See the field.
Understand
the force.

1. Use Coulomb’s law for the magnitude of each force F=k|qQ|/r². Because the test charge is negative, each force points toward its positive source charge.
2. At (a) x=0.5 m, the two forces oppose. Magnitudes from +3 μC and +12 μC are 2.1576×10⁻⁴ N and 8.6304×10⁻⁴ N, respectively; the difference points right.
3. At (b) x=−0.5 m, both forces point right. Add contributions at 0.5 m and 1.5 m.
4. At (c), with test charge at (1,0.5) m, +12 μC attracts it directly downward, while +3 μC attracts it diagonally down-left. Resolve the diagonal force into components: Fₓ≈−3.86×10⁻⁵ N and Fᵧ≈−8.82×10⁻⁴ N. Then take the vector magnitude and angle.

1. Forces: gravity mg downward, electric force qE to the right, and tension T along the string. The rightward deflection tells us the electric force is toward the right plate.
2. Resolve tension: T cos θ = mg, T sin θ = qE. Divide these equations to remove T: tan θ = qE/(mg).
3. Compute the electric force Fₑ = mg tan θ = (0.00150)(9.81)tan(20°) = 5.36×10⁻³ N.
4. The field is E = Fₑ/q = 1.07×10⁴ N/C; then ΔV = Ed = (1.07×10⁴)(0.0400) = 428 V. Since field points right, potential decreases rightward.

1. Use the gradient relation 𝐄 = −∇V; differentiate with respect to one variable at a time while holding the others fixed.
2. Components: Eₓ = −4xy, Eᵧ = −2x²+3z, E𝓏 = 3y−5.
3. At the origin 𝐄 = −5.0 ẑ V/m: magnitude 5.0 V/m, direction −z.
4. Calculate the endpoints: V(0,0,0)=0 and V(1,1,2)=2−6+10=6.0 V. Slow external work is Wₑₓₜ=qΔV=(2.0×10⁻⁶)(6.0).

1. Each face has area A=L²=0.040 m². Outward normal is +ẑ at top and −ẑ at bottom.
2. Uniform field: Φtop=E𝓏A = (5.0×10³)(0.040)=+200 N·m²/C; Φbottom=−200 N·m²/C.
3. The net flux through the closed cube is zero because the uniform field gives no enclosed charge; flux entering faces equals flux leaving faces.
4. For a central charge: Gauss’s law gives Φnet=q/ε₀≈339 N·m²/C. Cube symmetry divides it equally among six faces: Φface=q/(6ε₀)≈56.5 N·m²/C.

1. Use concentric thin spherical shells. A shell has volume dV=4πr²dr, so dQ=ρ(r)dV.
2. Integrate all charge: Q = (4πρ₀/R) ∫₀ᴿ r³dr = πρ₀R³.
3. For a Gaussian sphere with r<R, enclosed charge is qenc(r)=(4πρ₀/R)∫₀ʳ r′³dr′ = πρ₀r⁴/R. Use E(4πr²)=qenc/ε₀.
4. For r>R, enclose all Q. Both formulas agree at r=R, as they should.

1. (a) Charge. The rod length is 2L, so Q = 2λL.
2. (b) Geometry. For a source element at (x,0), dq = λ dx and the displacement to P is (a−x)x̂ + bŷ. Thus dE = kλ[(a−x)x̂ + bŷ] dx / [(a−x)²+b²]³ᐟ².
3. Integrate from −L to L: Eₓ = kλ{1/√[(a−L)²+b²] − 1/√[(a+L)²+b²]}. Eᵧ = (kλ/b){(L−a)/√[(L−a)²+b²] + (L+a)/√[(L+a)²+b²]}. Here b ≠ 0; a point on the charged rod is singular.
4. (c) Symmetry and limits. At a=0, Eₓ=0 and Eᵧ = kQ/[b√(L²+b²)]. For b ≫ L this tends to kQ/b². For 0 < b ≪ L, it tends to 2kλ/b = λ/(2πε₀b). The decay changes from 1/b² to 1/b.

Worked teaching notes based on the recovered assignment and matching solution PDF; assumptions and direction conventions are stated above.

1. The source-to-target displacement is r₁−r₂=(2,3,6) m, with length 7.0 m.
2. Use F₂→₁ = kQ₁Q₂(r₁−r₂)/|r₁−r₂|³. The product of charges is negative, so the force points opposite this displacement, toward Q₂.
3. With k=8.99×10⁹, the components are (−3.15×10⁻⁴, −4.72×10⁻⁴, −9.44×10⁻⁴) N. The magnitude is k|Q₁Q₂|/49 = 1.10×10⁻³ N.

Worked teaching notes based on the recovered assignment and matching solution PDF; assumptions and direction conventions are stated above.

1. (a) The arc length is R(π/2), so λ=2Q/(πR) and dq=(2Q/π)dφ.
2. (b) Reflection about the y-axis cancels Eₓ. At the centre, source and observation point share z=0, so E𝓏=0. A positive arc lies above the origin and pushes a positive test charge downward.
3. At (0,0,z), every element is √(R²+z²) away. Integrate k dq[−R cosφ x̂−R sinφ ŷ+z ẑ]/(R²+z²)³ᐟ². The integrals of cosφ, sinφ and 1 are 0, √2 and π/2 respectively.
4. (c) E(0,0,z)=kQ[−(2√2R/π)ŷ+z ẑ]/(R²+z²)³ᐟ². Set z=0 to obtain E(0)=−2√2 kQ/(πR²)ŷ.

Worked teaching notes based on the recovered assignment and matching solution PDF; assumptions and direction conventions are stated above.

1. For one ball, tension points along the thread, weight mg points down, and the repulsive Coulomb force points horizontally away from the other ball.
2. The charge separation is r=2l sinθ. Balance T cosθ=mg and T sinθ=kq²/r², giving mg tanθ=kq²/(4l²sin²θ).
3. Rearrange using k=1/(4πε₀): sin²θ tanθ=q²/(16πε₀mg l²). Solve for the magnitude of q; the angle alone does not determine whether both charges are positive or both negative.

Worked teaching notes based on the recovered assignment and matching solution PDF; assumptions and direction conventions are stated above.

1. Take +x right and +y up. The positive proton has aᵧ=−eE/mₚ. Horizontal motion has constant speed, so t=L/vₓ=0.120/(1.5×10⁷)=8.0 ns.
2. Using e=1.602×10⁻¹⁹ C and mₚ=1.673×10⁻²⁷ kg gives |aᵧ|≈3.83×10¹³ m/s².
3. The initial vertical velocity is zero: Δy=½aᵧt²≈−1.23×10⁻³ m. Neglect gravity and edge fields, as in the uniform-field model.

Worked teaching notes based on the recovered assignment and matching solution PDF; assumptions and direction conventions are stated above.

1. Field lines leave positive charges and end on the negative charge. Draw twice as many lines from +2 μC as from +1 μC, with the total terminating on −3 μC.
2. Near any one point charge, its own 1/r² field dominates and is approximately radial. Field lines cannot cross.
3. The net charge is zero, so the monopole 1/r² far-field term cancels. The charge distribution still has a nonzero dipole moment: far away the leading field is dipole-like and decays as 1/r³. Zero net charge does not mean zero field everywhere.
4. Use E(r)=k Σᵢ qᵢ(r−rᵢ)/|r−rᵢ|³ to plot the geometry. The notebook below provides the field-line plot.

Worked teaching notes based on the recovered assignment and matching solution PDF; assumptions and direction conventions are stated above.

1. (a) Equatorial line. At (x,0), horizontal components cancel and vertical components add downward: E=−2kQa ŷ/(x²+a²)³ᐟ². For |x| ≫ a, E≈−Qa ŷ/(2πε₀|x|³).
2. (b) Axial line. For y>a, Eᵧ=kQ[(y−a)⁻²−(y+a)⁻²]≈4kQa/y³=Qa/(πε₀y³). On the distant negative y-axis the field also points +y; write the general far-field magnitude using |y|³. The source’s y³ expression assumes the positive-axis branch.
3. (c) The dipole vector points from −Q to +Q: p=2Qa ŷ=8.0×10⁻⁹ ŷ C·m. Note that the separation is 2a.
4. (d) τ=p×Eext. Since ŷ×x̂=−ẑ, τ=−3.2×10⁻³ ẑ N·m, into the page.

Worked teaching notes based on the recovered assignment and matching solution PDF; assumptions and direction conventions are stated above.

1. Use the specified normal to fix the sign: Φ=E·n̂ A, with A=π(2.0)²=4π m².
2. (a) Only E𝓏 contributes: Φ=3000(4π)=3.77×10⁴ N·m²/C.
3. (b) Eₓ=0, so Φ=0.
4. (c) E·n̂=(4000+3000)/√2 N/C. Multiply by 4π to get 6.22×10⁴ N·m²/C. Reversing the normal reverses each nonzero flux.

Worked teaching notes based on the recovered assignment and matching solution PDF; assumptions and direction conventions are stated above.

1. (a) Thin spherical layers have dV=4πr²dr. Integrate Q=4πρ₀r₁∫[r₁,r₂] r dr=2πρ₀r₁(r₂²−r₁²).
2. Spherical symmetry makes E radial and constant in magnitude on each Gaussian sphere: E(4πr²)=Qenc/ε₀.
3. (b) For r<r₁, Qenc=0 and E=0.
4. (c) For r₁≤r≤r₂, Qenc=2πρ₀r₁(r²−r₁²), so E=[ρ₀r₁/(2ε₀)](1−r₁²/r²) r̂.
5. (d) For r>r₂, all Q is enclosed: E=Q r̂/(4πε₀r²). Outside, the field is identical to a point charge Q at the centre. The expressions agree at both boundaries; there are no separate surface-charge sheets here.

Worked teaching notes based on the recovered assignment and matching solution PDF; assumptions and direction conventions are stated above.

1. Transferred charge magnitude q=Ne=1.602×10⁻⁷ C. The left plate loses electrons and becomes +q; the right plate becomes −q. Convert area: 400 cm²=0.0400 m².
2. σleft=+q/A=+4.005×10⁻⁶ C/m²; σright has the opposite sign. In the stated large-plate model these charges reside on the inner faces.
3. The two sheet fields add between the plates: E=σ/ε₀≈4.52×10⁵ N/C, from left to right. Outside they cancel in this idealization.
4. Vleft−Vright=Ed≈9.05×10³ V. The left plate is at higher potential because potential decreases along E.

Worked teaching notes based on the recovered assignment and matching solution PDF; assumptions and direction conventions are stated above.

1. (a) E=−∇V. Differentiate one coordinate at a time: Eₓ=y²z−4y, Eᵧ=2xyz−4x, E𝓏=xy², in V/m.
2. (b) At (1,1,1), E=(−3x̂−2ŷ+ẑ) V/m. Its magnitude is √14≈3.74 V/m; its unit direction is (−3x̂−2ŷ+ẑ)/√14.
3. (c) V(0,0,0)=0 and V(1,1,1)=3 V. Hence Wext=qΔV=(2.0×10⁻⁶)(3)=+6.0×10⁻⁶ J. Work by the electric field has the opposite sign. Electrostatic work is independent of path.

Worked teaching notes based on the recovered assignment and matching solution PDF; assumptions and direction conventions are stated above.

1. (a) E=0 in the shell metal. A Gaussian surface with b<r<c must enclose zero net charge: qb=−5.0 μC. Since qb+qc=−3.0 μC, qc=+2.0 μC.
2. (b) E=0 for r<a; E=k(5.0 μC)r̂/r² for a<r<b; E=0 for b<r<c; E=k(2.0 μC)r̂/r² for r>c. Positive radial values mean outward.
3. (c) Vshell=k(2.0 μC)/c≈3.00×10⁵ V. Vsphere=Vshell+k(5.0 μC)(1/a−1/b)≈1.65×10⁶ V.
4. (d) V is constant inside each conductor. In the gap V(r)=Vshell+k(5.0 μC)(1/r−1/b); outside V(r)=k(2.0 μC)/r. V is continuous everywhere. E jumps at a, b and c because each charged surface gives a discontinuity σ/ε₀ in the normal field. Use the recovered notebook for the paired plots.
5. (e) Moving the inner sphere off-centre changes the gap field and makes the inner induced charge nonuniform. Fields still meet conducting surfaces normally. E remains zero in both conductors; the exterior field remains radial about the shell centre, determined by +2.0 μC. Gauss’s law remains valid, but without symmetry it alone cannot determine the gap field.

Worked teaching notes based on the recovered assignment and matching solution PDF; assumptions and direction conventions are stated above.

1. Choose a coaxial Gaussian cylinder of radius s and length L. The curved surface flux is E(2πsL)=λL/ε₀. Thus E(s)=λ/(2πε₀s).
2. Integrate from a to R: V₀ = ∫ₐᴿ E ds = [λ/(2πε₀)]ln(R/a).
3. Eliminate λ: E(s)=V₀/[s ln(R/a)]. Since ln(R/a)=ln(200)=5.30, the field at the wire is E(a)≈1.70×10⁶ V/m.
4. The field scales as 1/s: E(1 cm)≈1.7×10⁴ V/m and E(R)≈8.5×10³ V/m. The peak is below the nominal dry-air breakdown value 3×10⁶ V/m; real Geiger tubes use a gas with lower breakdown thresholds.
5. Use λ/V₀ = 2πε₀/ln(R/a) to obtain C/L≈10.5 pF/m.

1. Before insertion C₀=ε₀A/d. Inside a conductor E=0, so only the two air gaps contribute: their combined thickness is d−t. Hence C=ε₀A/(d−t), independent of the slab’s vertical position.
2. Isolated capacitor means Q fixed: U₀=Q²d/(2ε₀A); U=Q²(d−t)/(2ε₀A). Thus ΔU=−Q²t/(2ε₀A).
3. With bottom plate +Q and top −Q, the conducting slab has induced −Q/A on its bottom face and +Q/A on its top face. Edge fringing field pulls the slab into the plates.
4. During slow insertion with fixed Q, the field does positive mechanical work and your hand does negative work equal to ΔU (with the specified idealization).
5. With a battery attached, V is fixed and ΔUfield=½V²ΔC>0. The battery supplies V²ΔC=2ΔUfield; half goes to the field, half to pulling the slab in.

1. Series equivalent: Ceq=C₁C₂/(C₁+C₂)=1.333 μF. Each initially has Q=CeqV=1.333 mC.
2. Initial drops: V₁=Q/C₁=667 V, V₂=Q/C₂=333 V; total energy U₀=½CeqV²=0.667 J.
3. After reconnecting positive-to-positive, total charge on the positive node is 2Q, while Cparallel=C₁+C₂=6 μF. Common voltage Vf=2Q/(C₁+C₂)=444 V.
4. Final charges Q₁=C₁Vf=0.889 mC and Q₂=C₂Vf=1.78 mC.
5. Final energy Uf=½(C₁+C₂)Vf²=0.593 J. Difference 0.074 J (1/9≈11.1%) is dissipated mainly as Joule heating during charge redistribution.

1. Gauss gives radial field E(r)=Q/(4πε₀r²) between a and b. Integrate ΔV=∫ₐᵇE dr=Q(1/a−1/b)/(4πε₀).
2. Therefore C=Q/ΔV=4πε₀ab/(b−a) and U=Q²(1/a−1/b)/(8πε₀).
3. Check by field energy: integrate u=½ε₀E² over shells 4πr² dr. This yields the same expression.
4. For b→∞, energy within radius r is a fraction 1−a/r of the total. Hence at 2a it is 1/2; 90% is inside r=10a.
5. For the dome, Q=4πε₀aV≈5.0 μC; U=½QV≈0.75 J.