Gravitational, Electric and Magnetic Fields: Unit Review

F = Gm₁m₂/r² = (6.67 × 10⁻¹¹)(70)(70)/(1.0)² = 3.3 × 10⁻⁷ N, far too small to notice.

F ∝ 1/r², so it becomes 1/9 as large.

F = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(7.35 × 10²²)/(3.84 × 10⁸)² = 2.927 × 10³⁷ / 1.475 × 10¹⁷ = 1.98 × 10²⁰ N.

r = 6.37 × 10⁶ + 4.0 × 10⁵ = 6.77 × 10⁶ m. g = GM/r² = 3.982 × 10¹⁴ / (6.77 × 10⁶)² = 8.69 N/kg, about 89% of the surface value.

g ∝ 1/r², so r = √2 R_E = 1.414(6.37 × 10⁶) = 9.01 × 10⁶ m. Altitude = 9.01 × 10⁶ − 6.37 × 10⁶ = 2.64 × 10⁶ m (about 2640 km).

g = (6.67 × 10⁻¹¹)(6.42 × 10²³)/(3.39 × 10⁶)² = 4.282 × 10¹³ / 1.149 × 10¹³ = 3.73 N/kg.

Near the surface (over a small region) they are parallel and evenly spaced, pointing straight down: a uniform field. From far away they are radial, pointing to Earth's centre and spreading apart as r increases, so the field weakens.

r = 6.97 × 10⁶ m. v = √(GM/r) = √(3.982 × 10¹⁴ / 6.97 × 10⁶) = 7.56 × 10³ m/s.

They and the station are in free fall together, both falling around Earth with the same acceleration, so nothing pushes on them. They are not out of gravity.

r³ = GMT²/(4π²) = (3.982 × 10¹⁴)(86 400)²/(4π²) = 7.53 × 10²² m³, so r = 4.22 × 10⁷ m (about 35 800 km above the surface).

Kepler's third law with T in years and a in AU: T² = a³, so T = 1.52^1.5 = 1.87 years.

At perihelion (closest to the Sun). Equal areas in equal times means it must cover more distance per unit time when it is close.

M = 4π²r³/(GT²) = 4π²(1.50 × 10¹¹)³ / ((6.67 × 10⁻¹¹)(3.156 × 10⁷)²) = 1.332 × 10³⁵ / 6.644 × 10⁴ = 2.0 × 10³⁰ kg.

Stars far from a galaxy's centre orbit about as fast as stars closer in (flat rotation curves). The visible mass alone would make their speeds fall off, so extra unseen mass must be present.

v = √(2GM/R) = √(2(6.67 × 10⁻¹¹)(7.35 × 10²²)/1.74 × 10⁶) = √(5.635 × 10⁶) = 2.37 × 10³ m/s.

ΔEg = GMm(1/R − 1/r) = (3.982 × 10¹⁴)(1000)(1/6.37 × 10⁶ − 1/6.77 × 10⁶) = (3.982 × 10¹⁷)(9.275 × 10⁻⁹) = 3.69 × 10⁹ J.

r_s = 2GM/c² = 2(6.67 × 10⁻¹¹)(1.99 × 10³⁰)/(3.00 × 10⁸)² = 2.95 × 10³ m (about 3 km).

F = kq₁q₂/r² = (8.99 × 10⁹)(2.0 × 10⁻⁶)²/(0.30)² = 0.40 N, repulsive.

F ∝ q₁q₂/r² = (2 × 2)/2² = 1: the force stays the same.

n = q/e = 3.2 × 10⁻⁹ / 1.60 × 10⁻¹⁹ = 2.0 × 10¹⁰ electrons.

E = kQ/r² = (8.99 × 10⁹)(5.0 × 10⁻⁹)/(0.50)² = 1.8 × 10² N/C, pointing away from the charge.

F = qE = (1.60 × 10⁻¹⁹)(2.0 × 10³) = 3.2 × 10⁻¹⁶ N, opposite to the field direction.

Each charge is 0.10 m away: E = (8.99 × 10⁹)(4.0 × 10⁻⁹)/(0.10)² = 3.60 × 10³ N/C. Both fields point toward the negative charge, so they add: 7.2 × 10³ N/C toward the −4.0 nC charge.

Any three: they start on positive and end on negative charges (or at infinity); they never cross; they are closer together where the field is stronger; the number of lines is proportional to the charge; they meet conductor surfaces at right angles.

V = kQ/r = (8.99 × 10⁹)(3.0 × 10⁻⁹)/0.25 = 1.1 × 10² V (108 V).

Ee = kq₁q₂/r = (8.99 × 10⁹)(2.0 × 10⁻⁶)(−3.0 × 10⁻⁶)/0.10 = −0.54 J. Negative: they attract, so work is needed to pull them apart.

W = qΔV = (2.0 × 10⁻⁶)(50) = 1.0 × 10⁻⁴ J.

E = ΔV/d = 120/(4.0 × 10⁻³) = 3.0 × 10⁴ N/C (or V/m).

Straight, parallel and evenly spaced, from the positive plate to the negative plate: a uniform field.

ΔV stays the same and d doubles, so E = ΔV/d halves.

eΔV = ½mv², so v = √(2eΔV/m) = √(2(1.60 × 10⁻¹⁹)(500)/9.11 × 10⁻³¹) = 1.33 × 10⁷ m/s.

500 eV, which is (500)(1.60 × 10⁻¹⁹) = 8.0 × 10⁻¹⁷ J. (The same as the electron: energy depends on charge, not mass.)

a = eE/m = (1.60 × 10⁻¹⁹)(1.0 × 10⁴)/9.11 × 10⁻³¹ = 1.756 × 10¹⁵ m/s². t = 0.050/2.0 × 10⁷ = 2.5 × 10⁻⁹ s. y = ½at² = ½(1.756 × 10¹⁵)(2.5 × 10⁻⁹)² = 5.5 × 10⁻³ m (5.5 mm), toward the positive plate.

qE = mg with E = V/d: q = mgd/V = (3.2 × 10⁻¹⁵)(9.8)(0.010)/980 = 3.2 × 10⁻¹⁹ C, which is 2 electrons' worth.

Charges come in whole multiples of e = 1.60 × 10⁻¹⁹ C. 2.4 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 1.5, not a whole number, so there must be a measurement error.

The electric force qE balances gravity, so it points up. Electrons are negative, so the force is opposite to E: the field must point down (the top plate is positive).

B = μ₀I/(2πr) = (4π × 10⁻⁷)(10)/(2π(0.050)) = 4.0 × 10⁻⁵ T, close to Earth's field.

F = ILB sin θ = (5.0)(0.20)(0.30)(1) = 0.30 N.

B = μ₀NI/L = (4π × 10⁻⁷)(500)(2.0)/0.25 = 5.0 × 10⁻³ T.

They attract. Each wire sits in the other's field, and the motor-principle hand rule gives a force toward the other wire.

F = qvB = (1.60 × 10⁻¹⁹)(2.0 × 10⁶)(0.40) = 1.3 × 10⁻¹³ N. r = mv/(qB) = (1.67 × 10⁻²⁷)(2.0 × 10⁶)/((1.60 × 10⁻¹⁹)(0.40)) = 5.2 × 10⁻² m (5.2 cm).

The magnetic force is always perpendicular to the velocity, so it does no work and the kinetic energy stays constant. It only changes direction.

v = E/B = 3.0 × 10⁴ / 0.15 = 2.0 × 10⁵ m/s.

m = 23(1.66 × 10⁻²⁷) = 3.82 × 10⁻²⁶ kg. r = mv/(qB) = (3.82 × 10⁻²⁶)(1.0 × 10⁵)/((1.60 × 10⁻¹⁹)(0.50)) = 4.77 × 10⁻² m. They land a diameter away: 2r = 9.5 cm.

f = qB/(2πm) = (1.60 × 10⁻¹⁹)(1.0)/(2π(1.67 × 10⁻²⁷)) = 1.5 × 10⁷ Hz (15 MHz).

The time for each half circle, πm/(qB), does not depend on speed: a faster particle moves on a bigger circle and takes the same time. (This holds until relativistic effects matter.)

Electromagnetic: photon. Strong: gluons. Weak: W⁺, W⁻ and Z⁰ bosons.

Any two: it acts only on moving charges; it is perpendicular to the velocity (and the field), not along the field; it does no work; its field lines form closed loops (no monopoles).

Mass is always positive, so gravity always adds up; large objects are almost exactly neutral, so electric forces cancel; the nuclear forces have very short range.

Δd = cΔt = (3.00 × 10⁸)(5.0 × 10⁻⁹) = 1.5 m.

t = 2h/c = 2(3.58 × 10⁷)/3.00 × 10⁸ = 0.24 s.

Benefit (any): fast, low-latency internet in remote and northern communities; backup in emergencies. Concern (any): space debris and collision risk; streaks on telescope images; pollution from launches and re-entries.