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Omniguides by NOX
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Chapter 1 Solutions

These worked solutions are authored for the reviewed prompts shown here. Any added geometry or assumption is stated. Original chapter problems supplied no answer key, so these solutions are separate from the archived manuscript.

Original problem 1 · worked solution

Optional original-source extension. Prerequisites: vector addition, trigonometry and regular-pentagon geometry. This item makes no AP coverage claim.

Five identical point charges q are fixed at the vertices of a regular pentagon inscribed in a circle of radius R. Find the force on one charge and its initial direction of motion if released from rest.

Authored clarification: All five charges have the same signed charge; the circle radius is R and all positions are fixed while computing the instantaneous force.

  1. Symmetry cancels tangential components.

    Take the chosen charge at (R,0); the other four lie at angles 2πj/52\pi j/5. Pair angles j and 5−j.

  2. Use Coulomb’s law for each separation.

    dj=2Rsin⁡(πj/5),Fj=kq24R2sin⁡2(πj/5)d_j=2R\sin(\pi j/5),\qquad F_j={kq^2\over4R^2\sin^2(\pi j/5)}.

  3. Project each repulsive force onto the radial direction.

    Fr,j=Fjsin⁡(πj/5)=kq24R2csc⁡(πj/5)F_{r,j}=F_j\sin(\pi j/5)={kq^2\over4R^2}\csc(\pi j/5).

  4. Sum the four radial components.

    F⃗=kq22R2[csc⁡(π/5)+csc⁡(2π/5)]r^\vec F={kq^2\over2R^2}\left[\csc(\pi/5)+\csc(2\pi/5)\right]\hat r.
Answer

The force points radially outward. Identical masses released symmetrically from rest initially accelerate outward; their force changes as they separate.

Check

The result has units N, is positive for either common charge sign, and decreases as R⁻².

Practice prompt · Original reading location

Archived original prompt
  • Five point charges of charge q are equally spaced along the edge of a circle of radius r. Derive an expression for the force experienced by any one of the charges. If the charges were not held in place how would you expect them to move?

    • Original problem 2 · worked solution

    Explain superposition and show from Coulomb’s law that electric fields add vectorially.

    Authored clarification: Use classical electrostatics with fixed source charges; the test charge does not disturb them.

    1. Write the force from each source charge.

      F⃗i=kqtqir⃗−r⃗i∣r⃗−r⃗i∣3\vec F_i=kq_tq_i{\vec r-\vec r_i\over|\vec r-\vec r_i|^3}.

    2. Superposition adds the independent interactions.

      F⃗net=∑iF⃗i\vec F_{\rm net}=\sum_i\vec F_i.

    3. Divide by the common test charge.

      E⃗=F⃗netqt=∑ikqir⃗−r⃗i∣r⃗−r⃗i∣3=∑iE⃗i\vec E={\vec F_{\rm net}\over q_t}=\sum_i kq_i{\vec r-\vec r_i\over|\vec r-\vec r_i|^3}=\sum_i\vec E_i.
    Answer

    Electric fields add as vectors, so components rather than magnitudes alone must be summed.

    Check

    The test charge cancels; the resulting field depends only on the source charges and observation point.

    Practice prompt · Original reading location

    Archived original prompt
  • Explain the superposition principle and use it in conjunction with coulombs law to show that it also applies to electric fields.

    • Original problem 3 · worked solution

    A prescribed electric field is E⃗(x,y)=(Ax+By)x^+Byy^\vec E(x,y)=(Ax+By)\hat x+By\hat y. Find the force on a point charge q at (x,y).

    Authored clarification: A and B are constants with units chosen so each field component has units N/C. This prescribed field need not be electrostatic when B≠0.

    1. Electric force is charge times local field.

      F⃗=qE⃗\vec F=q\vec E.

    2. Multiply each component by the signed charge.

      F⃗=q(Ax+By)x^+qByy^\vec F=q(Ax+By)\hat x+qBy\hat y.
    Answer

    Fx=q(Ax+By),Fy=qByF_x=q(Ax+By),\quad F_y=qBy. Negative q reverses the force relative to the field.

    Check

    Each component has units C·N/C=N; at q=0 both components vanish.

    Practice prompt · Original reading location

    Archived original prompt
  • An electric field is described by the equation E = <Ax + By, By>. If a particle of charge q is placed in this field what would the force on that particle be?

    • Original problem 4 · worked solution

    Optional original-source extension. Prerequisites: point-charge vector fields and coordinate geometry. This item makes no AP coverage claim.

    Protons are at (a,−a) and (−a,a), while electrons are at (a,a) and (−a,−a). With a=1 m, describe the field-line pattern and give the electric field at an arbitrary point (x,y) away from a charge.

    Authored clarification: Coordinates are in metres; particles are fixed ideal point charges ±e. Field lines are a qualitative representation of the vector sum, not particle trajectories.

    1. Assign signed charges and source positions.

      Let r⃗1=(a,−a), r⃗2=(−a,a), r⃗3=(a,a), r⃗4=(−a,−a)\vec r_1=(a,-a),\ \vec r_2=(-a,a),\ \vec r_3=(a,a),\ \vec r_4=(-a,-a) with q₁=q₂=e and q₃=q₄=−e.

    2. Apply field superposition.

      E⃗(r⃗)=ke[r⃗−r⃗1∣r⃗−r⃗1∣3+r⃗−r⃗2∣r⃗−r⃗2∣3−r⃗−r⃗3∣r⃗−r⃗3∣3−r⃗−r⃗4∣r⃗−r⃗4∣3]\vec E(\vec r)=ke\left[{\vec r-\vec r_1\over|\vec r-\vec r_1|^3}+{\vec r-\vec r_2\over|\vec r-\vec r_2|^3}-{\vec r-\vec r_3\over|\vec r-\vec r_3|^3}-{\vec r-\vec r_4\over|\vec r-\vec r_4|^3}\right].

    3. Use equal charge magnitudes in the drawing.

      Draw equal numbers of lines emerging from each proton and terminating on the electrons, with reflection symmetries and no crossings. Direction at every point is tangent to the vector sum.
    Answer

    The displayed vector sum is the complete field expression; at the origin the field is zero by symmetry.

    Check

    Each term is singular only at its own point charge. The total charge is zero, so the far field has no monopole term.

    Practice prompt · Original reading location

    Archived original prompt
  • Two electrons and two protons are arranged on a grid as follows. The two electrons are at (1, 1) and (-1, -1) .The two protons are at (1, -1) and (-1, 1). Draw a field line diagram for this configuration. Find an expression for the electric field at an arbitrary point on the grid.

    • Original problem 5 · worked solution

    Optional original-source extension. Prerequisites: continuous line charge, trigonometric components and a definite integral. This item makes no AP coverage claim.

    A thin wire forms the upper semicircle x²+y²=R², y≥0 and carries uniform total charge Q. Find the field at its center.

    Authored clarification: The original omits the charge distribution. Here it is uniform, the length is L=πR, and Q is explicitly supplied.

    1. Convert total charge to line density.

      λ=Q/(πR),dq=λR dθ\lambda=Q/(\pi R),\quad dq=\lambda R\,d\theta.

    2. Use geometry at the center.

      dE⃗=−k dqR2(cos⁡θx^+sin⁡θy^),0≤θ≤πd\vec E=-{k\,dq\over R^2}(\cos\theta\hat x+\sin\theta\hat y),\quad0\le\theta\le\pi.

    3. Cancel horizontal components and integrate vertically.

      Ey=−kλR∫0πsin⁡θ dθ=−2kQπR2E_y=-{k\lambda\over R}\int_0^\pi\sin\theta\,d\theta=-{2kQ\over\pi R^2}.
    Answer

    E⃗=−2kQπR2y^\vec E=-{2kQ\over\pi R^2}\hat y. For positive Q it points away from the charged upper arc.

    Check

    The x component cancels. A change of charge sign reverses the direction, and units are N/C.

    Practice prompt · Original reading location

    Archived original prompt
  • A piece of wire of length L is curved into a half circle of radius r. What is the electric field at the center of the circle the half circle would be a a part of?

    • Original problem 6 · worked solution

    Optional original-source extension. Prerequisites: reflection symmetry, continuous surface charge and a double integral; the closed-form evaluation is supplied. This item makes no AP coverage claim.

    A square thin plate of side L in the xy plane has uniform total charge Q. Find its electric field on the positive z axis at height h>0.

    Authored clarification: The observation point is above the center, not an arbitrary off-axis point. Edge effects are included by integrating the finite square.

    Optional calculus setup with a supplied integral evaluation. You can interpret and check the stated result without carrying out the double integration.

    1. Use reflection symmetry.

      The x and y components cancel, leaving only EzE_z. Let a=L/2 and σ=Q/L².

    2. Set up the axial component of Coulomb’s field as a double integral.

      Ez=kσ∫−aa∫−aah dx dy(x2+y2+h2)3/2E_z=k\sigma\int_{-a}^a\int_{-a}^a {h\,dx\,dy\over(x^2+y^2+h^2)^{3/2}}.

    3. Use the supplied closed-form evaluation of this optional calculus integral.

      The following evaluation of the definite integral is supplied; the double integration is not derived here. Substitute a=L/2 and σ=Q/L², then check its limiting cases below.

      Ez=σπϵ0tan⁡−1 ⁣(a2hh2+2a2)E_z={\sigma\over\pi\epsilon_0}\tan^{-1}\!\left({a^2\over h\sqrt{h^2+2a^2}}\right).
    Answer

    E⃗=Ezz^\vec E=E_z\hat z with the stated expression; its sign follows Q.

    Check

    h→0+:Ez→σ/(2ϵ0);h≫L:Ez→kQ/h2h\to0^+: E_z\to\sigma/(2\epsilon_0);\quad h\gg L:E_z\to kQ/h^2.

    Practice prompt · Original reading location

    Archived original prompt
  • You are given a square plate with sides of length L with a uniformly distributed charge q. What is the electric field an arbitrary point above its center?

    • Original problem 7 · worked solution

    Two infinite nonconducting sheets at x=0 and x=d have uniform surface densities σ₁ and σ₂. Find the field between them for (a) σ₁=σ₂=σ and (b) σ₁=σ, σ₂=−σ.

    Authored clarification: The sheets are ideal infinite nonconductors in vacuum; σ is positive when identifying directions.

    1. Each infinite sheet produces a normal field on both sides.

      E⃗i=σi2ϵ0n^away,i\vec E_i={\sigma_i\over2\epsilon_0}\hat n_{\rm away,i}. Signed σᵢ handles negative charge.

    2. Between the sheets the outward normals are opposite.

      E⃗0<x<d=σ1−σ22ϵ0x^\vec E_{0<x<d}={\sigma_1-\sigma_2\over2\epsilon_0}\hat x.

    3. Substitute each charge arrangement.

      Equal densities give zero. Opposite densities give E⃗=σx^/ϵ0\vec E=\sigma\hat x/\epsilon_0.
    Answer

    (a) Zero between equally charged sheets. (b) Magnitude σ/ε₀ directed from the positive sheet to the negative sheet.

    Check

    The ideal-sheet field is independent of d; outside opposite equal sheets their fields cancel.

    Practice prompt · Original reading location

    Archived original prompt
  • Two infinite parallel plates are placed a distance x apart. What is the the value of the electric field between the plates if they have the same surface charge distribution? What is the value of the field between the plates if their charge distribution is equal but opposite?

    • Original problem 8 · worked solution

    Optional original-source extension. Prerequisites: cylindrical symmetry, continuous volume charge and Gauss’s law. This item makes no AP coverage claim.

    An infinitely long insulating cylinder has radius R and uniform volume charge density ρ. Find its radial electric field at distance s from its axis, inside and outside.

    Authored clarification: A finite total q is unsuitable for an infinite source. The chosen model is a uniform insulating cylinder; a conducting cable would have zero interior field at electrostatic equilibrium.

    1. Use a finite coaxial Gaussian cylinder.

      The curved side contributes flux E(2πsℓ)E(2\pi s\ell); end caps contribute zero.

    2. Inside, enclose only the smaller-radius material.

      Qenc=ρπs2ℓ⇒E=ρs2ϵ0(s<R)Q_{\rm enc}=\rho\pi s^2\ell\quad\Rightarrow\quad E={\rho s\over2\epsilon_0}\quad(s<R).

    3. Outside, enclose the whole cross section.

      Qenc=ρπR2ℓ⇒E=ρR22ϵ0s(s>R)Q_{\rm enc}=\rho\pi R^2\ell\quad\Rightarrow\quad E={\rho R^2\over2\epsilon_0s}\quad(s>R).
    Answer

    E⃗=ρs2ϵ0s^ (s<R),E⃗=ρR22ϵ0ss^ (s>R)\vec E={\rho s\over2\epsilon_0}\hat s\ (s<R),\qquad\vec E={\rho R^2\over2\epsilon_0s}\hat s\ (s>R).

    Check

    The expressions meet at s=R and the interior field vanishes at the axis.

    Practice prompt · Original reading location

    Archived original prompt
  • You are given an infinitely long cable ( aka a cylinder ) of radius r which carries a charge q. What is the electric field outside of the cable? What is the electric field inside the cable?
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