The Electric Field

You’ll be able to

Define the electric field and relate it to the force on a test charge
Compute the field of one or more point charges by superposition
Interpret electric field lines and the field of a dipole

From force-at-a-distance to a field

Rather than track forces between every pair of charges, we say a charge fills the space around it with an electric field E. Any other charge placed in that space feels a force from the local field. The field is defined by the force per unit charge on a small positive test charge q₀: E = F / q₀. Its units are newtons per coulomb (N/C), and it points the way a positive charge would be pushed.

Field from force / force from field

→E = →F / q₀ ⇔ →F = q→E

A positive charge feels a force along E; a negative charge feels a force opposite to E.

Field of a point charge

E = k·q / r² = q / (4πε₀ r²)

Directed radially outward for q > 0 and radially inward for q < 0. Like the force, it falls off as 1/r².

Superposition and field lines

Because forces superpose, so do fields: the total field at a point is the vector sum of the fields from every source charge, E = Σ k·qᵢ / rᵢ² (each a vector). We picture the result with field lines that start on positive charge and end on negative charge. The line direction gives E’s direction, and where lines crowd together the field is strong. Lines never cross — the field has one definite direction at each point.

Worked example

A +5.0 μC point charge sits at the origin. Find the electric field it produces at a point 0.30 m away.

1.Use E = k·q / r² with q = 5.0 × 10⁻⁶ C and r = 0.30 m.
2.Numerator: k·q = (8.99 × 10⁹)(5.0 × 10⁻⁶) = 4.50 × 10⁴ N·m²/C.
3.Denominator: r² = (0.30)² = 0.090 m².
4.E = 4.50 × 10⁴ / 0.090 = 5.0 × 10⁵ N/C, directed radially outward (away from the positive charge).

Answer: 5.0 × 10⁵ N/C, pointing away from the charge

Checkpoint

At a point in space the electric field is 200 N/C pointing east. What force acts on a −3.0 μC charge placed there?

Worked example

Two charges lie on the x-axis: +2.0 μC at x = 0 and −2.0 μC at x = 0.40 m. Find the net electric field at the midpoint, x = 0.20 m.

1.Each charge is r = 0.20 m from the midpoint. Compute each field magnitude: E = k(2.0 × 10⁻⁶)/(0.20)² = (8.99 × 10⁹)(2.0 × 10⁻⁶)/0.040 = 4.5 × 10⁵ N/C.
2.Direction from +2.0 μC: field points away from it, i.e. in the +x direction at the midpoint.
3.Direction from −2.0 μC: field points toward it, also in the +x direction at the midpoint.
4.Both fields point the same way (+x), so they add: Eₙₑₜ = 4.5 × 10⁵ + 4.5 × 10⁵ = 9.0 × 10⁵ N/C, +x.

Answer: 9.0 × 10⁵ N/C in the +x direction (from the positive toward the negative charge)

Checkpoint

Which statement about electric field lines is correct?

Tip

To find the field of a group of point charges: (1) draw each field vector at the target point, pointing away from positive and toward negative sources; (2) find each magnitude with kq/r²; (3) sum components. The field exists whether or not a test charge is actually there.

Answer the 2 checkpoints as you read.