Unit 8 · Topic 8.3

8.3 Electric Fields

An electric field describes what a charge does to the space around it: at each point it gives the force per unit charge that any charge placed there would feel. Fields from several charges add as vectors, and field-line diagrams let you read off direction and relative strength at a glance.

Key terms

electric field
test charge
superposition
field lines
vector field map
electrostatic equilibrium

What the field means

Place a small positive test charge q₀ at a point and measure the force on it. The electric field there is $\vec{E} = \dfrac{\vec{F}_E}{q_0}$ , in newtons per coulomb (N/C). A test charge is small enough that it doesn't push the other charges around.

Once you know the field, you can find the force on any charge q at that point: $\vec{F}_E = q\vec{E}$ . A positive charge is pushed along the field. A negative charge, like an electron, is pushed opposite to it.

Field of a point charge

From Coulomb's law, a point charge q makes a field of magnitude $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\lvert q \rvert}{r^2}$ at distance r. It points straight away from a positive charge and straight toward a negative charge.

The field is there whether or not a test charge is present. It belongs to the charge that makes it, so a charge never feels its own field.

Superposition

The net field at a point is the vector sum of the fields from every charge: $\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 + \cdots$ . Find each magnitude, draw each direction at the point (away from +, toward −), then add components.

Look for symmetry first. At the center of a square with four equal charges, the fields cancel in pairs. On the line through two charges, the field can be zero only where the two fields point in opposite directions with equal size. For like charges that's between them; for opposite charges of different sizes it's outside, on the side of the smaller charge. Two equal and opposite charges have no zero-field point on that line at all.

Field lines and vector field maps

Field lines start on positive charges and end on negative charges (or run off to, or come in from, infinity). At any point, the field is tangent to the line through that point. Where lines are crowded, the field is strong; where they spread out, it's weak. Lines never cross, because the field has one direction at each point.

The number of lines drawn from a charge should be proportional to its size: if a +2q charge has 16 lines, a −q charge has 8. A vector field map shows the same information with arrows on a grid: each arrow points along the field, and its length shows the strength.

Between two large, oppositely charged parallel plates, the lines are straight, parallel and evenly spaced, so the field is uniform. That setup returns in 9.2 and 10.3.

Conductors and insulators in equilibrium

Electrostatic equilibrium means no charge is moving. In a conductor, any field inside would push free electrons, so in equilibrium the field inside the material is zero, and any extra charge sits on the outer surface. Just outside the conductor, the field is perpendicular to its surface. Outside any sphere whose charge is spread symmetrically (conductor or not), the field is the same as if all the charge sat at the center as a point charge. An insulator can hold charge spread through its volume, so it can have a field inside. Unit 10 builds on this.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Field of a point charge and the force on an electron
A +4.0 nC point charge is fixed in place. Find the electric field 0.30 m away, and the force on an electron placed there.
Show the solutionHide the solution
1. Step 1: $E = \dfrac{(9.0\times10^{9})(4.0\times10^{-9})}{(0.30)^2} = \dfrac{36}{0.090} = 400$ N/C, pointing away from the charge.
2. Step 2: Force on the electron: $F = \lvert q \rvert E = (1.60\times10^{-19})(400) = 6.4\times10^{-17}$ N.
3. Step 3: The electron is negative, so the force points opposite to the field: toward the +4.0 nC charge.
Answer: 400 N/C away from the charge; 6.4 × 10⁻¹⁷ N on the electron, toward the charge
Example 2Calculator allowed
Where is the field zero? (like charges)
A +4.0 μC charge is at x = 0 and a +1.0 μC charge is at x = 0.30 m. Where on the x-axis is the net field zero?
Show the solutionHide the solution
1. Step 1: Both charges are positive, so their fields point away from them. Between the charges, the two fields point in opposite directions, so a zero is possible there. Outside, they point the same way and can't cancel.
2. Step 2: Set the magnitudes equal at position x between them: $\dfrac{k(4.0)}{x^2} = \dfrac{k(1.0)}{(0.30 - x)^2}$ .
3. Step 3: Take square roots: $\dfrac{2}{x} = \dfrac{1}{0.30 - x}$ , so 0.60 − 2x = x and x = 0.20 m.
4. Step 4: Check: the zero is closer to the smaller charge, as it must be.
Answer: x = 0.20 m
Example 3Calculator allowed
Where is the field zero? (opposite charges, classic trap)
Now the charge at x = 0.30 m is −1.0 μC (the +4.0 μC charge is still at x = 0). Where is the net field zero?
Show the solutionHide the solution
1. Step 1: Between the charges, the field from the + charge points right (away from it) and the field from the − charge also points right (toward it). They add, so there's no zero between them. This is the trap: solving the same equation as before gives x = 0.20 m, which is wrong here.
2. Step 2: To the left of x = 0, the point is always closer to the larger charge, so its field always wins. The zero must be to the right of the −1.0 μC charge.
3. Step 3: For x > 0.30 m: $\dfrac{4.0}{x^2} = \dfrac{1.0}{(x - 0.30)^2}$ , so 2(x − 0.30) = x and x = 0.60 m.
Answer: x = 0.60 m (0.30 m beyond the −1.0 μC charge)

Common mistakes

Drawing the field at a point as pointing toward a positive charge. Field vectors point away from positive and toward negative charges.
Forgetting that a negative charge feels a force opposite to the field.
Solving for a zero-field point without first checking which region the field directions allow. Sketch the directions before you write the equation.
Drawing field lines that cross, or that start on negative charges.

On the exam

Field-line and vector-map questions are common: rank the field strength at labeled points by line density, or pick the arrow showing the force on an electron.
For superposition in free response, draw each field vector at the point with labels, then show components. Symmetry arguments, stated in words, earn credit when you say which components cancel and why.

Connected topics

Videos

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