Source: Charges and Fields Lab (Part III), Dr. Howard
Difficulty: Intermediate | Prerequisites: Electric Fields and Field Lines study notes (Part 1), comfort with 2D vectors, Pythagorean theorem, basic trigonometry.
Knowing that electric fields exist is one thing. Calculating them at specific points is where the physics becomes quantitative and where exam marks live. This set of notes walks through the method for computing E-field vectors from point charges, both individually and via superposition. The procedure is mechanical once you have it down: find the distance, find the unit vector, plug into the formula, add components. If you can do this for two charges, you can do it for any number of them.
To find the electric field at a point from one or more charges: (1) find the displacement vector from the charge to the point, (2) compute the distance and unit vector, (3) plug into E = kQ/r² · r̂, (4) repeat for each charge, (5) add the vectors component by component. Sign of Q handles direction automatically for negative charges.
Displacement vector (r)
The vector from the source charge to the field point. If the charge is at position A and the field point is at position B, the displacement is B - A.
In simple terms, it is the arrow you draw from the charge to the spot where you want to know the field.
Resultant vector
The single vector that represents the sum of two or more individual vectors. In this context, E_net, the total electric field after superposition.
Component form
Expressing a vector as ⟨Eₓ, Eᵧ⟩ rather than as a magnitude and angle. Component form makes vector addition straightforward: just add the x-parts and the y-parts separately.
Null point (zero-field point)
A location where the net electric field from all charges sums to exactly zero. For two equal like charges, this is the midpoint between them.
Every single-charge field calculation follows the same recipe:
Write the formula: E = kQ/r² · r̂
Find the displacement vector from the charge to the field point and compute r (the distance) and r̂ (the unit vector).
Substitute the values of Q, k, and r.
Multiply the scalar result by r̂ to get the vector.
Write the answer in component form ⟨Eₓ, Eᵧ⟩ with units N/C.
Field point (0 m, 0.5 m)
Displacement: ⟨0, 0.5⟩. Distance r = 0.5 m. Unit vector r̂ = ⟨0, 1⟩.
E = (8.99 × 10⁹)(1 × 10⁻⁹) / (0.5)² · ⟨0, 1⟩ = 8.99/0.25 · ⟨0, 1⟩ = 35.96 · ⟨0, 1⟩ = ⟨0, 35.96⟩ N/C.
The field points straight up, away from the positive charge. This makes sense.
Field point (1 m, 0 m)
Displacement: ⟨1, 0⟩. Distance r = 1 m. Unit vector r̂ = ⟨1, 0⟩.
E = 8.99/1 · ⟨1, 0⟩ = ⟨8.99, 0⟩ N/C.
Farther away, weaker field (compare 35.96 at 0.5 m vs 8.99 at 1 m). The 1/r² dependence is visible.
Field point (1 m, 1 m)
Displacement: ⟨1, 1⟩. Distance r = √2 m. Unit vector r̂ = ⟨1/√2, 1/√2⟩ ≈ ⟨0.707, 0.707⟩.
E = 8.99/2 · ⟨0.707, 0.707⟩ = 4.495 · ⟨0.707, 0.707⟩ = ⟨3.178, 3.178⟩ N/C.
Field point (-0.5 m, -0.5 m)
Displacement: ⟨-0.5, -0.5⟩. Distance r = √0.5 ≈ 0.707 m. Unit vector r̂ = ⟨-0.707, -0.707⟩.
E = 8.99/0.5 · ⟨-0.707, -0.707⟩ = 17.98 · ⟨-0.707, -0.707⟩ = ⟨-12.71, -12.71⟩ N/C.
Negative components mean the field points into the third quadrant, away from the positive charge at the origin. Correct.
Field point (0 m, 1 m)
Displacement: ⟨0, 1⟩. Distance r = 1 m. Unit vector r̂ = ⟨0, 1⟩.
E = (8.99 × 10⁹)(-2 × 10⁻⁹) / 1² · ⟨0, 1⟩ = -17.98 · ⟨0, 1⟩ = ⟨0, -17.98⟩ N/C.
The negative sign on Q flips the direction: the field points downward, toward the negative charge. This is the key difference from the positive-charge case.
Field point (0.5 m, 0 m)
Displacement: ⟨0.5, 0⟩. Distance r = 0.5 m. Unit vector r̂ = ⟨1, 0⟩.
E = (8.99 × 10⁹)(-2 × 10⁻⁹) / 0.25 · ⟨1, 0⟩ = -71.92 · ⟨1, 0⟩ = ⟨-71.92, 0⟩ N/C.
Field points in the -x direction, toward the charge.
Field point (0 m, 0 m), the midpoint
Charge 1 at (-0.5, 0): displacement to origin is ⟨0.5, 0⟩, r = 0.5 m, r̂ = ⟨1, 0⟩. E₁ = 35.96 · ⟨1, 0⟩ = ⟨35.96, 0⟩.
Charge 2 at (0.5, 0): displacement to origin is ⟨-0.5, 0⟩, r = 0.5 m, r̂ = ⟨-1, 0⟩. E₂ = 35.96 · ⟨-1, 0⟩ = ⟨-35.96, 0⟩.
E_net = ⟨35.96 - 35.96, 0⟩ = ⟨0, 0⟩ N/C. Perfect cancellation by symmetry.
Field point (0 m, 1 m)
Charge 1 at (-0.5, 0): displacement to (0, 1) is ⟨0.5, 1⟩. Distance r = √(0.25 + 1) = √1.25 ≈ 1.118 m. r̂ = ⟨0.4472, 0.8944⟩.
E₁ = 8.99/1.25 · ⟨0.4472, 0.8944⟩ = 7.192 · ⟨0.4472, 0.8944⟩ ≈ ⟨3.216, 6.432⟩.
Charge 2 at (0.5, 0): displacement to (0, 1) is ⟨-0.5, 1⟩. Same distance by symmetry. r̂ = ⟨-0.4472, 0.8944⟩.
E₂ = 7.192 · ⟨-0.4472, 0.8944⟩ ≈ ⟨-3.216, 6.432⟩.
E_net = ⟨0, 12.864⟩ N/C. The x-components cancel (symmetry), the y-components add. The net field points straight up.
Field point (0 m, 0 m), the midpoint
Charge 1 (+1 nC): displacement ⟨0.5, 0⟩, E₁ = ⟨35.96, 0⟩ N/C (away from positive charge).
Charge 2 (-1 nC): displacement ⟨-0.5, 0⟩, but Q is negative, so E₂ = -35.96 · ⟨-1, 0⟩ = ⟨35.96, 0⟩ N/C (toward negative charge, which is also the +x direction).
E_net = ⟨71.92, 0⟩ N/C. Both fields point in the same direction. This is why unlike charges reinforce between them.
Electric field from a point charge: E = kQ / r² · r̂
Coulomb's constant: k = 8.99 × 10⁹ N·m²/C²
Distance: r = √(Δx² + Δy²)
Unit vector: r̂ = ⟨Δx/r, Δy/r⟩
Superposition: E_net = E₁ + E₂ + ... + Eₙ (vector sum, add components separately)
Magnitude of result: |E_net| = √(Eₓ² + Eᵧ²)
Direction of result: θ = arctan(Eᵧ / Eₓ), measured from the +x axis
Forgetting the sign of Q for negative charges. The negative sign is not cosmetic. It flips the direction of the field vector. If you drop it, your field points away from a negative charge instead of toward it.
Using the wrong r. The distance r is measured from the source charge to the field point, not from the origin (unless the charge happens to sit at the origin). When charges are at positions other than (0, 0), compute the displacement first.
Adding magnitudes instead of vectors. Two fields of 10 N/C do not automatically give 20 N/C. They give 20 N/C only if they point in the same direction. In general, you must add x-components and y-components separately.
Confusing the unit vector direction. r̂ always points from the charge toward the field point. If you reverse it (pointing from the field point toward the charge), every answer will be backwards.
⚠️ The five-step method (formula, displacement, distance and unit vector, substitute, component form) will carry you through every point-charge field problem. Practise it until it is automatic.
⚠️ Expect at least one problem with two charges where you must add field vectors. The examiners love placing the field point at the midpoint or on the perpendicular bisector because symmetry zeros out a component, and students who add magnitudes instead of vectors get it wrong.
⚠️ Sign errors on Q are the single most common mark lost. Double-check: positive charge, field points away. Negative charge, field points toward.
⚠️ When working with charges not at the origin, the displacement vector is (field point) minus (charge position), not the other way round. Getting this backwards flips every answer.
True or false: To find the unit vector, divide the displacement vector by the distance. (True.)
Fill in the blank: If a -3 nC charge is at the origin, the electric field at any point will point ____ the origin. (Toward.)
True or false: When two equal positive charges are 1 m apart, the net field at a point 2 m above the midpoint is zero. (False. The y-components add; only the x-components cancel by symmetry.)
Fill in the blank: The electric field strength falls off as 1/____. (r².)
True or false: At the midpoint between a +1 nC and a -1 nC charge, the net field is zero. (False. The fields reinforce; E_net is nonzero and points from + toward -.)
Q: A -2 nC charge is at the origin. Find the electric field at (0.5 m, -1 m).
A: Displacement: ⟨0.5, -1⟩. Distance r = √(0.25 + 1) = √1.25 ≈ 1.118 m. Unit vector r̂ ≈ ⟨0.4472, -0.8944⟩. E = (8.99 × 10⁹)(-2 × 10⁻⁹)/1.25 · ⟨0.4472, -0.8944⟩ = -14.384 · ⟨0.4472, -0.8944⟩ ≈ ⟨-6.43, 12.87⟩ N/C. The negative Q flips the direction: the field points toward the charge (upper-left relative to the field point), which is correct for a negative source.
Q: Two +1 nC charges sit at (-0.5, 0) and (0.5, 0). Find E at (-0.5, -1).
A: Charge 1 at (-0.5, 0): displacement to (-0.5, -1) is ⟨0, -1⟩, r = 1 m, r̂ = ⟨0, -1⟩. E₁ = 8.99 · ⟨0, -1⟩ = ⟨0, -8.99⟩.
Charge 2 at (0.5, 0): displacement to (-0.5, -1) is ⟨-1, -1⟩, r = √2 m, r̂ = ⟨-1/√2, -1/√2⟩. E₂ = 8.99/2 · ⟨-0.707, -0.707⟩ = ⟨-3.178, -3.178⟩.
E_net = ⟨-3.178, -12.168⟩ N/C.
Q: A +1 nC charge is at (-0.5, 0) and a -1 nC charge is at (0.5, 0). Find E at (0, 1).
A: Charge 1 (+1 nC): displacement ⟨0.5, 1⟩, r = √1.25, r̂ ≈ ⟨0.4472, 0.8944⟩. E₁ = 7.192 · ⟨0.4472, 0.8944⟩ ≈ ⟨3.216, 6.432⟩.
Charge 2 (-1 nC): displacement ⟨-0.5, 1⟩, r = √1.25, r̂ ≈ ⟨-0.4472, 0.8944⟩. E₂ = -7.192 · ⟨-0.4472, 0.8944⟩ ≈ ⟨3.216, -6.432⟩.
E_net = ⟨6.432, 0⟩ N/C. By the symmetry of the dipole, the y-components cancel and the net field points in the +x direction (from + toward -).
Q: Why does the field at the midpoint of two unlike charges point from the positive charge toward the negative, rather than cancelling?
A: Both charges contribute a field that points in the same direction at the midpoint. The positive charge pushes a test charge away (toward the negative), and the negative charge pulls the test charge toward itself (also toward the negative). Same direction means reinforcement, not cancellation.
electric field vector, E-field calculation, kQ/r², unit vector, r-hat, component form, vector addition, superposition, resultant field, null point, zero-field point, point charge field, positive charge, negative charge, dipole field, Coulomb's constant, field magnitude, field direction, PHY-222, Classical Physics II, PhET charges and fields, electric field at a point