Difficulty: Intermediate | Prerequisites: Charge and Coulomb's law notes
Once you know charges exert forces on each other (Coulomb's law), the next step is to describe the force a charge would feel at any point in space, even before you put a charge there. That is the electric field. From the field, you can derive the electric potential, which tells you about the energy landscape. Fields and potentials are the tools you will use for everything from parallel plate capacitors to ECG readings of the heart. You should be comfortable with Coulomb's law and vector addition before tackling this material.
The electric field describes the force per unit charge at every point in space. Electric potential describes the potential energy per unit charge. The field points from high potential to low potential, and work done by the electric field on a charge equals the charge times the potential difference. These ideas connect force, energy, and voltage in a single framework.
Electric field (E)
The force per unit positive test charge at a point in space. Units: N/C or V/m (these are equivalent). The field is a vector: it has both magnitude and direction.
Think of it as a map of "how hard and which way" a positive charge would be pushed if you placed one there.
Electric field lines
Imaginary lines that show the direction of the electric field at every point. They point away from positive charges and toward negative charges. Closer spacing means a stronger field.
Uniform electric field
An electric field that has the same magnitude and direction at every point. The field between two large parallel plates (far from the edges) is approximately uniform.
Equipotential line (or surface)
A line or surface along which the electric potential is constant. No work is done moving a charge along an equipotential. Equipotential lines are always perpendicular to electric field lines.
Think of equipotentials as contour lines on a topographic map: they show "elevation" (voltage), and the field points "downhill."
Electric potential (V)
The electric potential energy per unit charge at a point. Units: volts (V) = joules per coulomb. A scalar quantity (no direction).
Potential difference (delta V)
The difference in electric potential between two points: delta V = V_final - V_initial. Also called voltage. This is what determines how much work the electric field does on a charge moving between those points.
Work done by the electric field
W = -q(delta V) = -q(V_final - V_initial). When a positive charge moves from high potential to low potential, the field does positive work (the charge gains kinetic energy). When it moves from low to high, the field does negative work.
In simple terms, charges "want" to roll downhill in potential, and the field does the pushing.
Electric dipole moment
A vector quantity describing a pair of equal and opposite charges separated by a small distance. Points from the negative charge toward the positive charge. Used to model the electrical activity of the heart in ECG analysis.
Between two large parallel plates with equal and opposite charge Q, the electric field is approximately uniform and given by E = Q / (epsilon_0 x A), where A is the area of each plate and epsilon_0 = 8.85 x 10^-12 C^2/(N m^2).
The field points from the positive plate toward the negative plate.
A charged particle floating in equilibrium between the plates: the electric force on it balances gravity. Set qE = mg and solve for the unknown (mass, charge, or field strength).
Example: a +35 nC bead floats between plates with Q = 240 nC, plate area 0.12 m x 0.12 m = 0.0144 m^2, separation 5.0 mm. E = 240 x 10^-9 / (8.85 x 10^-12 x 0.0144) = about 1.88 x 10^6 V/m. Then mg = qE, so m = qE/g = (35 x 10^-9)(1.88 x 10^6)/9.8 = about 6.7 x 10^-4 kg = 0.67 g.
The electric field at a point is perpendicular to the equipotential surface passing through that point.
The magnitude of E equals the rate of change of potential with distance: E = |delta V| / d, where d is the perpendicular distance between equipotential lines.
The field points in the direction of decreasing potential.
Example: equipotential lines separated by 1.0 cm with a potential difference of 200 V between them give E = 200 / 0.01 = 2.0 x 10^4 V/m. The direction is perpendicular to the equipotential lines, pointing from higher V toward lower V.
For a parallel plate capacitor with charge Q on plates of area A, the field magnitude at any point between the plates is E = Q / (epsilon_0 x A). The direction is from the +Q plate to the -Q plate.
Example: Q = 0.55 uC, A = 2.0 m^2. E = 0.55 x 10^-6 / (8.85 x 10^-12 x 2.0) = 3.1 x 10^4 V/m, pointing from the positive plate toward the negative plate.
For a uniform field: delta V = -E x d (along the field direction), or equivalently |delta V| = E x d.
The sign depends on direction: moving in the direction of E means potential decreases; moving against E means potential increases.
Example: E = 3.1 x 10^4 V/m, distance between points A and B is 2.5 cm = 0.025 m, both between the plates. delta V = E x d = 3.1 x 10^4 x 0.025 = 780 V. The point closer to the positive plate is at higher potential.
The work done by the electric field on a charge q moving through a potential difference: W_e = -q(delta V) = -q(V_final - V_initial).
A positive charge moving from low V to high V: the field does negative work (the charge loses kinetic energy, or something else must supply energy).
Example: a +15 nC charge moves from V = -25 V to V = +350 V. delta V = 350 - (-25) = 375 V. W = -(15 x 10^-9)(375) = -5.625 x 10^-6 J = -5.6 uJ. The field does negative work because the positive charge moved to higher potential.
If the electric field inside a cell membrane is known and the potential difference across the membrane is known, the thickness is d = |delta V| / E.
Example: E = 1.5 x 10^7 V/m, delta V = 90 mV = 0.090 V. d = 0.090 / (1.5 x 10^7) = 6.0 x 10^-9 m = 6.0 nm.
The heart generates an electric dipole moment that changes direction during each heartbeat.
ECG leads measure voltage differences at different points on the body surface.
V_A = V_3 - V_2 and V_B = V_2 - V_1: the magnitude of each depends on how the dipole moment projects onto the line connecting the measurement points. The lead aligned more closely with the dipole vector records a larger voltage difference.
A charged particle in a uniform electric field experiences a constant force F = qE.
If the particle moves opposite to the force, it decelerates. Use the work-energy theorem: W_net = delta KE.
Alternatively, use kinematics: v^2 = v0^2 + 2a x d, where a = qE/m.
Example: a bead of mass 0.015 kg, charge -750 uC, initial velocity 250 m/s, field 3 x 10^4 N/C. The force on the bead is F = qE = (750 x 10^-6)(3 x 10^4) = 22.5 N (direction depends on charge sign and field direction). Using v^2 = v0^2 - 2(F/m)d with v = 0: d = v0^2 m / (2F) = (250)^2 (0.015) / (2 x 22.5) = 937.5 / 45 = about 21 m.
If a particle slows down (speed decreases), the net work on it is negative. By the work-energy theorem, W = delta KE = (1/2)mv_f^2 - (1/2)mv_i^2 < 0.
The potential difference the particle moves through is related to its charge and the work: delta V = -W / q. Two particles with the same mass and different speed changes experience different amounts of work, and therefore move through different potential differences (unless the charge also differs).
E = \frac{Q}{\varepsilon_0 A}Electric field between parallel plates. Q = charge on one plate, A = plate area, epsilon_0 = 8.85 x 10^-12 C^2/(N m^2).
E = \frac{|\Delta V|}{d}Electric field magnitude from equipotential spacing. d = perpendicular distance between equipotential lines.
W_e = -q \Delta V = -q(V_f - V_i)Work done by the electric field on charge q.
d = \frac{|\Delta V|}{E}Thickness of a membrane (or any region) from field strength and potential difference.
F = qE \quad \text{and} \quad a = \frac{qE}{m}Force and acceleration of a charged particle in a uniform electric field.
v^2 = v_0^2 + 2a \cdot dKinematics for constant acceleration. Use with a = qE/m to find stopping distance or final speed.
The electric field inside a cell membrane (roughly 10^7 V/m across a few nanometres) drives the ion transport that keeps every cell in your body alive. ECG machines exploit the heart's changing dipole moment: electrodes on the skin measure potential differences caused by the heart's electrical activity, and the pattern reveals arrhythmias, blockages, and other conditions.
Students often confuse the electric field (a vector, N/C) with the electric potential (a scalar, V). The field tells you about force, the potential tells you about energy. They are related but not the same thing.
Students sometimes think the electric field points from low potential to high potential. It does not. The field always points in the direction of decreasing potential.
Students frequently get the sign wrong on work done by the field. Remember: W = -q(delta V). If a positive charge moves to higher potential, the field does negative work on it.
Students sometimes assume a uniform field means the potential is the same everywhere. It does not. A uniform field means the potential changes linearly with distance.
Parallel plate field calculations (E = Q / epsilon_0 A) appear in multiple forms: finding the field, finding the mass of a floating bead, finding the voltage between two points.
Equipotential-to-field problems require you to read a diagram, find the spacing and potential difference, and get the field magnitude and direction.
Work done by the field is a favourite exam topic. Watch the sign of q and the direction of delta V.
Membrane thickness problems are a quick calculation but easy to lose marks on if you mix up units (mV vs V, nm vs m).
Conceptual questions about the work-energy theorem applied to charged particles test whether you understand the relationship between speed changes, work, and potential.
True or False: The electric field always points perpendicular to equipotential surfaces. (True)
Fill in the blank: The SI unit of electric potential is the ___, which equals one joule per coulomb. (volt)
True or False: A positive charge moving in the direction of the electric field gains potential energy. (False: it loses potential energy.)
True or False: The electric field between parallel plates depends on the plate separation. (False: E = Q / epsilon_0 A has no dependence on plate separation.)
Fill in the blank: The work done by the electric field on a charge q through a potential difference delta V is W = ___. (-q delta V)
Q: A glass bead with charge +35 nC floats in equilibrium between two parallel plates (square, side length 0.12 m, separation 5.0 mm, charge Q = 240 nC). What is the mass of the bead?
A: E = Q/(epsilon_0 A) = 240 x 10^-9 / (8.85 x 10^-12 x 0.0144) = 1.88 x 10^6 V/m. At equilibrium qE = mg, so m = qE/g = (35 x 10^-9)(1.88 x 10^6) / 9.8 = 6.7 x 10^-3 g, which rounds to about 6.7 x 10^-4 kg. (Note: check your answer against the available options; this corresponds to roughly 0.67 g or, depending on rounding, the closest listed answer.)
Q: A +15 nC charge moves from V = -25 V to V = +350 V. How much work is done on the charge by the electric field?
A: delta V = 350 - (-25) = 375 V. W = -q(delta V) = -(15 x 10^-9)(375) = -5.6 uJ. The field does negative work because the positive charge moved to higher potential.
Q: A cell membrane has an internal field of 1.5 x 10^7 V/m and a potential difference of 90 mV across it. How thick is the membrane?
A: d = delta V / E = 0.090 / (1.5 x 10^7) = 6.0 x 10^-9 m = 6.0 nm.
Q: Equipotential lines are spaced 1.0 cm apart, with a potential difference of 200 V between adjacent lines. What is the magnitude of the electric field at point A?
A: E = delta V / d = 200 / 0.01 = 2.0 x 10^4 V/m. The direction is perpendicular to the equipotential lines, from higher to lower potential.
Q: A parallel plate capacitor has Q = 0.55 uC on plates of area 2.0 m^2. Points A and B are between the plates, separated by 2.5 cm. What is delta V = V_A - V_B?
A: E = 0.55 x 10^-6 / (8.85 x 10^-12 x 2.0) = 3.1 x 10^4 V/m. delta V = E x d = 3.1 x 10^4 x 0.025 = 780 V. (A is closer to the positive plate, so V_A > V_B.)
Q: In Case 1, particle A (mass m) slows from 2v to v over distance x. Is the work done on particle A positive, negative, or zero?
A: Negative. The particle slowed down, so its kinetic energy decreased. By the work-energy theorem, W = delta KE = (1/2)m(v^2) - (1/2)m(2v)^2 = (1/2)m(v^2 - 4v^2) = -(3/2)mv^2 < 0.
Q: In Case 2, particle B (same mass m) slows from 4v to v over the same distance x in the same field. Is |delta V| for particle A greater than, less than, or equal to |delta V| for particle B?
A: Less than. Particle B has a larger change in kinetic energy: delta KE_B = (1/2)m(v^2 - 16v^2) = -(15/2)mv^2, compared to -(3/2)mv^2 for A. Since both particles travel the same distance in the same field, the work done is the same in magnitude for both if they have the same charge. But |delta V| = |W|/|q|, and if both have the same charge, B's larger |W| over the same distance means |delta V| for A is less than for B. (If the charges differ, the answer depends on the charge ratio, but the exam expects the same-charge comparison.)
Electric fields connect directly to capacitors (the next topic): a capacitor stores energy in the electric field between its plates. The potential difference across a capacitor drives current in circuits. The work-energy ideas here are the bridge between electrostatics and circuit analysis, where voltage differences drive charge flow.
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