Unit 10 · Topic 10.3

10.3 Electric Fields

An electric field describes the force a charge would feel at each point in space, per unit of charge. Fields point away from positive charges and toward negative ones, and fields from several charges add as vectors. Inside a conductor at equilibrium, the field is zero.

Key terms

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

What a field is

Instead of saying charge A pushes on charge B across empty space, we say charge A creates an electric field everywhere around it, and charge B responds to the field where it sits. The field at a point is the force on a small test charge q placed there, divided by that charge:

$\vec{E} = \frac{\vec{F}_E}{q}$

A test charge is small enough that it doesn't disturb the charges making the field. The unit is N/C (the same as V/m). Once you know E, the force on any charge is $\vec{F}_E = q\vec{E}$ . A positive charge feels a force along the field, and a negative charge feels a force opposite to it.

This works just like the gravitational field from Physics 1, where g = F/m and near Earth g is 9.8 N/kg.

Field of a point charge

Combine E = F/q with Coulomb's law and the field of a point charge Q at distance r is:

$E = \frac{k\lvert Q \rvert}{r^2}$

It points straight away from a positive charge and straight toward a negative one. Like the force, it falls off as 1/r².

When several charges are present, the net field at a point is the vector sum of the fields from each one. Find each field's size and direction, split into components if needed, and add.

Picturing fields

A vector field map draws arrows at many points, with each arrow's length showing the field's strength and its direction showing the field's direction.

A field line diagram is a simpler version. Lines start on positive charges and end on negative charges (or run off to infinity). The field at any point is tangent to the line there, and lines are packed closer together where the field is stronger. Field lines never cross, because the field can only point one way at each point.

Between two large, oppositely charged parallel plates, the lines are evenly spaced and parallel: the field is uniform (see 10.6).

Conductors and insulators

In a conductor at electrostatic equilibrium (no charges moving), the first three points below always hold:

Any extra charge sits on the outer surface, spread out as far as the charges can get from each other.
The field inside the conductor's material is zero. If it weren't, free electrons would keep moving.
Just outside the surface, the field is perpendicular to the surface. A sideways component would push charges along the surface.
Outside a uniformly charged sphere, the field is the same as if all the charge sat at a point at its center.
In an insulator, extra charge can be spread through the inside as well as the surface, and the field inside can be nonzero. You only need to describe this, not calculate it.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Field and force on an electron
Find the electric field 0.30 m from a +2.0 nC point charge. Then find the force on an electron placed there.
Show the solutionHide the solution
1. Step 1: $E = \dfrac{k\lvert Q \rvert}{r^2} = \dfrac{(9.0 \times 10^9)(2.0 \times 10^{-9})}{(0.30)^2} = 200$ N/C, pointing away from the positive charge.
2. Step 2: Force on the electron: F = |q|E = (1.60 × 10⁻¹⁹)(200) = 3.2 × 10⁻¹⁷ N.
3. Step 3: The electron is negative, so the force points opposite to the field: toward the +2.0 nC charge.
Answer: E = 200 N/C away from the charge; F = 3.2 × 10⁻¹⁷ N toward the charge
Example 2Calculator allowed
Field between opposite charges
A +4.0 nC charge and a −4.0 nC charge are 0.20 m apart. Find the electric field at the midpoint.
Show the solutionHide the solution
1. Step 1: Each charge is 0.10 m from the midpoint, and the charges have equal size, so each makes a field of $\dfrac{(9.0 \times 10^9)(4.0 \times 10^{-9})}{(0.10)^2} = 3600$ N/C.
2. Step 2: Directions: the positive charge's field points away from it, toward the negative charge. The negative charge's field points toward itself, which is the same direction.
3. Step 3: Same direction, so they add: 3600 + 3600 = 7200 N/C, pointing toward the negative charge.
Answer: 7200 N/C toward the negative charge
Example 3Calculator allowed
Adding fields at right angles
Two +3.0 nC charges sit at (0, 0) and (0.40 m, 0). Find the net electric field at point P = (0, 0.30 m).
Show the solutionHide the solution
1. Step 1: From the charge at the origin: distance 0.30 m, $E_1 = \dfrac{(9.0 \times 10^9)(3.0 \times 10^{-9})}{(0.30)^2} = 300$ N/C, pointing straight up (+y), away from the charge.
2. Step 2: From the charge at (0.40, 0): distance $\sqrt{0.40^2 + 0.30^2} = 0.50$ m, so $E_2 = \dfrac{(9.0 \times 10^9)(3.0 \times 10^{-9})}{(0.50)^2} = 108$ N/C, pointing away from that charge, toward P. From the charge to P is 0.40 m left and 0.30 m up, so the direction is (−0.8, +0.6).
3. Step 3: Components of E₂: x = −0.8 × 108 = −86.4 N/C; y = 0.6 × 108 = 64.8 N/C.
4. Step 4: Net: $E_x$ = −86.4 N/C, $E_y$ = 300 + 64.8 = 364.8 N/C. Size = $\sqrt{86.4^2 + 364.8^2} \approx 375$ N/C.
5. Step 5: Direction: up and to the left, at an angle of about 13° to the left of the +y axis.
Answer: About 375 N/C, pointing up and slightly left (about 13° left of straight up)

Common mistakes

Drawing the force on a negative charge in the same direction as the field. Negative charges feel a force opposite to E.
Adding field magnitudes from charges in different directions. Fields are vectors; use components.
Thinking the field inside a charged conductor is strong because the conductor is charged. At equilibrium it's zero inside.
Drawing field lines that cross, or that start on negative charges.

On the exam

Expect to sketch field vectors or field lines for a few charges, or to find where the net field is zero. For two like charges that point is between them, closer to the smaller one; for two opposite charges of different sizes it's outside the pair, beyond the smaller one.
Questions about conductors often ask you to justify why the field inside is zero or why it's perpendicular at the surface; explain in terms of free charges that would otherwise move.

Connected topics

Videos

AP Physics 2 - Unit 10 - Lesson 3 - Electric Fields
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Check yourself

4 questions on 10.3 Electric Fields. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

What is the magnitude of the electric field 0.30 m from a +4.0 nC point charge?

Question 2 of 4Calculator allowed

An electron is in a uniform electric field of $2.0 \times 10^4$ N/C that points east. What is the electric force on the electron?

Question 3 of 4Calculator allowed

Charges +Q and −Q are a distance d apart. Which describes the electric field at the point midway between them?

Question 4 of 4Calculator allowed

A metal sphere of radius 0.10 m carries a charge of +2.0 nC spread evenly over its surface. What is the electric field magnitude at a point 0.30 m from the center of the sphere?

0 of 4 answered