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04-BS-9 · December 2018

Question 4 of 8: Minimum Midpoint Charge to Cancel Mutual Repulsion of Two Point Charges

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

Notes on this paper

National Exams — December 2018 — 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 — Gauss's law and boundary conditions for graded (spatially-varying) dielectrics, magnetic-circuit (reluctance) analysis of a partially-filled long solenoid, superposition of infinite current sheets and Ampère's law, Coulomb's-law equilibrium of collinear point charges, Faraday's law and motional EMF, the Biot–Savart law for arc segments, and the Lorentz force in a velocity selector; Young & Freedman, University Physics with Modern Physics — Snell's law and plane-wave reflection/refraction at a dielectric interface.

Question 4: Minimum Midpoint Charge to Cancel Mutual Repulsion of Two Point Charges (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
Two point charges $+Q$, separation $R$$R=1\times10^{-10}$ m
Charge $-q$located exactly at the midpoint

Find. The minimum $|q|$ such that the net force on each $+Q$ charge is zero.

+Q −q +Q R/2 R/2
Three collinear charges: +Q, −q, +Q, with −q exactly at the midpoint (distance R/2 from each +Q).

Approach. Set the Coulomb attraction from $-q$ on one $+Q$ equal to the Coulomb repulsion from the other $+Q$, and solve for $q$; the common separation dependence cancels.

  1. Distances. Full separation between the two $+Q$ charges is $R$; each $+Q$ is a distance $r=R/2$ from $-q$ at the midpoint.
  2. Force expressions. Repulsion between the two $+Q$ charges: $$F_{QQ}=\frac{kQ^2}{R^2}$$ Attraction from $-q$ on one $+Q$: $$F_{Qq}=\frac{kQq}{r^2}=\frac{kQq}{(R/2)^2}=\frac{4kQq}{R^2}$$
  3. Equilibrium condition. Setting $F_{Qq}=F_{QQ}$ (net force zero) and cancelling the common factor $kQ/R^2$: $$\frac{4kQq}{R^2}=\frac{kQ^2}{R^2}\ \Rightarrow\ 4q=Q$$ $$\boxed{q_{\min}=\frac{Q}{4}}$$ — independent of $R$, so the given $1\times10^{-10}$ m separation does not enter the final ratio.
QuantityResult
$q_{\min}$$Q/4$ (one quarter of $Q$)
Dependence on separation $R$none — ratio only