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AP Physics 1 · guided topic map

Fluids, density and pressure for AP Physics 1

Fluids, density and pressure for AP Physics 1, organized into 1 syllabus topic and 5 mapped concept guides.

Syllabus topics
1
Mapped concept guides
5
Educational level
AP Physics 1: Algebra-Based

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8

Fluids

AP Physics 1: Algebra-Based

5 guides
  1. 01Density and fluid properties10–15%
  2. 02Pressure in fluids10–15%
  3. 03Atmospheric pressure and manometers10–15%
  4. 04Buoyancy and Archimedes' principle10–15%
  5. 05Continuity, flow, and fluid calculus10–15%

Diagrams

Fluids, density and pressure as AP Physics 1 draws it

The figures from the AP Physics 1 practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Fig. 7.1Electric Charges, Fields, and Gauss's Law · Electric PotentialAP
An insulating sphere whose charge density rises with distance from the centreinsulating spherePrRρ(r) = ρ₀r/R for r ≤ R

Figure comment

Fig. 7.1A large circle represents the insulating sphere. Inside it, faint concentric rings are drawn at intervals that get smaller towards the outside, indicating that the charge density increases with distance from the centre. A dimension line from the centre out to the surface, tick-marked at both ends, is labelled R. A second, shorter dimension line runs from the centre to a small marked point P inside the sphere and is labelled r. Below the sphere the density is written as a function of radius, proportional to r divided by R. No Gaussian surface is drawn.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

8 Electric Charges, Fields, and Gauss's Law9 Electric PotentialOpen AP question · AP Physics C: Electricity and Magnetism Q7
Guided questions 5 parts

Reading cue. The rings crowd towards the surface, so the density rises with radius: r runs from the centre out to P, and the charge inside r grows faster than r³.

  1. aDetermine Determine the volume charge density at the centre of the sphere, at the point P marked at radius r, and at the surface, in terms of ρ₀.

    recall3 marks

    Check answer 3 marks
    1. zero at the centre
    2. ρ₀r/R at P
    3. ρ₀ at the surface, which is why the rings are drawn closest together there

  2. bDetermine Determine the fraction of the sphere's total charge that lies within a radius R/2 of the centre.

    routine4 marks

    Check answer 4 marks
    1. q(r) = ∫₀^r (ρ₀s/R)4πs² ds = πρ₀r⁴/R
    2. so q ∝ r⁴, and the total charge is Q = πρ₀R³
    3. q(R/2)/Q = (1/2)⁴ = 1/16, that is 6.25%
    4. much less than the 1/8 a uniform sphere would give, because the charge is pushed outwards

  3. cSketch Sketch a graph of the magnitude of the electric field against distance from the centre, from the centre out to 3R, marking the position and the value of the maximum.

    demanding5 marks

    Check answer 5 marks
    1. for r < R, E = q(r)/4πε₀r² = ρ₀r²/4ε₀R, drawn as a curve rising from the origin with increasing gradient, not a straight line
    2. for r > R, E = Q/4πε₀r², falling as 1/r²
    3. the two expressions agree at r = R, so the curve has no step in it
    4. maximum at the surface, r = R, of value ρ₀R/4ε₀
    5. at r = 3R the field has fallen to one ninth of the maximum

  4. dDerive Using an energy density of ½ε₀E², derive an expression for the total electric energy stored in the field outside the sphere, and explain why the same calculation cannot be carried through for a point charge.

    top of the paper5 marks

    Check answer 5 marks
    1. dU = ½ε₀E²(4πr² dr) with E = Q/4πε₀r² for r > R
    2. U = (Q²/8πε₀)∫_R^∞ dr/r² = Q²/8πε₀R
    3. substituting Q = πρ₀R³ gives U = πρ₀²R⁵/8ε₀
    4. for a point charge the lower limit becomes r = 0 and the integral diverges
    5. so a point charge implies infinite field energy, which is why a finite size has to be assumed in such calculations

Transfer challenge

A very long solid insulating cylinder of radius a carries a charge density that varies with distance from its axis as ρ(r) = ρ₀r/a. Derive expressions for the magnitude of the electric field inside and outside the cylinder, and verify that the two agree at the surface.

Check answer 5 marks
  1. Gaussian surface taken as a coaxial cylinder of radius r and length ℓ, with flux only through the curved surface
  2. charge enclosed for r < a: ℓ∫₀^r (ρ₀s/a)2πs ds = 2πρ₀ℓr³/3a
  3. E(2πrℓ) = q_enc/ε₀ gives E = ρ₀r²/3ε₀a for r < a
  4. for r > a the enclosed charge is fixed, giving E = ρ₀a²/3ε₀r
  5. both expressions give ρ₀a/3ε₀ at r = a