AP Physics C: Electricity and Magnetism FRQ Practice

1. Electric Field and Potential

Question:

A charge of +5.0 μC is placed at the origin. A second charge of -2.0 μC is placed 0.4 meters away from the first charge along the x-axis.

(a) Calculate the electric field at a point 0.2 meters to the right of the first charge (along the x-axis).

(b) Determine the electric potential at this point.

(c) Calculate the force on a +1.0 μC charge placed at this point.

Response Guidelines:

• Part (a):
  The electric field due to a point charge is given by:
  E = kₑ |q| / r²
  Where:
  kₑ = 8.99 × 10⁹ N·m²/C² (Coulomb's constant)
  q = +5.0 μC = 5.0 × 10⁻⁶ C (charge at the origin)
  r = 0.2 m (distance from the first charge to the point)
  E₁ = (8.99 × 10⁹ × 5.0 × 10⁻⁶) / (0.2)² ≈ 1.12 × 10⁶ N/C
  The electric field from the second charge (at x = 0.4 m) is directed toward the charge because it is negative. The distance from this second charge to the point is 0.2 m.
  E₂ = (8.99 × 10⁹ × 2.0 × 10⁻⁶) / (0.2)² ≈ 4.50 × 10⁶ N/C
  The net electric field is the vector sum of these two fields. Since the first charge is positive and the second is negative, they point in opposite directions. Therefore,
  E_net = E₁ - E₂ ≈ 1.12 × 10⁶ - 4.50 × 10⁶ ≈ -3.38 × 10⁶ N/C (toward the second charge).
• Part (b):
  The electric potential due to a point charge is given by:
  V = kₑ q / r
  For the first charge at the origin:
  V₁ = (8.99 × 10⁹ × 5.0 × 10⁻⁶) / 0.2 ≈ 2.25 × 10⁵ V
  For the second charge at x = 0.4 m:
  V₂ = (8.99 × 10⁹ × -2.0 × 10⁻⁶) / 0.2 ≈ -8.99 × 10⁴ V
  The total potential is the sum of these two:
  V_total = V₁ + V₂ ≈ 2.25 × 10⁵ - 8.99 × 10⁴ ≈ 1.35 × 10⁵ V
• Part (c):
  The force on a charge placed in an electric field is given by:
  F = qE
  Where:
  q = +1.0 μC = 1.0 × 10⁻⁶ C (charge placed at the point)
  E = -3.38 × 10⁶ N/C (electric field at the point)
  F = (1.0 × 10⁻⁶) × (-3.38 × 10⁶) ≈ -3.38 N (directed toward the second charge).

2. Magnetic Field and Force

Question:

A proton moves with a velocity of 3.0 × 10⁶ m/s in a magnetic field of 0.1 T. The velocity is perpendicular to the magnetic field.

(a) Calculate the magnetic force on the proton.

(b) If the proton is moving in a circular path, calculate the radius of the path.

(c) Determine the period of the proton's motion.

Response Guidelines:

• Part (a):
  The magnetic force on a charged particle is given by:
  F = qvB sin(θ)
  Where:
  q = 1.6 × 10⁻¹⁹ C (charge of a proton)
  v = 3.0 × 10⁶ m/s (velocity of the proton)
  B = 0.1 T (magnetic field strength)
  θ = 90° (since the velocity is perpendicular to the field)
  F = (1.6 × 10⁻¹⁹) × (3.0 × 10⁶) × (0.1) × sin(90°) ≈ 4.8 × 10⁻¹³ N
• Part (b):
  The radius of the circular path is given by:
  r = mv / qB
  Where:
  m = 1.67 × 10⁻²⁷ kg (mass of a proton)
  v = 3.0 × 10⁶ m/s
  q = 1.6 × 10⁻¹⁹ C
  B = 0.1 T
  r = (1.67 × 10⁻²⁷ × 3.0 × 10⁶) / (1.6 × 10⁻¹⁹ × 0.1) ≈ 3.13 × 10⁻² m
• Part (c):
  The period of the motion is the time it takes for the proton to complete one full revolution, which is given by:
  T = 2πr / v
  Using the radius calculated in part (b) and the velocity:
  T = (2π × 3.13 × 10⁻²) / (3.0 × 10⁶) ≈ 6.5 × 10⁻⁵ s

3. Capacitors and Energy Storage

Question:

A parallel-plate capacitor has a capacitance of 4.0 μF. The plates are separated by a distance of 0.02 m and have an area of 1.0 m². The capacitor is connected to a 12 V battery.

(a) Calculate the charge stored on each plate of the capacitor.

(b) Determine the energy stored in the capacitor.

(c) If the battery is disconnected and the plates are pulled apart to a distance of 0.04 m, calculate the new capacitance.

Response Guidelines:

• Part (a):
  The charge stored on a capacitor is given by:
  Q = C V
  Where:
  C = 4.0 μF = 4.0 × 10⁻⁶ F (capacitance)
  V = 12 V (voltage across the capacitor)
  Q = (4.0 × 10⁻⁶) × (12) = 4.8 × 10⁻⁵ C
• Part (b):
  The energy stored in a capacitor is given by:
  U = (1/2) C V²
  U = (1/2) × (4.0 × 10⁻⁶) × (12)² ≈ 2.9 × 10⁻⁴ J
• Part (c):
  The capacitance of a parallel-plate capacitor is given by:
  C = ε₀ A / d
  Where:
  ε₀ = 8.85 × 10⁻¹² F/m (permittivity of free space)
  A = 1.0 m² (area of the plates)
  d = 0.04 m (new distance between the plates)
  C = (8.85 × 10⁻¹² × 1.0) / 0.04 ≈ 2.2 × 10⁻¹² F = 2.2 pF

Grading Rubric for AP Physics C: Electricity and Magnetism FRQs

1. Electric Field and Potential

• Part (a):
  0-1 points: Incorrect or incomplete calculation of electric field.
  2 points: Correct calculation of electric field with proper formula and units.
• Part (b):
  0-1 points: Incorrect or incomplete calculation of electric potential.
  2 points: Correct calculation of electric potential with proper formula and units.
• Part (c):
  0-1 points: Incorrect or incomplete calculation of the force on a charge.
  2 points: Correct calculation of force with proper formula and units.

2. Magnetic Field and Force

• Part (a):
  0-1 points: Incorrect or incomplete calculation of magnetic force.
  2 points: Correct calculation of magnetic force with proper formula and units.
• Part (b):
  0-1 points: Incorrect or incomplete calculation of radius of circular motion.
  2 points: Correct calculation of radius with proper formula and units.
• Part (c):
  0-1 points: Incorrect or incomplete calculation of period of motion.
  2 points: Correct calculation of period with proper formula and units.

**3

. Capacitors and Energy Storage**

• Part (a):
  0-1 points: Incorrect or incomplete calculation of charge.
  2 points: Correct calculation of charge with proper formula and units.
• Part (b):
  0-1 points: Incorrect or incomplete calculation of energy.
  2 points: Correct calculation of energy with proper formula and units.
• Part (c):
  0-1 points: Incorrect or incomplete calculation of new capacitance.
  2 points: Correct calculation of capacitance with proper formula and units.