Fiveable
💡AP Physics C: E&M
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FRQ 1 – Mathematical Routines
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Unit 8: Electric Charges, Fields, and Gauss's Law
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Practice FRQ 1 of 20
1. A solid insulating sphere of radius R=0.12 mR = 0.12\ \text{m}R=0.12 m is centered at the origin and has a uniform volume charge density ρ=+5.0×10−6 C/m3\rho = +5.0× 10^{-6}\ \text{C/m}^3ρ=+5.0×10−6 C/m3. The sphere is embedded in a homogeneous, linear dielectric material with permittivity ε=2.5 ε0\varepsilon = 2.5\,\varepsilon_0ε=2.5ε0​. A point particle of mass m=0.020 kgm = 0.020\ \text{kg}m=0.020 kg and charge q=−3.0×10−6 Cq = -3.0× 10^{-6}\ \text{C}q=−3.0×10−6 C is located on the positive xxx-axis at x=0.30 mx = 0.30\ \text{m}x=0.30 m, as shown in Figure 1. The sphere and the particle are stationary and the dielectric fills all space.

Figure 1. Uniformly charged insulating sphere at the origin and a point particle on the +x-axis; concentric spherical Gaussian surface of radius r.

A clean, black-and-white physics setup diagram with a coordinate axis and three concentric/collinear elements. Layout and axes: - Draw a horizontal x-axis across the center of the figure, with an arrow on the right end indicating the positive x-direction. - Label the axis as “x (m)” centered below the axis. - Mark and label the origin at the center of the sphere with the text “0” at the axis intersection. Charged insulating sphere (the source charge): - Center a solid sphere at the origin. Depict it as a circle (2D cross-section) centered exactly on the origin. - Label the sphere “insulating sphere”. - Add a clear radius marker: a straight line from the origin to the sphere surface on the +x side, ending exactly at the sphere boundary, with the label “R = 0.12 m” placed above that radius line. - Inside or adjacent to the sphere, add a text label for the charge density: “ρ = +5.0×10⁻⁶ C/m³”. Dielectric medium: - Add a note in an open area (upper-left quadrant) reading: “dielectric fills all space, ε = 2.5ε₀”. (This is text only; do not draw a boundary for the dielectric.) Concentric Gaussian surface: - Draw a dashed circle centered on the same origin as the sphere to represent a spherical Gaussian surface. - The dashed circle must be larger than the sphere boundary so that it is visually distinct. - Label this dashed circle “Gaussian surface” with a leader arrow pointing to the dashed boundary. - Add a radial dimension marker from the origin to the dashed circle on the +y direction (straight up from the origin) and label it “r”. (Do not assign a number to r in Figure 1.) Point particle on the +x-axis: - Place a small filled dot on the +x-axis to the right of the sphere, clearly outside both the sphere and the dashed Gaussian surface. - Next to this dot, place a label “point particle”. - Directly beside the dot, list its properties as visible text on two lines: “q = −3.0×10⁻⁶ C” and “m = 0.020 kg”. - Add a horizontal distance marker along the x-axis from the origin to the particle: a double-headed arrow exactly spanning from the origin mark to the particle dot, labeled “0.30 m” centered above the arrow. Relative spacing constraints to enforce numerical meaning: - The particle must be drawn at a distance that is visually consistent with being 2.5 times the sphere radius (since 0.30 m divided by 0.12 m equals 2.5). Ensure the particle appears noticeably farther from the origin than the sphere’s edge, with the origin-to-particle dimension arrow longer than the origin-to-sphere-edge radius marker. Styling: - Sphere outline: solid medium-thick line. - Gaussian surface: dashed medium line. - Dimension arrows: thin line with clear arrowheads. - No gridlines. No additional decorative elements.

Figure 2. Axes for a student sketch of electric-field magnitude E versus radial distance r from the center of the sphere.

A blank set of Cartesian axes (no curve pre-drawn) for graphing the magnitude of electric field E as a function of radial distance r. Axes (required exact formatting): - Horizontal axis: labeled “r (m)”. Range starts at 0 on the left and ends at 0.36 on the right. - Horizontal tick marks: every 0.06 m, labeled exactly: 0, 0.06, 0.12, 0.18, 0.24, 0.30, 0.36. - Add a small vertical tick emphasis and an annotation directly above the tick at 0.12 reading “R = 0.12 m”. - Add a small vertical tick emphasis at 0.30 (keep the numeric tick label “0.30” on the axis). - Vertical axis: labeled “E (N/C)”. Range starts at 0 at the origin and ends at 1.2×10⁴ at the top. - Vertical tick marks: every 2.0×10³ N/C, labeled exactly: 0, 2×10³, 4×10³, 6×10³, 8×10³, 1.0×10⁴, 1.2×10⁴. Origin and arrows: - The origin at the intersection of the axes is labeled “0”. - Place arrowheads on the positive ends of both axes (right end of r-axis and top end of E-axis). Blank-graph requirement: - Do not draw any field curve; only the axes, tick marks, tick labels, and the annotation “R = 0.12 m”. (For the renderer: keep the plotting region empty so students can sketch.)
A.
i. Using Gauss’s law, derive an expression for the magnitude E(r)E(r)E(r) of the electric field produced by the uniformly charged insulating sphere for the region 0<r<R0 < r < R0<r<R. Express your answer in terms of ρ\rhoρ, ε\varepsilonε, rrr, and physical constants, as appropriate.
ii. Derive an expression for the magnitude E(r)E(r)E(r) of the electric field produced by the sphere for the region r>Rr > Rr>R. Express your answer in terms of ρ\rhoρ, RRR, ε\varepsilonε, rrr, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
iii. On the axes shown in Figure 2, sketch a graph of the magnitude of the electric field produced by the sphere, EEE, as a function of rrr from r=0r = 0r=0 to r=0.36 mr = 0.36\ \text{m}r=0.36 m.
B. Derive an expression for the electric flux ΦE\Phi_EΦE​ through this Gaussian surface. Express your answer in terms of ρ\rhoρ, RRR, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Figure 3. Gaussian surface of radius r = 0.20 m centered at the origin encloses the insulating sphere but not the point particle at x = 0.30 m.

A black-and-white diagram similar in style to Figure 1, emphasizing which objects are enclosed by a particular Gaussian surface. Layout and axes: - Draw a horizontal x-axis with an arrow on the right end for +x. - Label it “x (m)” below the axis. - Mark the origin with the label “0”. Insulating sphere: - Draw the insulating sphere centered exactly at the origin as a circle (cross-section). - Add a radius marker from the origin to the sphere boundary on the +x side labeled “R = 0.12 m”. - Label inside/near the sphere: “ρ = +5.0×10⁻⁶ C/m³”. Gaussian surface (specified radius): - Draw a dashed circle centered exactly at the origin representing the spherical Gaussian surface. - This dashed circle must be larger than the sphere so that the entire insulating sphere is clearly inside it. - Add a radial dimension marker from the origin to the dashed circle (for example, straight upward) labeled exactly “r = 0.20 m”. - Label the dashed circle “Gaussian surface” with a leader arrow pointing to the dashed boundary. Point particle location (outside the Gaussian surface): - Place a small filled dot on the +x-axis to the right of the dashed circle. - Label next to it: “point particle”. - Add two lines of text next to the dot: “q = −3.0×10⁻⁶ C” and “m = 0.020 kg”. - Add a double-headed horizontal distance arrow from the origin to the particle dot labeled “0.30 m”, centered above the arrow. Enclosure relationship must be visually unambiguous: - Ensure the dashed Gaussian surface radius marker labeled 0.20 m is visibly shorter than the origin-to-particle distance arrow labeled 0.30 m, so the particle is clearly outside the dashed circle. - Ensure the sphere of radius 0.12 m is clearly inside the dashed circle of radius 0.20 m. Styling: - Sphere boundary: solid medium-thick line. - Gaussian surface: dashed medium line. - Dimension arrows: thin line with arrowheads. - No gridlines. No additional objects.
Pep