Chapter 3 Solutions

Original problem 1 · worked solution

Authored clarification: No external fields are present. The sphere’s symmetry gives uniform surface charge.

1. Use electrostatic equilibrium inside the metal.

   $\vec E=0\quad(r<R)$. Consequently V is constant inside.

2. Apply Gauss’s law outside.

   $\vec E={kQ\over r^2}\hat r\quad(r>R)$.

3. Integrate from infinity and match the boundary.

   $V(r)=kQ/r\quad(r\ge R),\qquad V(r)=kQ/R\quad(r\le R)$.

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Practice prompt · Original reading location

Original problem 2 · worked solution

Optional original-source extension. Prerequisites: infinite-sheet superposition and conductor electrostatic equilibrium. This item makes no AP coverage claim.

Authored clarification: The source does not specify finite versus infinite geometry, grounding, net slab charge or sheet density. This symmetric infinite-sheet model fixes those conditions; a different electrical connection can change the answer.

1. Symmetry makes the gap fields equal in magnitude and opposite in direction.

   $\vec E_{+}=E\hat z,\quad\vec E_{-}=-E\hat z$.

2. Use a finite pillbox across the insulating sheet.

   $2EA=\sigma A/\epsilon_0\quad\Rightarrow\quad E=\sigma/(2\epsilon_0)$.

3. Use conductor equilibrium and neutrality.

   Field inside each metal slab is zero. Each inner slab face has −σ/2 and each outer face +σ/2 under this symmetric isolated-neutral model.

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Practice prompt · Original reading location

Original problem 3 · worked solution

Authored clarification: Merely placing capacitors next to one another does not specify a circuit. Here their corresponding terminals are connected in parallel; different connections give different results.

1. Parallel units share the same voltage.

   $Q_{\rm total}=\sum_i C_iV$.

2. Divide total charge by common voltage.

   $C_{\rm eq}=\sum_i C_i=4(2\,\mathrm{mF})=8\,\mathrm{mF}$.

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Practice prompt · Original reading location

Original problem 4 · worked solution

Optional original-source extension. Prerequisites: the ideal parallel-plate model and the meaning of an unconstrained maximum. This item makes no AP coverage claim.

Authored clarification: No minimum electrode spacing, dielectric, conductor thickness or breakdown limit was supplied. The task is to diagnose whether a maximum is defined.

1. Capacitance depends on more than the enclosing volume.

   $C_{\rm plate}\simeq\epsilon A/d$ shows explicit dependence on gap d and permittivity ε.

2. Construct a counterexample to a finite geometric maximum.

   Keep two plates with area below L² inside the cube. In the ideal model let d approach zero while keeping them separated; C grows without bound.

3. Identify the missing comparison constraints.

   A meaningful design comparison must specify dielectric, minimum gap, available area/material, conductor thickness and voltage/breakdown constraints.

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Practice prompt · Original reading location

Original problem 5 · worked solution

Optional original-source extension. Prerequisites: series and parallel capacitance for two distinct dielectric layouts. This item makes no AP coverage claim.

Authored clarification: The original “fourth” is ambiguous. Both physically distinct layouts are solved explicitly.

1. Area portions are parallel capacitors.

   $C_a=C_0[3/4+K/4]=7C_0/4$.

2. Thickness portions are series dielectric layers.

   $C_b={\epsilon_0A\over(3d/4)+(d/4)/K}={C_0\over3/4+1/(4K)}={16\over13}C_0$.

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Practice prompt · Original reading location

Original problem 6 · worked solution

Authored clarification: Energy density refers to the nearly uniform gap field, and A is one plate’s facing area.

1. Use the parallel-plate capacitance.

   $C=\epsilon_0A/d$.

2. Express stored energy.

   $U={Q^2\over2C}={Q^2d\over2\epsilon_0A}$.

3. Divide by the gap volume Ad.

   $u={U\over Ad}={Q^2\over2\epsilon_0A^2}=\tfrac12\epsilon_0E^2$.

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Practice prompt · Original reading location

Original problem 7 · worked solution

Optional original-source extension. Prerequisites: long-wire magnetic fields and force on a current-carrying wire. This item makes no AP coverage claim.

Authored clarification: The original does not say whether wires are electrically joined or merely placed adjacent. This problem supplies two distinct insulated long wires; joining them would require a circuit specification.

1. The first wire creates a circular magnetic field.

   $B=\mu_0I/(2\pi s)$ at the second wire.

2. Apply force on a current element.

   $F/\ell=IB=\mu_0I^2/(2\pi s)$.

3. Use the right-hand rule.

   Opposite parallel currents repel, so each wire is pushed away from the other. Equal and opposite forces give zero net internal force on the pair.

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Practice prompt · Original reading location

Original problem 8 · worked solution

Authored clarification: The ideal source holds the voltage at 20 V while the total resistance changes. The reviewed wording asks for both powers and their ratio.

1. Use the fixed-voltage power expression.

   $P=V^2/R$.

2. Evaluate each resistance.

   $P_1=20^2/1=400\,\mathrm W,\qquad P_2=20^2/10=40\,\mathrm W$.

3. Take the ratio.

   $P_2/P_1=R_1/R_2=1/10$.

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Practice prompt · Original reading location