Electric Field Above a Charged Ring with a Central Point Charge: 04-BS-9, December 2015, Q1

Question 1: Electric Field Above a Charged Ring with a Central Point Charge (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
Ring radius $R$	$0.5\times10^{-10}$ m
Ring charge (uniform on circumference)	$-e$
Point charge at centre	$+e$
Observation height on axis $z$	$0.5\times10^{-10}$ m above centre

Find. The magnitude and direction of $\vec E$ at the point on the ring's axis, height $z$ above the centre.

Charged ring ($-e$, radius $R$) with point charge $+e$ at its centre; field evaluated on the axis at height $z=R$ above the centre.

Approach. Superpose the axial field of the uniformly charged ring with the field of the point charge at the centre — both lie along the same axis, so they combine algebraically rather than as a full vector sum.

Field of the centre point charge at height $z$. Directed straight up (away from the positive charge): $$E_{\text{point}}=\frac{ke}{z^2}=\frac{(8.9918\times10^9)(1.6\times10^{-19})}{(0.5\times10^{-10})^2}$$ $$E_{\text{point}}=\boxed{5.755\times10^{11}\ \text{V/m, upward}}$$
Axial field of the charged ring. A ring of total charge $Q$ and radius $R$ produces, on its own axis at height $z$, a field $E_{\text{ring}}(z)=\dfrac{kQz}{(R^2+z^2)^{3/2}}$, directed away from the ring if $Q>0$. Here $Q=-e$ and, since this problem sets $z=R$, the denominator simplifies to $(2R^2)^{3/2}=2\sqrt2\,R^2$: $$E_{\text{ring}}=\frac{k(-e)R}{2\sqrt2\,R^2}=-\frac{ke}{2\sqrt2\,R^2}=-\frac{5.755\times10^{11}}{2.828}$$ $$E_{\text{ring}}=\boxed{-2.035\times10^{11}\ \text{V/m (i.e. }2.035\times10^{11}\text{ V/m toward the ring, downward)}}$$
Superpose the two axial contributions. Both act along the same vertical line, point up positive: $$E_{\text{net}}=E_{\text{point}}+E_{\text{ring}}=5.755\times10^{11}-2.035\times10^{11}$$ $$E_{\text{net}}=\boxed{3.720\times10^{11}\ \text{V/m, directed vertically upward}}$$

Quantity	Result
$E_{\text{point}}$ (centre charge alone)	$5.755\times10^{11}$ V/m, up
$E_{\text{ring}}$ (ring alone)	$2.035\times10^{11}$ V/m, down
$E_{\text{net}}$ at P	$3.720\times10^{11}$ V/m, vertically upward

Concept & Theory

Topic. Superposition of a point-charge field with the axial field of a uniformly charged ring.

Key relations. Axial field of a charged ring, $E_{\text{ring}}(z)=\dfrac{kQz}{(R^2+z^2)^{3/2}}$; point-charge field $E=\dfrac{kq}{r^2}$.

Why this works. By the ring's rotational symmetry, every off-axis component of its field cancels around the circumference, leaving a purely axial resultant — exactly like the point charge's own field at a point directly above it. Because both contributions lie on the same line, they add as signed scalars rather than requiring a full vector sum.

Common pitfall. Treating the ring's on-axis field as a simple $kQ/z^2$ point-charge formula (valid only far from the ring, $z\gg R$) instead of the exact $kQz/(R^2+z^2)^{3/2}$ expression, which is essential here since $z=R$ places the point well within the near-field region; and forgetting the ring's charge is negative, so its contribution opposes rather than adds to the centre charge's field.

Source. Sadiku, Elements of Electromagnetics, Ch. 4 (Coulomb's law, fields of continuous charge distributions — ring/line charge).

Question 1 of 8: Electric Field Above a Charged Ring with a Central Point Charge

Question 1: Electric Field Above a Charged Ring with a Central Point Charge (20 marks)