Parallel-Plate Capacitor and a Levitated Charged Sphere (15 marks)