Difficulty: Intermediate | Prerequisites: Coulomb's Law, vector components, basic trigonometry
Most real charge configurations involve more than one source charge. To find the electric field at any point, you need to add up the individual contributions from every charge, and because the electric field is a vector, that means adding components. This is the principle of superposition, and it is the workhorse technique for nearly every electric field calculation you will do in this course. The specific problem here (four charges at the corners of a rectangle, one charge at the centre) is a classic setup designed to drill the full workflow: find distances, find angles, decompose into x and y, and add.
The electric field from multiple point charges is the vector sum of each individual field contribution. You break each contribution into x and y components using trigonometry, add the components separately, then recombine to find the total field. The force on a charge sitting in that field is simply F = qE.
Superposition principle
The total electric field at any point is the vector sum of the fields produced by each individual source charge. In simple terms, each charge contributes independently, and you just stack all the contributions on top of each other.
Component decomposition
The process of splitting a vector into its x and y (and sometimes z) parts. Think of it as asking: "How much of this vector points east-west, and how much points north-south?"
Resultant field
The single vector you get after adding all the individual field components together. This is the actual field that a charge at that location would experience.
Four charges sit at the corners of a rectangle with sides a (short) and 3a (long).
Three corners carry charge q; one corner carries charge 2q.
A fifth charge Q sits at the centre.
The distance from any corner to the centre is the same for all four corners (by rectangle symmetry):
r = √(5/2) × a
This comes from the Pythagorean theorem applied to half-sides a/2 and 3a/2.
Draw a sketch. Label the coordinate axes. Mark the angle θ that the diagonal makes with the horizontal. Skipping this step makes the problem far harder than it needs to be.
Each charge q at distance r produces a field magnitude:
|E_q| = kq / r²
The charge 2q at the same distance produces:
|E_{2q}| = 2kq / r²
At this stage you have magnitudes only. Direction comes next.
The rectangle has half-sides a/2 (vertical) and 3a/2 (horizontal).
The angle θ between the diagonal and the horizontal satisfies:
sin θ = (a/2) / r = 1/√10
cos θ = (3a/2) / r = 3/√10
tan θ = 1/3
These exact algebraic forms keep your answer clean. Resist the urge to convert to decimals until the very end.
Resolve each charge's field contribution into x and y components using sin θ and cos θ.
Pay careful attention to signs: a field that points in the negative-x direction gets a minus sign on its x-component.
Because the three identical charges q have some symmetry, several components cancel or combine neatly. The single 2q charge breaks the full symmetry and is what gives the net field a non-zero result.
After summing all contributions:
E_x = −6kq / (5√10 × a²)
E_y = −2kq / (5√10 × a²)
With q = 3 µC and a = 2 cm:
|E| = 2.7 × 10⁷ N/C
Both components are negative, so the total field points into the third quadrant (down and to the left in standard orientation).
Once you know E at the centre, the force on charge Q placed there is:
F = QE (vector equation)
With Q = 4 µC:
F_x = −102 N, F_y = −34 N
The force direction matches the field direction (since Q is positive).
Quantity | Expression | Notes |
|---|---|---|
Distance, corner to centre | r = √(5/2) × a | From Pythagorean theorem on half-sides |
Field from charge q | |E_q| = kq / r² | Magnitude only; direction from geometry |
Field from charge 2q | |E_{2q}| = 2kq / r² | Twice the field of q at same distance |
Trig ratios | sin θ = 1/√10, cos θ = 3/√10 | Exact; do not approximate early |
Net x-component | E_x = −6kq / (5√10 a²) | Negative means pointing in −x direction |
Net y-component | E_y = −2kq / (5√10 a²) | Negative means pointing in −y direction |
Force on Q | F = QE | Vector; each component multiplied by Q |
Superposition is how engineers calculate the electric field inside any device with multiple electrodes: circuit boards, capacitor arrays, electrostatic precipitators used in smokestacks, even the deflection plates inside old oscilloscopes. The geometry changes, but the method (sum the vector contributions) is identical.
"I can just add the magnitudes of the individual fields." You cannot. Electric fields are vectors. You must add components separately, then find the magnitude of the resultant. Simply adding magnitudes ignores direction and gives the wrong answer.
"Symmetry means the field is zero at the centre." Only if the charge configuration is fully symmetric. Here, one corner has 2q instead of q, which breaks the symmetry and produces a net field.
"I need a calculator for sin θ and cos θ." For standard geometry problems with integer side ratios, you can express trig functions algebraically. Keeping exact forms avoids rounding errors and often simplifies the algebra.
"The force on Q depends on Q's contribution to the field." Q does not exert a force on itself. The field at the centre is produced by the four corner charges only; Q then sits in that external field.
⚠️ Setting up the geometry (distances, angles) is where most marks are lost. Always draw and label a diagram before writing any equations.
⚠️ Sign errors in components are the single most common mistake. Track directions carefully: does this component point in the +x or −x direction?
⚠️ Exam problems often ask for the field in general variables first, then for numerical values. Keep your algebra clean; plug in numbers only at the end.
⚠️ Know the difference between the field at a point (due to external charges) and the force on a charge placed at that point: F = qE.
True or false: The electric field at the centre of a square with four identical charges at the corners is zero. (True, by symmetry)
If one of those four identical charges is removed, is the net field at the centre zero? (No)
The distance from the corner of a rectangle with sides a and b to its centre is ______. (√(a² + b²) / 2)
True or false: You need a calculator to find sin θ for a right triangle with legs 1 and 3. (False; sin θ = 1/√10)
If the net electric field at a point is E and you place a charge Q = −2 µC there, the force on Q points ______ to E. (opposite/antiparallel)
Q: Two charges, +q and +q, sit at opposite corners of a square of side a. What is the magnitude of the electric field at the centre of the square?
A: Each charge is at distance r = a√2/2 from the centre. Each produces a field kq/r² = 2kq/a² directed away from itself. Because the two charges are at opposite corners, their fields at the centre point in the same direction (both push away). The net field is 2 × 2kq/a² = 4kq/a² along the diagonal from the charges toward the empty diagonal.
Q: In the rectangle problem, why are the x-component and y-component of the net field both negative?
A: The 2q charge sits in the upper-right corner. Its field at the centre points away from it, toward the lower-left (negative x, negative y). Because 2q produces a larger field than a single q charge, its contribution dominates, pulling the resultant into the third quadrant.
Q: You calculate E at a point and then place a charge Q there. Does placing Q change the value of E at that point?
A: In principle, Q creates its own field, but the "electric field at the centre due to the four corner charges" is defined as the field from those external sources only. Q does not contribute to the field that acts on itself. (In practice, Q could redistribute mobile charges elsewhere, but for fixed point charges this does not apply.)
This problem is pure application of the superposition principle, which returns throughout the course: in calculating electric potential (a scalar sum, much easier than vector addition), in magnetic fields from multiple wires, and eventually in electromagnetic wave superposition.
The trig and component-addition workflow here is identical to what you will need for continuous charge distributions (integrals replace sums, but the logic is the same).
superposition principle, electric field components, vector addition, point charge field, Coulomb's law, component decomposition, resultant field, rectangle charge configuration, E = kq/r², force on a charge in a field, P212, PHYS 212, electrostatics, trig in physics, field calculation