Charges and Fields

Theory — Electric Field and Electric Potential

Coulomb's Law

Two point charges exert equal and opposite forces on each other along the line joining them. The magnitude of the force is proportional to the product of the charges and inversely proportional to the square of the distance between them.

Coulomb's Law F = k · |q₁ · q₂| / r²
k = 8.99 × 10⁹ N·m²/C² (Coulomb constant)
1 nC = 10⁻⁹ C

Like charges repel · opposite charges attract

The Electric Field

Rather than describing the force between every pair of charges, we say each charge sets up an electric field in the space around it. The field is the force that a small positive test charge would feel per unit charge. It is a vector — it has both magnitude and direction at every point.

Electric Field of a Point Charge E = F / q_test = k · q / r²

Direction: away from a positive charge,
toward a negative charge
Units: newtons per coulomb (N/C) = volts per metre (V/m)

E falls off as 1/r² → E · r² = k·q = constant

Superposition

When several charges are present, the total field at any point is the vector sum of the fields from each individual charge. Fields add as arrows: you find the field from each charge separately, then add them head-to-tail. This is the single most important idea in the lab — it lets a few simple point charges build up complex field patterns.

Superposition Principle E_total = E₁ + E₂ + E₃ + … (vector sum)

Field Lines

Field lines are a way of drawing the field. The line points in the direction of E at every point, and the lines are packed more densely where the field is stronger. Field lines begin on positive charges and end on negative charges (or run off to infinity). They never cross, because the field has only one direction at each point.

Electric Potential

The electric potential V is the electric potential energy per unit charge. Unlike the field, potential is a scalar — just a number at each point, with no direction — so potentials add by ordinary addition, not vector addition. This makes V much easier to compute than E.

Potential of a Point Charge V = k · q / r

Total: V_total = Σ (k·qᵢ / rᵢ) (scalar sum, keep the sign of each q)
Units: volts (V) = joules per coulomb (J/C)

V falls off as 1/r → V · r = k·q = constant

Equipotential Surfaces

An equipotential is a line (in 2-D) or surface (in 3-D) along which the potential is the same everywhere. Because no work is done moving a charge along an equipotential, the electric field can have no component along the equipotential. The field must therefore always be perpendicular to the equipotential lines. The field also points from high potential toward low potential — "downhill."

Field–Potential Relationship E = − dV / dr (field points down the steepest slope of V)
For a point charge: −d/dr (k·q/r) = k·q/r² = E ✓

Field lines ⟂ equipotential lines, everywhere

Field — vector

Has direction. Adds as arrows (vector sum). Falls off as 1/r². Drawn as field lines and arrows. Points away from + and toward −.

Potential — scalar

Just a number (with sign). Adds by ordinary addition. Falls off as 1/r. Drawn as equipotential lines. High near +, low near −.

Quantity	Symbol	Point charge	Distance law	Units
Force	F	k q₁q₂/r²	1/r²	N
Field	E	k q/r²	1/r²	N/C (V/m)
Potential	V	k q/r	1/r	V (J/C)

Instructions — Running the Virtual Experiment

Four experiments build from a single charge to multi-charge fields. Record every reading in your lab notebook and complete the data table in the example report.

Experiment 1 — Distance Laws (Measurements tab)

Open Simulation → Measurements. A single +2 nC charge is fixed at the centre and a movable sensor (voltage probe) sits on the axis to its right.

Click each distance button (5, 10, 15, 20, 25 cm) to place the sensor, then click Record reading. The measured field E and potential V are logged into the table.

After taking all five readings, click Compute k·q & check. The table fills the columns E·r² and V·r. Confirm that both columns are constant and that they are equal to each other — this common value is k·q.

In your notebook, state the value of k·q you obtained and confirm it matches k·q = (8.99×10⁹)(2×10⁻⁹) = 17.98 N·m²/C. This verifies E ∝ 1/r² and V ∝ 1/r at the same time.

Experiment 2 — Single-Charge Field Map (Field Explorer tab)

Open Field Explorer, click Clear, then Add + charge. Turn on Field vectors and Equipotentials.

Confirm the field arrows point radially outward and the equipotential lines form concentric circles. Drag the sensor around a single equipotential circle and verify V stays constant while the field direction always crosses it at 90°.

Experiment 3 — The Dipole (Field Explorer tab)

Click the Dipole preset (one + and one − charge). Turn on Field lines.

Observe that field lines leave the + charge and arrive at the − charge. Drag the sensor to the point exactly halfway between the two charges (on the perpendicular bisector) and record the potential — it should read V ≈ 0, because the equal-and-opposite contributions cancel.

Experiment 4 — Two Like Charges & the Null Point (Field Explorer tab)

Click the Two + charges preset. Drag the sensor to the midpoint between them.

Record the field magnitude at the midpoint. By symmetry the two fields cancel, so you should find a null point (neutral point) where |E| ≈ 0 — even though the potential there is not zero (potentials add, they don't cancel for like charges). Note both readings.

Simulation — Electric Field & Potential Mapper

positive charge

negative charge

field vector

sensor

Add charges

Display

Field vectors Equipotential lines Potential heatmap Field lines

Sensor readout

Potential V— V

Field |E|— N/C

Direction—

Nearest r— cm

+2 nC source (fixed)

sensor

Distance r is read along the ruler from the charge to the sensor.

Place sensor at distance r

Sensor readout

Distance r— cm

Field |E|— N/C

Potential V— V

r (cm)	E measured (N/C)	V measured (V)	E·r² (N·m²/C)	V·r (V·m)
No readings yet — place the sensor and click "Record reading".

Team Questions

Question 1. A +2 nC point charge sits in empty space. Using E = kq/r², calculate the magnitude of the electric field at a distance of 0.10 m from the charge. (Type just the number in N/C, e.g. 1798)

Question 2. For the same +2 nC charge, calculate the electric potential at that same point 0.10 m away, using V = kq/r. (Type just the number in volts, e.g. 179.8)

Question 3. In which direction does the electric field point in the space around an isolated negative charge — toward the charge or away from it? (Answer "toward" or "away")

Question 4. Why must electric field lines always cross equipotential lines at right angles? Answer in terms of the work done moving a charge along an equipotential. (One word for the key idea — what is the component of E along an equipotential?)

Question 5. Two identical +1 nC charges are placed 0.20 m apart. Using superposition, what is the magnitude of the net electric field at the exact midpoint between them? (Type just the number in N/C)

Question 6. For those same two +1 nC charges 0.20 m apart, what is the electric potential at the midpoint? Remember potentials add as scalars and do not cancel. (Type just the number in volts, e.g. 179.8)

Question 7 — Challenge. A +q charge and a +4q charge are placed a distance d apart. The null point (where the net field is zero) lies on the line between them. Is the null point closer to the smaller charge (+q) or the larger charge (+4q)? (Answer "smaller" or "larger")

Example Lab Report

Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission.

Charges and Fields: Electric Field and Electric Potential of Point Charges

Physics | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To map the electric field and electric potential produced by point charges and to verify the distance laws for each: that the field of a point charge falls off as 1/r² and the potential falls off as 1/r. The lab also confirms the principle of superposition for multiple charges and the geometric rule that field lines are everywhere perpendicular to equipotential lines.

Theory

A point charge q produces an electric field of magnitude E = kq/r² directed radially (outward for a positive charge, inward for a negative one) and an electric potential V = kq/r, where k = 8.99 × 10⁹ N·m²/C². The field is a vector and obeys superposition by vector addition; the potential is a scalar and obeys superposition by ordinary signed addition.

E = k·q / r²  (vector, 1/r²)
V = k·q / r   (scalar, 1/r)
E = − dV/dr  (field is the negative slope of potential)

Multiplying through by the appropriate power of r gives two quantities that should be constant for a single charge: E·r² = kq and V·r = kq. Because both equal kq, the two products must also equal each other — a strong internal check on the measurements.

Calculations — Sample: +2 nC charge at r = 0.10 m

k·q (reference): (8.99 × 10⁹)(2 × 10⁻⁹) = 17.98 N·m²/C

Field: E = kq/r² = 17.98 / (0.10)² = 17.98 / 0.0100 = 1798 N/C

Potential: V = kq/r = 17.98 / 0.10 = 179.8 V

Check: E·r² = 1798 × 0.0100 = 17.98 ✓ and V·r = 179.8 × 0.10 = 17.98 ✓ (both equal k·q)

Results Table — Single +2 nC Charge (k·q = 17.98 N·m²/C)

r (cm)	r (m)	E measured (N/C)	V measured (V)	E·r² (N·m²/C)	V·r (V·m)
5	0.05	7192	359.6	17.98	17.98
10	0.10	1798	179.8	17.98	17.98
15	0.15	799.1	119.9	17.98	17.98
20	0.20	449.5	89.9	17.98	17.98
25	0.25	287.7	71.9	17.98	17.98

Superposition checks: at the midpoint of two equal +1 nC charges 0.20 m apart, the measured field was ≈ 0 N/C (the two fields cancel) while the potential was ≈ 179.8 V (the two potentials add). For the dipole, the potential on the perpendicular bisector was ≈ 0 V.

Discussion

The measured field decreased by a factor of four each time the distance doubled (1798 N/C at 10 cm versus 449.5 N/C at 20 cm), exactly as the inverse-square law predicts. The product E·r² was constant at 17.98 for every distance, and the product V·r was also constant at 17.98. The fact that these two independently measured products agree confirms both distance laws at once and recovers the value of k·q directly from the data.

The superposition results behaved as the theory requires. For two like charges the field vectors at the midpoint point in opposite directions and cancel, giving a null point, yet the scalar potentials simply add and remain large and positive. For the dipole the potential vanishes on the perpendicular bisector because the positive and negative contributions are equal in size and opposite in sign. Throughout the field map, the field arrows crossed the equipotential lines at right angles, confirming E = −dV/dr: the field points along the steepest downhill direction of the potential, which is always perpendicular to a line of constant potential.

Conclusion

The experiment successfully mapped the electric field and potential of point charges and verified their distance laws. The field obeyed E ∝ 1/r² and the potential obeyed V ∝ 1/r, with E·r² and V·r both constant and equal to k·q = 17.98 N·m²/C. Superposition was confirmed for both vector fields and scalar potentials, and equipotential lines were shown to be everywhere perpendicular to the field. These results validate the point-charge model of the electric field and the relationship between field and potential.

Practice Questions

Show all work and include units in your answers.

Question 1

A −3 nC point charge is isolated in space. Find the magnitude and direction of the electric field 0.15 m from the charge, and the electric potential at that point.

Hint: E = k|q|/r² (direction toward a negative charge); V = kq/r (keep the negative sign for V).

Question 2

At a point 0.20 m from a point charge the field is 900 N/C. What is the field at 0.40 m and at 0.10 m from the same charge? Do not solve for the charge first — use the inverse-square scaling directly.

Hint: doubling r divides E by 4; halving r multiplies E by 4.

Question 3

Two charges, +4 nC and −4 nC, are placed 0.10 m apart (a dipole). Calculate the electric potential at the point exactly halfway between them. Then explain why the field there is not zero even though the potential is.

Hint: V is a scalar sum, so the +4 and −4 contributions cancel. The field vectors, however, point the same way between the charges and add.

Question 4

Two +5 nC charges are 0.30 m apart. Find the net electric field and the potential at the midpoint between them.

Hint: by symmetry the two field vectors cancel at the midpoint (E = 0), but the potentials add. Each charge is 0.15 m from the midpoint.

Question 5

A test charge is moved along a path that follows an equipotential line from point A to point B. How much work is done by the electric force? If it is then moved directly from B to a point at higher potential, is positive or negative work done by the field?

Hint: W = qΔV. Along an equipotential ΔV = 0. Moving to higher potential means ΔV > 0.

Question 6 — Challenge

A +q charge is at x = 0 and a +4q charge is at x = 0.30 m. Locate the point on the x-axis between them where the net electric field is zero.

Hint: set kq/x² = k(4q)/(0.30−x)². Take the square root of both sides to get (0.30−x) = 2x, then solve. The null point sits closer to the smaller charge.