This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Electric potential due to multiple point charges is found using superposition: V = Σ(kqi/ri), where the potential is a scalar sum of contributions from each charge. In this scenario, we calculate V = k(q₁/r₁ + q₂/r₂ + q₃/r₃) with the given distances from point P to each charge. Choice A is correct because V = (8.99×10⁹)[(2.0×10⁻⁹/0.30) + (-2.0×10⁻⁹/0.30) + (1.0×10⁻⁹/0.20)] = (8.99×10⁹)[0 + 5.0×10⁻⁹] = 44.95 ≈ 4.50×10¹ V, where the equal positive and negative charges at equal distances cancel. Choice B incorrectly doubles the result, possibly counting the third charge twice. Students should remember that potential is a scalar quantity, so we add algebraically (not vectorially) and must include the sign of each charge. Setting up the calculation systematically and checking that opposite charges at equal distances cancel helps avoid errors.

This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. In a uniform electric field, the potential difference is ΔV = -Ed when moving in the field direction, where d is the displacement along the field. In this scenario, E = 1.2×10³ N/C points upward, and we move from A to B (upward) by d = 0.50 m. Choice A is correct because ΔV = V_B - V_A = -Ed = -(1.2×10³)(0.50) = -600 V = -6.0×10² V. Choice B has the wrong sign, forgetting that potential decreases in the field direction. Students should remember that electric field points from high to low potential. When moving with the field (A to B upward), potential decreases, giving negative ΔV.

This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. For potential difference in a uniform field, we use ΔV = -E⃗·Δr⃗, where the dot product accounts for the relative directions of field and displacement. In this scenario, E⃗ points in -x direction while displacement from A to B is in +x direction (d = 0.060 m), making them antiparallel. Choice B is correct because VB - VA = -E⃗·Δr⃗ = -(-800)(0.060) = +48.0 = +4.80×10¹ V, where the negative signs cancel since field and displacement are opposite. Choice A incorrectly keeps the negative sign, missing that antiparallel vectors give negative dot product. Students should carefully evaluate the dot product: when field and displacement are opposite, E⃗·Δr⃗ is negative, but the minus sign in ΔV = -E⃗·Δr⃗ makes the result positive. Drawing vectors and explicitly writing E⃗·Δr⃗ = |E||Δr|cos(180°) = -|E||Δr| helps track signs.

This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. In parallel-plate capacitors, the uniform field E = σ/ε₀ creates potential difference ΔV = Ed between plates. In this scenario, σ = 3.5×10⁻⁷ C/m² and d = 1.5×10⁻³ m determine the potential difference. Choice A is correct because E = σ/ε₀ = (3.5×10⁻⁷)/(8.85×10⁻¹²) = 3.95×10⁴ N/C, then ΔV = Ed = (3.95×10⁴)(1.5×10⁻³) = 59.3 V ≈ 5.9×10¹ V. Choice C might result from an error in powers of ten. Students should connect this to capacitance: C = ε₀A/d and Q = σA give C = Q/V, confirming the relationship. Always verify units: (C/m²)/(C²/N·m²) = N/C for field strength.

This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. For a uniform electric field, the potential difference between two points is given by ΔV = -E·d when moving in the direction of the field, where E is the field magnitude and d is the displacement. In this scenario, the field E = 2.50×10³ V/m points in +x direction, and we move from A to B along +x by d = 0.040 m. Choice A is correct because VB - VA = -Ed = -(2.50×10³)(0.040) = -100 V = -1.00×10² V, as moving in the direction of the field decreases potential. Choice B incorrectly has a positive sign, forgetting that potential decreases in the field direction. Students should carefully apply the negative sign in ΔV = -E·d and remember that electric field points from high to low potential. Drawing a diagram showing field direction and displacement helps avoid sign errors.

This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Electric potential V = kQ/r is directly proportional to the charge Q when distance r remains constant. In this scenario, the charge is doubled from Q to 2Q while maintaining the same distance r = 0.25 m. Choice A is correct because if V = kQ/r initially, then with doubled charge: V_new = k(2Q)/r = 2(kQ/r) = 2V. Choice B incorrectly inverts the relationship, while choice C suggests quadrupling. Students should recognize that potential has a linear relationship with charge. This proportionality makes superposition possible: doubling the source charge doubles the potential at any fixed point.

This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. The change in electric potential energy is ΔU = qΔV, where q is the charge and ΔV is the potential difference through which it moves. In this scenario, an electron (q = -1.60×10⁻¹⁹ C) moves through ΔV = +250 V. Choice B is correct because ΔU = qΔV = (-1.60×10⁻¹⁹)(+250) = -4.00×10⁻¹⁷ J. Choice A has the wrong sign, forgetting the electron's negative charge. Students should remember that when a negative charge moves to higher potential, it loses potential energy (ΔU < 0). The work done by the field equals -ΔU, so the field does positive work on the electron.

This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Electric potential due to a point charge is V = kQ/r, where the sign of Q determines the sign of the potential - negative charges create negative potentials. In this scenario, Q = -6.0×10⁻⁹ C at distance r = 0.15 m from point P. Choice B is correct because V = (8.99×10⁹)(-6.0×10⁻⁹)/(0.15) = -53.94×10⁰/0.15 = -359.6 ≈ -3.60×10² V, with the negative sign from the negative charge. Choice A incorrectly has a positive sign, forgetting that negative charges create negative potentials. Students should remember that the sign of the potential directly follows the sign of the source charge, unlike electric field which always points away from positive charges. Keeping the charge sign explicit in calculations helps avoid sign errors.

This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. For parallel plates with uniform field, the potential difference magnitude is simply ΔV = Ed, where E is field strength and d is plate separation. In this scenario, E = 6.0×10⁴ N/C and d = 2.0×10⁻³ m allow direct calculation. Choice A is correct because ΔV = Ed = (6.0×10⁴)(2.0×10⁻³) = 12.0×10¹ = 1.2×10² V. Choice B might result from dropping a power of ten, while choice D incorrectly adds exponents. Students should remember that parallel-plate capacitors create uniform fields, making V = Ed exact. Always verify units: (N/C)(m) = (J/C) = V, confirming dimensional consistency.

This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. The work done by an electric field on a charge is W = -qΔV, where ΔV is the potential difference and the negative sign accounts for the field doing work against potential energy. In this scenario, an electron moves upward in an upward field, so ΔV = -Ed = -(1.20×10⁴)(0.015) = -180 V, and W = -(-1.60×10⁻¹⁹)(-180) = -2.88×10⁻¹⁷ J. Choice A is correct because the calculation yields negative work, meaning the field opposes the electron's upward motion (since the field exerts downward force on negative charges). Choice B has the wrong sign, forgetting that upward field pushes electrons downward. Students should carefully track signs: negative charge times negative potential difference gives positive potential energy change, but work by field is negative of this. Drawing force and displacement vectors helps visualize that work is negative when force opposes motion.