AP Unit 10 Solutions

A has an electron deficit, so qA = +(2.0 × 10¹²)(1.60 × 10⁻¹⁹) = +3.2 × 10⁻⁷ C. B has an excess, so qB = −1.6 × 10⁻⁷ C.

A neutron carries zero charge. Net ordinary-object charge changes by integer multiples of e. The 0.001 m diameter is 1/200 of the separation, so treating each bead as a point is reasonable here.

FE = (8.99 × 10⁹)(3.2 × 10⁻⁷)(1.6 × 10⁻⁷)/(0.20)² = 0.0115 N.

The opposite signs attract, so A is pulled toward B and B toward A with equal 0.0115 N forces. The combined change gives factor (2 × 2)/3² = 4/9.

Reversing only B makes both charges positive; the forces become repulsive with the same magnitude, still along their connecting line.

$F_E/F_g=ke^2/(Gm_p^2)=1.24\times10^{36}$. The common r² cancels.

Proton electric forces repel while their gravitational forces attract. Nearly neutral planets have cancellations of positive and negative contributions; gravitational mass contributions add. A table’s normal force is an effective contact-force description of many microscopic electromagnetic interactions.

ε = 5ε₀, so FE,medium = FE,vacuum/5.

Permittivity characterizes a medium’s polarization response to an electric field. Bound electron distributions shift or dipoles partly align, giving charge separation and an opposing response without changing the sample’s net charge. Composition and molecular arrangement set how easily these configurations respond. In an insulator charge cannot move freely through the whole sample; a conductor has mobile carriers.

Total rod-plus-cloth charge remains zero, so Qcloth = +12 nC. N = 12 × 10⁻⁹/(1.60 × 10⁻¹⁹) = 7.5 × 10¹⁰ electrons move to the rod.

The rod’s field redistributes the can’s electrons, but no charge crosses the insulated can boundary, so its net charge stays zero.

The negative rod repels can electrons away, leaving the near region relatively positive and the far region relatively negative.

Bound electrons in the insulator shift slightly away from the rod. The nearer positive part is attracted more strongly than the farther negative part is repelled in the nonuniform field.

Hold the rod nearby, close the grounding switch, then open the switch while keeping the rod in place. Finally remove the rod.

Electrons enter from Earth while grounded because of attraction toward the rod. Net sphere charge changes during the Earth connection, then remains fixed after isolation.

The isolated two-sphere system has +6 nC total charge. The rod has been removed and the identical spheres share a symmetric environment. Their common equilibrium potential therefore corresponds to equal charges, giving +3 nC each.

The formerly −4 nC sphere changes by +7 nC and the formerly +10 nC sphere by −7 nC. Electrons move from the former to the latter.

Each source is 0.141 m from P, so each field magnitude is (8.99 × 10⁹)(2.0 × 10⁻⁹)/0.020 = 899 N/C. Both arrows point away from the respective positive source at 45° above horizontal.

Horizontal components cancel; vertical components add: E = 2(899)sin45° = 1.27 × 10³ N/C upward. Scaled arrow lengths encode magnitude; closer field lines indicate larger magnitude under a consistent drawing convention.

With only the right source negative, its 899 N/C contribution points from P down and right toward that source. The left positive source still gives an up-and-right arrow. Vertical components cancel; horizontal components add, producing 1.27 × 10³ N/C right.

F⃗ = qE⃗ gives magnitude (3.0 × 10⁻⁹)(1.27 × 10³) = 3.81 × 10⁻⁶ N downward.

The test charge must not appreciably rearrange the sources or their charge distribution; otherwise it changes the field being measured.

The +3.0 nC test charge has the same 3.81 μN force magnitude upward, in the same direction as the original upward field.

At 0.030 m the observation point is in the conducting material, so E = 0. At 0.10 m it is outside the spherically symmetric source, so E = kQ/r² = 4.50 × 10³ N/C.

The graph is zero for r < R, jumps to the outside-surface value kQ/R² = 1.80 × 10⁴ N/C, then falls as 1/r². Surface field is radial outward and perpendicular to the surface.

The conductor’s excess +5.0 nC charge resides on its surface; freely mobile charge has redistributed until the internal field vanishes.

A nonzero field in the metal would exert force on mobile carriers, which would redistribute instead of remaining in electrostatic equilibrium.

No. Excess insulating charge can remain in the interior and on the surface; the internal field may be nonzero. This is a qualitative statement, not a request to calculate its general internal field.

Ui = kq₁q₂/0.10 = −5.39 × 10⁻⁷ J; Uf = kq₁q₂/0.25 = −2.16 × 10⁻⁷ J.

ΔU = +3.24 × 10⁻⁷ J. Wext = +3.24 × 10⁻⁷ J and WE = −3.24 × 10⁻⁷ J.

The curve is negative, rises toward zero as r grows, and has inverse-distance magnitude.

Doubling both distances halves Ui and Uf, so ΔU also halves to +1.62 × 10⁻⁷ J.

The positive-positive pair contributes +kq²/a; each positive-negative pair contributes −kq²/a. Thus U = −kq²/a.

U = −(8.99 × 10⁹)(4.0 × 10⁻⁹)²/0.20 = −7.19 × 10⁻⁷ J.

The remaining positive-positive pair still has +kq²/a. The two pairs involving the removed charge tend to zero, so Uf = +kq²/a.

ΔU = Uf − Ui = 2kq²/a = +1.44 × 10⁻⁶ J. For a slow move Wext = ΔU.

V = k(6.0 − 2.0) × 10⁻⁹/0.20 = 179.8 V.

U = qtestV = (3.0 × 10⁻⁹)(179.8) = 5.39 × 10⁻⁷ J. Potential has no direction and no vector components.

ΔU = qΔV = (2.0 × 10⁻³)(9) = +0.018 J. Chemical energy supplies this internal charge-separation energy.

After the spheres touch and reach equilibrium, their potentials are equal. Their charges need not be equal because geometry affects the charge–potential relation.

The arrows are horizontal toward increasing x, perpendicular to the contours, because V decreases rightward.

In the first region E ≈ 6 V/0.010 m = 600 N/C. In the farther region E ≈ 6 V/0.020 m = 300 N/C. Draw the first arrows twice as long on a common scale.

ΔU = qΔV = (−2.0 × 10⁻⁶)(12 − 24) = +24 × 10⁻⁶ J.

The electric force points left, opposite the rightward field. Along 18 V, ΔV = 0, so electric work is zero.

C = κε₀A/d = 3.54 × 10⁻¹⁰ F. Q = CΔV = 3.54 × 10⁻⁹ C; the plates carry +Q and −Q.

E = ΔV/d = 6.67 × 10³ N/C from positive plate to negative plate. U = ½CΔV² = 1.77 × 10⁻⁸ J.

The capacitor-voltage graph is a straight line through the origin with slope 1/C; its triangular area is ½QfinalΔVfinal = 17.7 nJ. This is stored field energy and minimum quasistatic assembly work. The fixed-voltage battery graph is horizontal at 10 V; delivering final Q = 3.54 nC supplies QΔV = 35.4 nJ.

Doubling A doubles C; fixed battery ΔV then doubles Q and U. C′ = 708 pF and U′ = 35.4 nJ.

Initially Q₀ = C₀V₀ = 4.0 nC and U₀ = ½C₀V₀² = 40 nJ. Both cases give C′ = 800 pF.

Isolated: Q′ = 4.0 nC, V′ = 5 V, E′/E₀ = 1/4, U′ = 10 nJ. Connected: V′ = 20 V, Q′ = 16 nC, E′/E₀ = 1, U′ = 160 nJ.

The dielectric polarizes and its induced field opposes the field from the free plate charges. In the connected case extra free charge compensates, maintaining the net field V/d.

The force is upward: |a| = |q|E/m = 2.5 m/s². Horizontal velocity stays constant.

t = 0.20/4.0 = 0.050 s. Upward deflection y = ½(2.5)(0.050)² = 0.00313 m.

ΔV = −200 V. ΔU = (+1.60 × 10⁻¹⁹)(−200) = −3.20 × 10⁻¹⁷ J.

Kf = −ΔU = 3.20 × 10⁻¹⁷ J = 200 eV. v = √(2Kf/mp) = 1.96 × 10⁵ m/s.

Draw a 200 eV decrease in U and a 200 eV increase in K; their sum stays fixed.

ΔU = +3.20 × 10⁻¹⁷ J and ΔK = 0. Thus Wext = ΔK + ΔU = +3.20 × 10⁻¹⁷ J.

The electric force does negative work that is balanced by the agent’s positive work, so kinetic energy does not change.

At the stipulated one-dimensional turning point the only velocity component becomes zero, so Kf = 0. Conservation gives 0 + qVmax = ½mv₀² + qV₀.

$V_{\max}=V_0+\frac{mv_0^2}{2q}$. A proposed larger potential would require negative total kinetic energy and is inaccessible. Without the stated one-dimensional assumptions this expression is an energy upper bound, not a guaranteed attained turn.

For both particles under the same stipulated one-dimensional uphill conditions, ΔVturn = mv₀²/(2q) is inversely proportional to positive charge. The +2q particle reaches half the potential rise of the +q particle.

A negative charge moving to higher V has ΔU = qΔV < 0 and, with only electric work, gains kinetic energy.

The points lie on a line of slope (2.0 − 8.0)/0.060 = −100 V/m, so Eavg,x = +100 N/C.

δE/E ≈ 0.10/6.0 + 0.002/0.060 = 0.050, so Eavg ≈ (100 ± 5) N/C. Keep electrode voltage and geometry fixed; assume the sampled region approximates a stable, smooth potential map.

Draw vertical 8, 6, 4, and 2 V lines at the listed x values, with equal rightward field arrows normal to them.

F = qE gives 2.0 × 10⁻⁴ N left; ΔU = q(2 − 8) = +12 μJ. With only electric work, K would decrease by 12 μJ if the endpoint were reachable.

Plot C in farads against 1/d in m⁻¹. The points are (1000, 354 × 10⁻¹²), (500, 177 × 10⁻¹²), (250, 88.5 × 10⁻¹²), with slope 3.54 × 10⁻¹³ F·m.

κ = slope/(ε₀A) = (3.54 × 10⁻¹³)/[(8.85 × 10⁻¹²)(0.020)] = 2.0. With the supply disconnected and plates discharged, set a measured uniform spacer gap and use a capacitance meter across the plates. Change d and record C at each gap. Keep overlap area and dielectric fixed, repeat each gap measurement after reassembly, and consider uneven separation or fringing as systematic errors.

Initially Q = (177 pF)(10 V) = 1.77 nC, E = 5000 N/C, and U = 8.85 nJ. Disconnecting preserves Q. At 4.0 mm, C halves to 88.5 pF, so V doubles to 20 V.

E = Q/(κε₀A) remains 5000 N/C. U = Q²/(2C) doubles to 17.7 nJ. A slow external plate-separating agent supplies ΔU = +8.85 nJ.

The two fields have equal magnitude kQ/a² and opposite directions, so Enet = 0. The potentials add: V = 2kQ/a.

The extra interaction energy is qV = 2kQq/a, from the two source–test pairs. The existing source–source energy is unchanged.

At the origin the interaction energy involving q is 2kQq/a; far away it approaches zero. Thus ΔK = +2kQq/a.

$v_f=\sqrt{v_i^2+4kQq/(ma)}$. By symmetry the field along the y axis has no x component and points outward for y > 0, so a small positive upward kick lets it move outward.