Electric potential (V) is the electric potential energy per unit charge at a point in space—basically a “voltage” map that tells you how much potential energy a 1-C test charge would have there (CED 10.5.A.1). Mathematically for point charges: V = (1/4πε0) Σ qi/ri (scalar superposition; CED 10.5.A.2). Electric potential energy (UE) is the actual energy a specific charge q has in that potential: UE = qV. So V is a property of the field/space; UE depends on the test charge you put there. The potential difference ΔV = ΔUE/q tells you how much UE per charge changes when you move a charge between points (CED 10.5.A.3). Remember: V is scalar, adds by superposition; the electric field points toward decreasing V and relates by |E| = |ΔV/Δr| (CED 10.5.B). For AP practice and worked examples, check the Topic 10.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma), unit overview (https://library.fiveable.me/ap-physics-2-revised/unit-10), and plenty of practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Electric potential V at a point is defined as the electric potential energy per unit charge: V = U_E / q. If you put a test charge q into a field, the electric potential energy U_E scales with q (U_E ∝ q), so when you divide by q that factor cancels. That leaves V determined only by the source charges and the position in space. In formula form for point charges: V(r) = (1 / 4πε0) Σ (qi / ri)—notice no test-charge q appears. Practically, that means V is a property of the field (a scalar) at each location—you can use it to get ΔV between two points (ΔV = ΔU_E / q) and predict work per unit charge without worrying about which test charge you use. This is why conductors come to the same potential on their surfaces and why equipotential lines depend only on position and source charges (see Topic 10.5 in the CED). Want extra practice problems or the AP-aligned study guide for Topic 10.5? Check the Fiveable study guide (https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma) and the Unit 10 overview (https://library.fiveable.me/ap-physics-2-revised/unit-10).

Electric potential (V) and electric field (E) are related but different ways to describe the same electrostatic situation. - What each is: V is a scalar that gives electric potential energy per unit charge at a point (units J/C = volts). E is a vector that gives the force per unit charge (units N/C or V/m) and tells you the direction a positive test charge would accelerate. (CED 10.5.A.1) - How they connect: E points in the direction of decreasing V, and for one-dimensional changes |E| ≈ |ΔV/Δr|; more generally E = −∇V. So V tells you how much energy a charge has; E tells you how it will move. (CED 10.5.B.1, 10.5.B.2.iii) - Calculation & superposition: V from point charges is scalar superposition V = (1/4πε0) Σ qi/ri (CED 10.5.A.2). That’s often easier than summing vector E when symmetry helps. - Visual tools: Equipotential lines are perpendicular to E; no work is done moving along an equipotential (CED 10.5.B.2.i–iv). For AP exam focus: be ready to compute V for up to four point charges, use ΔV = ΔUE/q, and translate between E maps and equipotentials (see the Topic 10.5 study guide for examples: (https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma)). For extra practice, check the unit page (https://library.fiveable.me/ap-physics-2-revised/unit-10) and the practice problem bank (https://library.fiveable.me/practice/ap-physics-2-revised).

Use scalar superposition: the total electric potential at a point is the sum of the potentials from each point charge. For N point charges, V = (1 / 4πε0) Σ (qi / ri), where ri is the distance from charge i to the point, and 1/4πε0 = k ≈ 8.99×10^9 N·m^2/C^2. Because V is a scalar you add algebraically (signs matter: negative charges give negative contributions). Usually take V(∞)=0, so ri are finite distances from that point. To get potential difference use ΔV = ΔUE / q (or ΔV = V(B) − V(A)); the electric field relates to the potential gradient: |E| ≈ |ΔV/Δr| for average field. AP only expects you to handle configurations of up to four point charges (or more with symmetry)—remember to plug numbers, keep units, and show the sum explicitly on the exam. For a quick review, check the Topic 10.5 study guide (Fiveable) here: https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma. For more practice, use Fiveable’s AP Physics 2 practice problem set (https://library.fiveable.me/practice/ap-physics-2-revised).

Saying electric potential is a scalar means it has only a magnitude (a number) at each point in space—no direction attached. That matters because: you add potentials algebraically (scalar superposition), so for point charges V = (1/4πε0) Σ qi/ri (CED 10.5.A.2). You don’t have to add vector components like you do for E-fields, which makes calculating V for multiple charges easier. Even though V is scalar, the electric field is a vector: E points in the direction of decreasing V and equals the (negative) spatial gradient of V (CED 10.5.B.1, 10.5.B.2.iii). Practical consequences on the AP: use scalar superposition to find potentials for up to four point charges and then get E from ΔV/Δr or equipotential maps (CED limits). For extra practice on problems and the topic guide, check the Topic 10.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma) and lots of practice questions (https://library.fiveable.me/practice/ap-physics-2-revised).

Equipotential lines are perpendicular to electric field lines because along an equipotential the potential V is constant, so no work is done by the field when you move a test charge along that line. Mathematically, the change in potential dV = −E · dl. If dV = 0 (along an equipotential), then E · dl = 0, so the electric field E has zero component tangent to the equipotential—i.e., E is perpendicular to the line. Equivalently, E = −∇V points in the direction of greatest decrease of V, so it must be normal to surfaces (or lines) of constant V. This is exactly what the CED says: isolines (equipotentials) are perpendicular to field vectors and the field points toward decreasing potential (10.5.B.2.i–iii). For more practice and review on equipotentials and E–V relationships, see the Topic 10.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma) and lots of practice questions at (https://library.fiveable.me/practice/ap-physics-2-revised).

Think of electric potential (V) as “electric potential energy per coulomb” at a point. If a test charge q has electric potential energy U, then V = U/q. Voltage is just a name people use for an electric potential difference (ΔV) between two points—the amount of potential energy per unit charge changed when you move a charge between those points: ΔV = ΔU/q. So voltage = “how much energy per charge you gain or lose” moving between two spots. Useful quick links and formulas from the AP CED you should know: - Potential of point charges: V = (1/4πε0) Σ qi/ri (scalar superposition)—AP expects up to a few point charges (see Topic 10.5) (study guide: https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma). - Relation to field: |E| = |ΔV/Δr| and E points toward decreasing V. - Batteries create a fixed ΔV by separating charges; conductors in contact share the same V. For more practice, see the unit overview (https://library.fiveable.me/ap-physics-2-revised/unit-10) and dozens of practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

If you move a test charge anywhere along an equipotential line, the electric potential at every point on that line is the same—so ΔV = 0. By 10.5.A.3 and 10.5.B.2, that means the change in electric potential energy per unit charge is zero, so ΔU_E = qΔV = 0. The electric field is always perpendicular to equipotentials, so there’s no component of E along the path; the electric field does no work on the charge (W_field = 0). In other words, you don’t need to do work against the electric field to move the charge along an equipotential (assuming quasi-static motion). For more on equipotential lines and their relation to E, check the Topic 10.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma) and try practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

A battery uses chemical (redox) reactions to separate charge and create an electric potential difference (emf). Inside, oxidation at one electrode releases electrons (making that terminal negative) while reduction at the other consumes electrons (making that terminal positive). Those chemical reactions do work to move charges against the internal electric field, so the battery stores electric potential energy per unit charge (V = ΔU_E/q). When you connect a circuit, electrons flow out the negative terminal, through the load doing work, and back into the positive terminal; the internal chemistry then restores the charge separation. On the AP framework this is exactly the chemical-origin case for potential difference in 10.5.A.3.i—remember emf comes from non-electrostatic (chemical) work that maintains ΔV. For a clear AP-aligned review, see the Topic 10.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma) and more unit practice at (https://library.fiveable.me/ap-physics-2-revised/unit-10).

Think of V as electric potential energy per unit charge (CED 10.5.A.1). If a positive test charge moves in the direction of the electric field, the electric force does positive work on it, so its electric potential energy U decreases. Since V = U/q, V also decreases. Mathematically the local relation is E = −dV/dr (CED 10.5.B.1): the minus sign means the field points where V drops fastest. So for a positive charge the field and force are toward lower V; for a negative charge the force is opposite the field (but the field still points toward decreasing V). That’s why equipotential lines are always perpendicular to field lines (CED 10.5.B.2.i–ii): moving along an equipotential doesn’t change V, so the field has no component along it. Want more practice visualizing this with maps of equipotentials and field vectors? Check the Topic 10.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma) and the unit overview (https://library.fiveable.me/ap-physics-2-revised/unit-10).

Electric potential (V) is a scalar that tells you the electric potential energy per unit charge at a single point in space—basically "how much energy a +1 C would have" at that location. For point charges use V = (1/4πε0) Σ qi/ri (CED 10.5.A.1–A.2). Electric potential difference (ΔV) is the change in that quantity between two points: ΔV = ΔUE/q (CED 10.5.A.3). It’s what matters for work and circuits—the field does work equal to q·ΔV when a charge moves. ΔV is also what a battery produces (charge separation). Quick links: review Topic 10.5 (Fiveable study guide) for worked examples (https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma), check the whole unit overview (https://library.fiveable.me/ap-physics-2-revised/unit-10), and practice many problems (https://library.fiveable.me/practice/ap-physics-2-revised). Also remember: equipotential surfaces have the same V everywhere (conductors in contact share the same potential) and E points in the direction of decreasing V with |E| ≈ |ΔV/Δr| (CED 10.5.B).

When two conductors touch, charges flow until there’s no electric field along or inside them. In conductors free electrons move very easily; they redistribute until the electric field everywhere inside the connected conductor is zero. Because the electric field is the spatial rate of change of potential (|E| = |ΔV/Δr| from the CED 10.5.B.1), E = 0 means ΔV = 0 between any two points in the conductor—so every point is at the same electric potential (an equipotential). Practically, that’s why charge spreads until potentials equalize; if any two points had different V, an E-field would exist and charges would move, changing V until it doesn’t. This idea is tested on the AP (Topic 10.5—conductor at equipotential), so be ready to connect “E = 0 inside conductor” to “ΔV = 0 → same potential.” For a quick refresher, check the Topic 10.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma) and try practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Equipotential maps show lines (isolines) of constant electric potential. Key rules to predict motion: - Electric field points perpendicular to equipotentials toward decreasing V; its magnitude ≈ |ΔV/Δr| (closer lines → stronger field) (CED 10.5.B.1–B.2). - A positive test charge feels force in the direction of the electric field (so it moves toward lower potential). A negative charge feels force opposite the field (moves toward higher potential). - No work is done if a charge moves exactly along an equipotential (because E has no component along the line). - If you know V-values, find the direction of decreasing V to get E direction; use line spacing to compare speeds/accelerations qualitatively (tighter spacing → larger acceleration). - Remember conductors are at uniform potential (CED 10.5.A.4). This is exactly the kind of reasoning AP asks you to use on diagrams (Topic 10.5 and 10.5.B in the CED). For practice, review the Topic 10.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma) and try more problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

V = kq/r gives the electric potential V (volts) at a point a distance r from a point charge q. Key things it tells you: - V is electric potential energy per unit charge at that point: U = q_test·V. - It’s a scalar (no direction) and comes from taking work per unit charge to bring a test charge from infinity (where V = 0) to distance r. - Sign matters: if q is positive V is positive (work must be done against the field to bring a positive test charge in); if q is negative V is negative. - For multiple charges use scalar superposition: V_total = (1/4πε0) Σ qi/ri (AP CED 10.5.A.2). - Related to field: E_r = −dV/dr, which for a point charge gives |E| = k|q|/r^2 and points toward decreasing V (CED 10.5.B). Use V = kq/r for up to a few point charges on the AP exam (Unit 10). For review see the Topic 10.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma) and more practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Short answer: the sign of electric potential V at a point comes from the signs of the source charges (V = (1/4πε0) Σ qi/ri). Positive source charges give positive V; negative source charges give negative V. Potential is scalar so you just add contributions. How to think about signs in problems: - V is energy per unit test charge. If a positive test charge q moves to a point of lower V, its electric potential energy U = qV decreases (ΔU = q ΔV). For a negative test charge the sign flips: moving to lower V increases its U. - The electric field points toward decreasing V (E = −∇V). So a positive charge “falls” to lower V (loses U), and a negative charge “climbs” to lower V (gains U). - For potential difference use ΔV = V(b) − V(a) = ΔU/q. Be careful which point is first/second. - Use scalar superposition for up to 4 point charges on the AP (k = 1/4πε0) as in the CED (10.5.A.2). Conductors share the same V (10.5.A.4). For more examples and worked practice aligned to Topic 10.5, check the Fiveable study guide (https://library.fiveable.me/ap-physics-2-revised/unit-2/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma) and the unit overview (https://library.fiveable.me/ap-physics-2-revised/unit-10). For lots of practice problems go to (https://library.fiveable.me/practice/ap-physics-2-revised).