Difficulty: Intermediate | Prerequisites: Coulomb's Law study notes, integral calculus (for continuous distributions), basic trigonometry
Once you know how charges exert forces on each other (Coulomb's Law), the next question is: what does a charge do to the space around it? The answer is the electric field. Rather than thinking about force between specific pairs of charges, the field gives you the force per unit charge at every point in space. This is the framework for nearly everything that follows in E&M: Gauss's Law, electric potential, capacitors, and eventually Maxwell's equations all speak the language of fields. If Coulomb's Law was "two charges talking to each other," the electric field is "one charge broadcasting to everyone."
The electric field at a point is the force a positive test charge would feel there, divided by the size of that test charge. For point charges, you compute E = kQ/r² and add contributions as vectors. For continuous charge distributions (lines, arcs, surfaces), you chop the object into tiny pieces, compute each piece's contribution, and integrate.
Electric field (E)
A vector field defined at every point in space. At a given location, E = F/q₀, where F is the force a small positive test charge q₀ would experience there. Units: newtons per coulomb (N/C), equivalently volts per metre (V/m).
Think of it as a map of "force per unit charge" that exists everywhere, even when no test charge is present.
Field due to a point charge
E = kQ / r², directed radially outward from a positive charge and radially inward toward a negative charge. The magnitude drops off as the inverse square of the distance, just like the Coulomb force.
In simple terms, positive charges "push" the field outward; negative charges "pull" it inward.
Superposition principle (fields)
The net electric field at any point is the vector sum of the fields from each individual source charge, calculated independently. Identical in spirit to force superposition, but now you are adding field vectors rather than force vectors.
Linear charge density (λ)
Charge per unit length along a one-dimensional object: λ = Q / L. Units: C/m. For a semicircular arc of radius a carrying total charge Q, the arc length is πa, so λ = Q / (πa).
Think of it as "how much charge is packed into each metre of the wire or arc."
Continuous charge distribution
Any object where the charge is spread out rather than concentrated at a point. You cannot use E = kQ/r² directly for the whole object; instead, you break it into infinitesimal pieces dq, compute dE from each, and integrate.
Equilibrium point (zero-field point)
A location where the net electric field is exactly zero because the contributions from different source charges cancel. For two positive charges on a line, the equilibrium lies between them. For charges of different sign or magnitude, the equilibrium can lie outside the segment joining them.
The magnitude is:
E = k |Q| / r²
Direction: radially away from Q if Q is positive; radially toward Q if Q is negative.
This is simply Coulomb's Law divided by the test charge: if F = kQq₀/r², then E = F/q₀ = kQ/r².
Example from the homework (Problem 7): Q = −1.3 μC at the centre of a semicircular arc, a = 6.1 cm.
For a test point on the axis, the point-charge component gives E = k(1.3 × 10⁻⁶) / (0.061)² ≈ 3,144,316 N/C.
The procedure mirrors force superposition:
Compute the magnitude of E from each charge at the field point.
Determine the direction of each E vector (away from positive, toward negative).
Decompose into x and y components using trigonometry.
Sum all x-components; sum all y-components.
Worked pattern from the homework (Problem 4):
Charges q₁ = −3.9 μC and q₃ = 3.7 μC arranged symmetrically, with field point P at distance d = 6.1 cm.
q₂ = 6.6 μC also contributes.
Each field magnitude: E = kQ / r². For the diagonal charges, r = √2 × d, and the field is at 45° to the axes.
Decompose: Eₓ = E cos 45°, E_y = E sin 45°.
The homework finds Eₓ from one charge ≈ −333,505 N/C (negative because the field from a negative charge points toward it) and the E from the positive charge ≈ 1,596,351 N/C.
The third charge contributes E ≈ 8,949,207 N/C.
Net field is the vector sum of all contributions.
For two charges on a line, the zero-field point is where the magnitudes of the two individual fields are equal and the directions are opposite.
Setup from the homework (Problem 5): q = 2 μC (negative, on the left), Q = 4 μC (positive, on the right), separated by d = 12 cm.
Let the equilibrium point be at distance a from Q (the larger charge). Then it is at distance (a + d) from q.
Setting the field magnitudes equal: kQ / a² = kq / (a + d)².
Cancel k: Q / a² = q / (a + d)².
Cross-multiply and take square roots: (a + d)√q = a√Q.
Solve for a: a = (√q / (√Q − √q)) × d.
Substituting: a = (√(2 × 10⁻⁶) / (√(4 × 10⁻⁶) − √(2 × 10⁻⁶))) × 0.12 ≈ 0.29 m.
Key insight: when the charges have different magnitudes, the equilibrium is not at the midpoint. It sits closer to the weaker charge, because a smaller field needs a shorter distance to compensate.
For two charges of the same sign, the equilibrium is between them. For opposite signs, it is on the far side of the weaker charge (outside the segment).
When charge is spread over a shape, you integrate contributions from infinitesimal elements dq.
For a semicircular arc of radius a and total charge Q (Problem 7):
Linear charge density: λ = Q / (πa).
An element dq = λ a dθ at angle θ produces a field dE = k dq / a² = kλ dθ / a at the centre.
By symmetry, one component (say Eₓ) cancels when you integrate over the full semicircle.
The surviving component is found by integrating dE_y = (kλ/a) sin θ dθ from 0 to π.
Result: E_y = 2kλ / a = 2kQ / (πa²).
From the homework: Q = −1.3 μC, a = 0.061 m.
λ = (−1.3 × 10⁻⁶) / (π × 0.061).
The point-charge field from Q at the centre and the integrated field from the arc are computed separately.
The integrated field ≈ −2,001,733 N/C (sign reflects that Q is negative, so the net field points toward the arc).
Problem 6 sketches a charge distribution along a vertical axis from −L/2 to +L/2, with the field evaluated at a point (a, 0) on the horizontal axis.
The standard result for the field at perpendicular distance a from the midpoint of a finite line charge of length L and total charge Q is:
E = kQ / (a √(a² + (L/2)²))
directed perpendicular to the line (for a uniformly charged segment, the parallel component cancels by symmetry at the midpoint).
This formula bridges the gap between the point-charge result (when a >> L, the line looks like a point) and the infinite-line result (when L >> a, E = 2kλ/a).
Electric field from a point charge
E = kQ / r²
Direction: away from +Q, toward −Q
Superposition
E_net = Σ Eᵢ (vector sum)
Linear charge density
λ = Q / L
For a semicircular arc of radius a: λ = Q / (πa)
Electric field at the centre of a semicircular arc
E = 2kλ / a = 2kQ / (πa²)
(directed along the axis of symmetry)
Zero-field condition for two charges on a line
kQ₁ / r₁² = kQ₂ / r₂²
which gives r₁ / r₂ = √(Q₁ / Q₂)
Electric-field calculations for continuous charge distributions are how engineers design cathode-ray tubes, ion traps, and particle accelerators. The field from a semicircular arc, for instance, is structurally similar to the field inside a bending magnet in a synchrotron. Understanding where the field is zero (equilibrium points) is central to the design of electrostatic traps used to confine charged particles for precision measurements.
Treating E = kQ/r² as the answer for extended objects. That formula works only for point charges. For a charged rod, ring, or arc, you must integrate. Using the point-charge formula for the total charge and some "average" distance will give the wrong result.
Forgetting that E is a vector. When adding fields from multiple charges, students often add the magnitudes. This is only correct if all the fields point in the same direction. If they do not, you must decompose into components.
Misidentifying where the zero-field point lies. For two same-sign charges, it is between them. For two opposite-sign charges, it is beyond the smaller one on the line connecting them, not between them. Mixing these up is one of the most common exam errors.
Dropping the symmetry argument for continuous distributions. The whole reason you can skip one component of the integral (because it cancels) is symmetry. If the geometry is not symmetric about the relevant axis, both components survive and you must integrate both.
⚠️ Exams frequently give you three or four point charges and ask for the net electric field at a specified location. This is a direct test of superposition with vector decomposition, identical in method to the force problems but with E = kQ/r² instead of F = kqQ/r².
⚠️ The zero-field equilibrium problem appears on nearly every introductory E&M exam. The algebra is simple, but the conceptual question of "which side of which charge?" trips many students.
⚠️ Continuous-distribution problems (arcs, rings, rods) typically appear on the second or third exam. The setup is the hard part: defining dq, choosing the integration variable, and invoking symmetry to kill one component. The integration itself is usually straightforward.
⚠️ Be precise with units. Electric field is in N/C. A common error is to leave a distance in centimetres instead of converting to metres, which inflates the answer by a factor of 10⁴ (because r² appears in the denominator).
T/F: The electric field at a point is zero only if there are no charges anywhere nearby.
False. The field can be zero at a point where contributions from multiple charges cancel exactly.
Fill in the blank: The electric field from a point charge is proportional to 1/r², so tripling the distance reduces the field to ______ of its original value.
One-ninth (1/9).
T/F: For a semicircular arc of charge, the electric field at the centre has components in both the radial and tangential directions.
False. By symmetry, one component cancels. Only the component along the axis of symmetry survives.
T/F: The zero-field point between two positive charges of equal magnitude is at the midpoint.
True. By symmetry, both fields have equal magnitude and opposite direction at the midpoint.
Fill in the blank: Linear charge density λ has units of ______.
Coulombs per metre (C/m).
Q: A charge of +5.0 μC is at the origin. What is the electric field at a point 0.10 m away, and in what direction does it point?
A: E = (8.99 × 10⁹)(5.0 × 10⁻⁶) / (0.10)² = 4.495 × 10⁶ N/C, directed radially away from the charge.
Q: Two charges, Q₁ = +8 μC at x = 0 and Q₂ = +2 μC at x = 0.30 m, lie on the x-axis. Where on the x-axis is the net electric field zero?
A: The equilibrium is between them, closer to Q₂ (the smaller charge). Let x be the distance from Q₁. Then k(8 μC)/x² = k(2 μC)/(0.30 − x)². Solving: x/(0.30 − x) = √(8/2) = 2, so x = 0.20 m. The zero-field point is at x = 0.20 m.
Q: A semicircular arc of radius 5.0 cm carries a total charge of +3.0 μC. What is the magnitude of the electric field at the centre of curvature?
A: λ = Q/(πa) = (3.0 × 10⁻⁶)/(π × 0.05). E = 2kλ/a = 2k Q/(πa²) = 2(8.99 × 10⁹)(3.0 × 10⁻⁶) / (π × 0.0025) ≈ 6.87 × 10⁶ N/C, directed along the symmetry axis, away from the arc (since Q is positive).
Q: A negative charge of −4.0 μC and a positive charge of +1.0 μC are separated by 0.20 m. On which side of which charge does the zero-field point lie?
A: For opposite-sign charges, the zero-field point is on the far side of the weaker charge (the +1.0 μC), beyond the segment joining them. It is not between them, because between them both fields point in the same direction (toward the negative charge).
Electric fields lead directly into electric potential (V = kQ/r), which is a scalar rather than a vector, making calculations easier in many situations. The field and potential are related by E = −dV/dr, so understanding E thoroughly makes the potential chapter far more intuitive.
The continuous-distribution work here (integrating dE from small charge elements) is the same technique you will use for Gauss's Law problems, except Gauss's Law provides a shortcut when symmetry is high (spheres, infinite planes, infinite cylinders). Gauss's Law does not eliminate the need to integrate; it replaces a difficult vector integral with an easier scalar one.
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