AP Physics multiple choice questions on electrostatics, electric fields, and potential. Each question includes detailed solutions at the bottom of the page.
Two point charges of +3 μC and -4 μC are placed 0.5 m apart. What is the magnitude of the electrostatic force between them? (k = 9×10⁹ N·m²/C²)
A) 0.216 N
B) 0.432 N
C) 0.648 N
D) 0.864 N
E) 1.080 N
What is the electric field at a point midway between a +5 μC charge and a -5 μC charge separated by 0.2 m?
A) 0 N/C
B) 4.5×10⁶ N/C toward positive
C) 4.5×10⁶ N/C toward negative
D) 9.0×10⁶ N/C toward positive
E) 9.0×10⁶ N/C toward negative
A parallel-plate capacitor has plates of area 0.2 m² separated by 0.001 m. What is its capacitance? (ε₀ = 8.85×10⁻¹² C²/N·m²)
A) 1.77 nF
B) 17.7 nF
C) 177 nF
D) 1.77 μF
E) 17.7 μF
A proton (q = +1.6×10⁻¹⁹ C) moves through a potential difference of 1000 V. What is its kinetic energy gain?
A) 1.6×10⁻¹⁶ J
B) 3.2×10⁻¹⁶ J
C) 4.8×10⁻¹⁶ J
D) 6.4×10⁻¹⁶ J
E) 8.0×10⁻¹⁶ J
Three identical point charges q are placed at the vertices of an equilateral triangle of side a. What is the magnitude of the net force on one charge due to the other two?
A) \( \dfrac{k q^2}{a^2} \)
B) \( \sqrt{2} \dfrac{k q^2}{a^2} \)
C) \( \sqrt{3} \dfrac{k q^2}{a^2} \)
D) \( 2 \dfrac{k q^2}{a^2} \)
E) \( 3 \dfrac{k q^2}{a^2} \)
B) 0.432 N
Coulomb's law: \( F = k \dfrac{|q_1 q_2|}{r^2} \)
\( F = \dfrac{(9 \times 10^9) \, |3 \times 10^{-6} \times -4 \times 10^{-6}|}{(0.5)^2} \)
\( = \dfrac{(9 \times 10^9) \, 12 \times 10^{-12}}{0.25} \)
\( = \dfrac{108 \times 10^{-3}}{0.25} = 0.432 \,\text{N} \)
D) 9.0×10⁶ N/C toward negative charge
At midpoint, distance to each charge: \( r = 0.1\,\text{m} \)
Field due to \( +5 \,\mu\text{C} \): \( E_1 = k \dfrac{q}{r^2} = \dfrac{9 \times 10^9 \times 5 \times 10^{-6}}{(0.1)^2} = 4.5 \times 10^6 \,\text{N/C} \) (away from + charge)
Field due to \( -5 \,\mu\text{C} \): \( E_2 = \dfrac{9 \times 10^9 \times 5 \times 10^{-6}}{(0.1)^2} = 4.5 \times 10^6 \,\text{N/C} \) (toward - charge)
Both fields point in the same direction (away from +, toward -), which is toward the negative charge.
Net field: \( E_{\text{net}} = E_1 + E_2 = 9.0 \times 10^6 \,\text{N/C} \) toward negative charge
A) 1.77 nF
\( C = \varepsilon_0 \dfrac{A}{d} = \dfrac{8.85 \times 10^{-12} \times 0.2}{0.001} \)
\( = \dfrac{1.77 \times 10^{-12}}{0.001} = 1.77 \times 10^{-9} \,\text{F} = 1.77 \,\text{nF} \)
A) 1.6×10⁻¹⁶ J
\( \Delta KE = q \Delta V = 1.6 \times 10^{-19} \times 1000 = 1.6 \times 10^{-16} \,\text{J} \)
C) \(\sqrt{3}\, k q^2 / a^2\)
Force from one charge: \( F = \dfrac{k q^2}{a^2} \) at 60° angle
Two forces of equal magnitude \( F \) at 60°
Resultant: \( F_{\text{net}} = \sqrt{F^2 + F^2 + 2 F^2 \cos 60^\circ} = \sqrt{2F^2 + 2 F^2 \cdot 0.5} = \sqrt{3F^2} = F \sqrt{3} = \sqrt{3} \dfrac{k q^2}{a^2} \)