Calculate electric field strength and the force on a charge in a field, and account for electric potential energy stored in the field of a system

Learning goal: Define electric field strength as force per unit charge, give its direction, add fields from several charges as vectors, draw field-arrow diagrams for single charges, dipoles, pairs of like charges and parallel plates, solve for the acceleration, final velocity and energy change of a charged particle in a constant field using Newton's laws and the work-energy relation, and explain when electric potential energy of a system increases or decreases, comparing it with the gravitational case.

Field strength: force per unit charge. The electric field of a charged object exists at every point around it whether or not anything is there to feel it. Its strength at a location is defined by what it would do to a charge placed there: E = F_e/q, the electric force per unit charge experienced by a charged object at that point, in newtons per coulomb (N/C). The equation can be rearranged to find the field from a measured force, the force on a known charge in a known field (F = qE), or the charge from a force and a field. The direction of the field at a point is the direction of the electric force on a positive charge placed there; a negative charge feels a force opposite to the field. Greater field strength means larger forces on any charge placed in it. Around a single point charge Q the field has magnitude E = k|Q|/r² (since the force has size F = k|Qq|/r² and E = F/|q|; k is the vacuum value, nearly the same in air), pointing radially outward from a positive charge and inward toward a negative one, and weakening with the square of the distance. Where several charges are present, the field at a point is the vector sum of the fields of the individual charges: superposition. This can be explored with a field simulation, placing charges and reading the field arrow at chosen points.

Field-arrow diagrams. In this course a field is represented by plotting a field arrow at a series of locations, each arrow pointing along the field there and drawn longer where the field is stronger. (Continuous field-line diagrams, in which lines never cross and their density shows strength, are a different representation and are not used here.) For a positive point charge the arrows point outward and shorten with distance; for a negative charge they point inward. For a dipole, a positive and a negative charge close together, the arrows leave the positive charge, curve around and enter the negative one, and the field is strongest between them. For two equal positive charges the arrows push away from both and cancel at the midpoint (for unequal like charges the zero point lies closer to the smaller charge), where the two contributions are equal and opposite; two negative charges give the mirror image. Between two large parallel plates carrying opposite charges the arrows are all the same length and point straight from the positive plate to the negative plate: a uniform field, the simplest field to calculate with, as the diagram shows.

A charged particle in a uniform field: Newton's laws and energy. In a uniform field a charge q feels a constant force F = qE, so it has a constant acceleration a = qE/m, and the kinematic equations apply. Worked model: a proton (q = 1.60 × 10⁻¹⁹ C, m = 1.67 × 10⁻²⁷ kg) is released from rest in a uniform field of 2.0 × 10⁴ N/C. Force: F = (1.60 × 10⁻¹⁹)(2.0 × 10⁴) = 3.2 × 10⁻¹⁵ N along the field. Acceleration: a = 3.2 × 10⁻¹⁵/1.67 × 10⁻²⁷ = 1.9 × 10¹² m/s². After moving 0.050 m: v² = 2ad = 2(1.9 × 10¹²)(0.050), v = 4.4 × 10⁵ m/s. The work done by the field over that distance, W = Fd = (3.2 × 10⁻¹⁵)(0.050) = 1.6 × 10⁻¹⁶ J, equals the proton's kinetic energy gain, ½mv² = 1.6 × 10⁻¹⁶ J, which is the work-energy check. An electron in the same field accelerates the opposite way and about 1800 times faster, because its mass is much smaller while the force has the same size.

Equilibrium of electric and gravitational forces: the oil-drop idea. Millikan measured the charge of the electron by suspending tiny charged oil drops between horizontal plates: when the upward electric force balanced the weight, qE = mg, so q = mg/E. Every measured charge was, within the measurement uncertainty, a whole-number multiple of 1.60 × 10⁻¹⁹ C, showing that charge comes in units of the electron charge e. Worked model: a drop of mass 2.0 × 10⁻¹⁵ kg hangs motionless in an upward field of 6.13 × 10⁴ N/C. q = (2.0 × 10⁻¹⁵)(9.8)/(6.13 × 10⁴) = 3.2 × 10⁻¹⁹ C, which is 2e. Because the field points up and the electric force on the drop must also point up, the charge is positive: the drop has lost two electrons. (In an apparatus where the field points down, a drop held still must be negative, carrying extra electrons.)

Electric potential energy belongs to the system. Suppose two charges that attract or repel each other are let go. Both speed up (or slow down), yet neither one hands its energy to the other; the energy comes from, or goes into, the field that surrounds the pair, stored there as electric potential energy. That energy belongs to the pair of interacting charges, not to either charge by itself. When the distance between the charges changes, the system's electric potential energy changes. Pushing two like charges closer together, or pulling two opposite charges farther apart, requires work against the electric force, and that energy is transferred into the field. Letting two opposite charges move closer together, or two like charges move farther apart, transfers energy out of the field, usually as kinetic energy. This mirrors gravity: lifting a mass (separating two attracting bodies) stores energy in the gravitational field, and the mass gains kinetic energy as it falls back. The difference is that gravity only attracts, whereas with like charges it is bringing them together, not separating them, that stores energy. Charging by friction pulls electrons away from the positive charges they were next to, and pulling opposite charges apart takes work; if nothing speeds up and no energy escapes, all of that work ends up as extra electric potential energy of the system.

Potential difference: energy per unit charge. Because the energy stored depends on how much charge is moved, it is convenient to divide it out. The potential difference ΔV between two points is the change in electric potential energy per unit charge moved between them, ΔV = ΔE/q, measured in volts (1 V = 1 J/C). In a uniform field, when a positive charge q moves a distance d in the direction of the field, the field does work qEd on it and the system's electric potential energy falls by qEd (ΔU = −qEd); the electric potential falls in the direction of the field, ΔV = −Ed, so the size of the potential difference is Ed; a field of 600 N/C across a 0.020 m gap corresponds to 12 V, and a +2.0 µC charge that moves across it from the positive plate toward the negative plate lowers the stored energy by (2.0 × 10⁻⁶)(12) = 2.4 × 10⁻⁵ J (ΔU = −2.4 × 10⁻⁵ J); moving it the other way, against the field, raises the stored energy by the same amount. This is the same potential difference that batteries and circuits are described with in the next lesson.

Misconception: the field is only there when something is in it. The field of a charged object is always present; the test charge merely reveals it. A second misconception is that a negative charge "has negative energy": energy belongs to the system of charges and is defined by changes in separation, not to the sign of one charge.

Investigation: fields and forces without high voltage. Materials: a field simulation on a computer; a charged balloon; a small ball of aluminum foil on a thread; a ruler. Steps: in the simulation, place one charge and read the field arrow at 1, 2 and 3 grid units; place a second charge of the same and then opposite sign and observe where the arrows cancel or strengthen; add a test charge and record its acceleration direction. Then hang the foil ball, bring the charged balloon near from several directions and note the direction and size of the deflection at 10, 5 and 3 cm. A result that supports the model: the simulated field falls to a quarter at double the distance; the arrows between opposite charges are longest; the foil ball always deflects along the line to the balloon and more strongly when closer. What the hands-on model cannot show: the field's numerical value, or the potential energy; the simulation shows field arrows but not the energy stored in the field. Safety: no static generators or high-voltage supplies; this lesson uses balloons and software only.

Worked example 1 — field from a measured force. A +4.0 µC test charge feels a 0.020 N force to the east at a point. E = F/q = 0.020/(4.0 × 10⁻⁶) = 5.0 × 10³ N/C, directed east (the force on a positive charge gives the field's direction). A −2.0 µC charge placed at the same point would feel F = qE = (2.0 × 10⁻⁶)(5.0 × 10³) = 0.010 N, directed west.

Worked example 2 — a proton in a uniform field. E = 2.0 × 10⁴ N/C, from rest, 0.050 m: F = 3.2 × 10⁻¹⁵ N, a = 1.9 × 10¹² m/s², v = 4.4 × 10⁵ m/s, ΔKE = W = Fd = 1.6 × 10⁻¹⁶ J. The system's electric potential energy fell by 1.6 × 10⁻¹⁶ J as the proton moved along the field, and the kinetic energy rose by the same amount.

Worked example 3 — an electron between plates. Plates 0.020 m apart with 12 V across them give E = ΔV/d = 600 N/C. An electron released at the negative plate: F = (1.60 × 10⁻¹⁹)(600) = 9.6 × 10⁻¹⁷ N toward the positive plate (opposite to the field), a = 9.6 × 10⁻¹⁷/9.11 × 10⁻³¹ = 1.1 × 10¹⁴ m/s², and at the positive plate v = √(2ad) = √(2 × 1.1 × 10¹⁴ × 0.020) = 2.1 × 10⁶ m/s. Energy check: the electron (q = −1.60 × 10⁻¹⁹ C) moves through a potential rise of ΔV = +12 V, so the system's potential energy changes by ΔU = qΔV = (−1.60 × 10⁻¹⁹)(12) = −1.9 × 10⁻¹⁸ J, and the kinetic energy gained is ΔK = −ΔU = +1.9 × 10⁻¹⁸ J = ½mv². The field did positive work on the electron; the system's potential energy fell.

Worked example 4 — balancing weight with an electric force. A drop of mass 2.0 × 10⁻¹⁵ kg is held stationary by an upward 6.13 × 10⁴ N/C field: qE = mg gives q = 3.2 × 10⁻¹⁹ C = 2e, positive because the electric force must point up, along the field. Doubling the field would make the electric force twice the weight, so the drop would accelerate upward at a = (qE − mg)/m = g = 9.8 m/s²; the equilibrium condition is what makes the measurement possible.

Worked example 5 — energy into or out of the field. Two protons are pushed from 1.0 nm apart to 0.5 nm apart: like charges brought closer, so work is done on the system and energy goes into the field. An electron and a proton drift from 1.0 nm to 0.5 nm apart: opposite charges approaching, so energy leaves the field and appears as kinetic energy. Compare gravity: a falling apple (attracting masses approaching) also releases field energy as kinetic energy.

Representation task: six field-arrow diagrams. On one page draw field arrows at at least eight points around (a) a single positive charge, (b) a single negative charge, (c) a dipole, (d) two equal positive charges, (e) two equal negative charges and (f) two parallel plates, making arrow length show relative strength and marking any point where the field is zero. Under each, write where a positive test charge would be pushed and whether moving it closer to the source stores energy in the field or releases it. Then add an energy bar chart for the proton worked example at the start and end of its 0.050 m run, with 'electric potential energy of the system' and 'kinetic energy' bars totaling the same.

Independent performance: a charged particle problem set with a simulation check. Solve, with a labeled diagram and units for each: (1) the field 0.30 m from a +5.0 µC point charge (5.0 × 10⁵ N/C) and the force there on a −2.0 µC charge (1.0 N, toward the positive charge); (2) an electron released from rest in a 500 N/C field, its acceleration (8.8 × 10¹³ m/s², opposite to the field) and its speed after 0.010 m; (3) the size and sign of the charge on a 3.0 × 10⁻¹⁵ kg drop held stationary by a 1.84 × 10⁵ N/C field that points downward (1.6 × 10⁻¹⁹ C in size, one elementary charge; negative, because the electric force must point up, against the field); (4) the potential difference and the energy change when a +3.0 µC charge moves 0.040 m along a 2000 N/C uniform field (a potential drop of 80 V in the direction of the field; ΔU = −2.4 × 10⁻⁴ J, out of the field). Success criteria: every field and force has a direction; the work-energy check is shown for (2); each answer in which a charge moves, (2) and (4), states whether energy went into or out of the field; and one paragraph compares the gravitational and electric cases, explaining the one situation (like charges) that has no gravitational parallel.