Electric Charges, Fields and Gauss's Law: Unit Diagnostic (Physics C: E&M Unit 8)

Answer: (A). The two forces are a Newton’s third law pair: equal in size, opposite in direction. Coulomb’s law uses the product of the charges, the same for both beads. Opposite signs attract, so A is pulled towards B.

• (B) assumes the larger charge feels three times the force.
• (C) makes the opposite assumption.
• (D) has the right size but treats the force as repulsive. Opposite signs attract.

If you missed this: read “Coulomb’s law” in the 8.1 study guide.

Answer: (A). After rubbing, the cloth carries +3.2 nC. Grounding lets electrons flow from Earth into the cloth until it is neutral: N = (3.2 × 10⁻⁹) ÷ (1.60 × 10⁻¹⁹) = 2.0 × 10¹⁰. Only the rod’s −3.2 nC is left in the system.

• (B) Conservation holds only while no charge crosses the boundary. Here charge entered the system from Earth.
• (C) has the right total but the wrong carrier. Protons stay in their nuclei.
• (D) gets the direction wrong. A positive cloth attracts electrons; it does not give them away.

If you missed this: read “Charge is conserved” and “Grounding” in the 8.2 study guide.

Answer: (A). Lines start on positive charges and end on negative ones, so X is positive and Y is negative. The number of lines is proportional to the size of the charge: 18 lines against 6 gives a ratio of 3.

• (B) compares the 12 lines that escape with the 6 that end on Y. The ratio must use all the lines at each charge.
• (C) reverses the direction rule.
• (D) Line diagrams do show relative sizes, through the number of lines at each charge.

If you missed this: read “Showing a field: maps and field lines” in the 8.3 study guide.

Answer: (A). In equilibrium the free charges have stopped moving, so there is no field inside the metal, and the excess charge sits on the surface. A field component along the surface would push surface charges sideways, so E just outside is perpendicular to it.

• (B) describes a charged insulator, not a conductor.
• (C) A field parallel to the face would make surface charge move, so the cube would not be in equilibrium.
• (D) The zero field comes from equilibrium, not from the shape, so it holds everywhere in the metal.

If you missed this: read “Fields of charged conductors and insulators” in the 8.3 study guide.

Answer: (A). Measure φ from the arc’s line of symmetry, so the arc runs from −30° to +30°. Each piece has dq = λR dφ at distance R from O. Components across the symmetry line cancel in pairs, so:

E = (kλ/R) ∫ cos φ dφ, from −30° to +30°, = (2kλ/R) sin 30° = kλ/R, pointing along the symmetry line, away from the arc.

• (B) is the semicircle result. A shorter arc changes the limits.
• (C) adds the magnitudes of all the dE vectors, kQ/R² with Q = λπR/3. It ignores cancellation, so it is about 5% too large.
• (D) uses cos 30° instead of sin 30° after integrating.

If you missed this: read “Arcs: a semicircle at its centre” in the 8.4 study guide.

(a) At rest, the electric force balances the weight: (5e)E = mg. E = (3.3 × 10⁻¹⁵)(9.8) ÷ [5(1.60 × 10⁻¹⁹)] = 4.0 × 10⁴ N/C. The force must point up. The droplet is negative, so the force is opposite to E: E points downward.

(b) The electric force is now 6eE = (6/5)mg, still upward, while the weight is mg. Net force = 0.2mg upward, so a = 0.2g = 2.0 m/s² upward.

(c) (2.4 × 10⁻¹⁹) ÷ (1.60 × 10⁻¹⁹) = 1.5. Charge comes only in whole-number multiples of e, so 1.5e cannot exist.

Check yourself: 1 mark each for the size of E, its direction with the reason, the acceleration with direction, and the “whole number of e” argument (4 in total).

If you missed this: for (a) and (b), read the 8.3 study guide; for (c), the 8.1 study guide.

Answer: (A). E is parallel to the four y- and z-faces, so their flux is zero. On the face at x = 2a, the outward normal is +î: Φ = c(2a)(a²) = +2ca³. On the face at x = a, the outward normal is −î: Φ = −c(a)(a²) = −ca³. Net: ca³. (By Gauss’s law this means the cube encloses a charge ε₀ca³.)

• (B) uses the rule for a uniform field. This field is stronger on the right, so more flux leaves than enters.
• (C) adds the sizes of the two fluxes. On a closed surface, entering flux is negative.
• (D) counts only the face where field lines leave.

If you missed this: read “Closed surfaces” in the 8.5 study guide.

Answer: (A). Gauss’s law: Φ = q_enc/ε₀ = (3q − 5q)/ε₀ = −2q/ε₀. The surface’s size and shape, and where the charges sit inside, do not matter; outside charges add nothing.

• (B) includes the outside charge: 3q − 5q + 4q = +2q.
• (C) adds the sizes of the enclosed charges and ignores the sign of −5q.
• (D) The field is weaker further out, but the area grows to make up for it exactly.

If you missed this: read “From flux to Gauss’s law” in the 8.6 study guide.

(a) A pillbox with end area A and its ends at +x and −x. By symmetry E is perpendicular to the slab, points away from the middle plane and has the same size at ±x. So each end has flux EA, and the curved side, parallel to E, has none.

(b) Inside: the box encloses a slice of thickness 2x, so q_enc = ρ(2x)A. Gauss’s law: 2EA = 2ρxA/ε₀, so E = ρx/ε₀. Outside: the enclosed slice is the whole thickness 2b, so 2EA = 2ρbA/ε₀ and E = ρb/ε₀, which does not depend on x.

(c) At x = 0.010 m (inside): E = (4.0 × 10⁻⁷)(0.010) ÷ (8.85 × 10⁻¹²) = 4.5 × 10² N/C. At x = 0.030 m (outside): E = (4.0 × 10⁻⁷)(0.020) ÷ (8.85 × 10⁻¹²) = 9.0 × 10² N/C. Check: viewed from outside, the slab is a sheet with σ = ρ(2b) = 1.6 × 10⁻⁸ C/m², and σ/(2ε₀) gives the same 9.0 × 10² N/C.

Check yourself: 1 mark each for the symmetric pillbox with a reason, the inside result, the outside result and both numbers (4 in total).

If you missed this: read “Choosing a Gaussian surface” and “Standard results” in the 8.6 study guide.