Question 1Level 2 · ExamSingle correct
Point charges +4q and −q (q>0) are fixed on the x-axis at x=0 and x=d respectively. The potential is taken to be zero at infinity. Considering only finite points on the x-axis, the point(s) where the electric field is zero and the point(s) where the electric potential is zero are ______. Exactly one option is correct.
AE=0 at x=2d; V=0 only at x=54d
BE=0 at x=32d; V=0 at x=54d and at x=34d
CE=0 at x=2d; V=0 at x=54d and at x=34d
DE=0 at x=34d; V=0 at x=54d and at x=2d
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Answer: C
Field: between the charges both fields point in +x (away from +4q, toward −q), so E=0 there; to the left of x=0 the larger charge is nearer, so E=0. For x>d: x24q=(x−d)2q⇒x=2(x−d)⇒x=2d.
Potential: V=0 needs r14=r21, i.e. r1=4r2 (distances from +4q and −q). Between the charges: x=4(d−x)⇒x=54d. For x>d: x=4(x−d)⇒x=34d. For x<0, r1<r2, impossible.
So E=0 only at x=2d; V=0 at x=4d/5 and x=4d/3. The points differ: zero field and zero potential are unrelated conditions. Answer (C).
Difficulty: HardQuestion 3Level 3 · AdvancedSingle correct
A point charge +4q is fixed at the origin and a point charge −q is fixed at x=a on the x-axis (q>0); there are no other charges. The field lines are symmetric about the x-axis. A field line leaves +4q making an angle α with the +x direction, where cosα=87, and it ends on −q. Near −q the line arrives making an angle β with the direction pointing from −q toward +4q (the −x direction). Also, of all field lines leaving +4q, those leaving at angles α<αmax to the +x direction end on −q, while the rest go to infinity. The pair (β,αmax) is ______. Exactly one option is correct.
A(30∘, 60∘)
B(60∘, 90∘)
C(90∘, cos−143)
D(60∘, 60∘)
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Answer: D
Near each point charge its own field dominates, so field lines start (end) radially and uniformly in solid angle. Consider the surface of revolution formed by rotating the chosen field line about the x-axis. No field line crosses it, so the flux through any cap bounded by it is the same at both ends.
Flux leaving +4q inside a cone of half-angle α: ε04q⋅21−cosα. Flux entering −q inside a cone of half-angle β about the −x direction: ε0q⋅21−cosβ.
Equating: 4(1−cosα)=1−cosβ. With cosα=7/8: 1−cosβ=21, so β=60∘.
The last line that still ends on −q arrives from the far side, β=180∘: 4(1−cosαmax)=2⇒cosαmax=21, αmax=60∘. (Check: the cone of 60∘ carries 4q⋅41=q worth of lines, exactly the number ending on −q.)
(A) halves the angle wrongly; (C) uses 1−cosβ=1 for the cap. Answer (D).
Difficulty: Peak challenge