Electric field strengthAQA A-Level Physics: Revision notes

An electric field is a region where a charged object experiences a force. We represent it with field lines. The direction of a line shows the direction of the force on a small positive test charge, so lines run away from positive charges and towards negative charges.

Rules for drawing field lines:

• Lines never cross.
• Lines meet the surface of a conductor at right angles.
• The closer the lines, the stronger the field.
• A point charge has a radial field (lines straight out of or into the charge).
• Between oppositely charged parallel plates the field is uniform: parallel, equally spaced lines (with slight bulging at the edges).

The electric field strength E at a point is the force per unit positive charge placed at that point:

$E = \frac{F}{Q}$

It is a vector, in the direction of the force on a positive charge. Its unit is N C⁻¹, which is equivalent to V m⁻¹.

Rearranged, the force on a charge Q in a field E is $F = QE$. For a negative charge the force is opposite to E.

Between two parallel plates with p.d. V and separation d, the field is uniform and

$E = \frac{V}{d}$

Derivation. Moving a charge Q across the gap against the field needs a constant force F = QE over a distance d, so the work done is $W = Fd = QEd$. The p.d. is the work done per unit charge, so $W = QV$. Equating gives $QEd = QV$, so $E = V/d$.

Worked example. A p.d. of 1500 V is applied across plates 0.030 m apart. $E = 1500 / 0.030 = 5.0 \times 10^4$ V m⁻¹. The force on an electron is $F = eE = 1.60 \times 10^{-19} \times 5.0 \times 10^4 = 8.0 \times 10^{-15}$ N.

A particle entering a uniform field at right angles to the field lines behaves like a projectile.

• Along the plates there is no force, so the velocity component is constant: $x = vt$.
• Across the plates the force $F = QE$ is constant, so the acceleration $a = QE/m$ is constant and the initial perpendicular velocity is zero: $y = \tfrac{1}{2}at^2$.

The path in the field is a parabola. After leaving the field the particle moves in a straight line. A positive particle is deflected in the direction of E, a negative one opposite to E.

Worked example. An electron enters at $3.0 \times 10^7$ m s⁻¹ between plates 0.040 m long, with $E = 2.0 \times 10^4$ V m⁻¹. Time in the field $t = 0.040 / 3.0 \times 10^7 = 1.33 \times 10^{-9}$ s. Acceleration $a = eE/m = 3.5 \times 10^{15}$ m s⁻². Deflection $y = \tfrac{1}{2}at^2 = 3.1 \times 10^{-3}$ m.

For a point charge Q (or a uniformly charged sphere, with r measured from its centre), the field strength at distance r is

$E = \frac{Q}{4\pi\varepsilon_0 r^2}$

where $\varepsilon_0 = 8.85 \times 10^{-12}$ F m⁻¹ is the permittivity of free space. E obeys an inverse square law: doubling r reduces E to one quarter. The field points away from a positive charge and towards a negative one.

Worked example. Q = +5.0 nC, r = 0.20 m: $E = \dfrac{5.0 \times 10^{-9}}{4\pi \times 8.85 \times 10^{-12} \times 0.20^2} = 1.1 \times 10^3$ V m⁻¹.

This mirrors Newton's law of gravitation, but electric forces can be attractive or repulsive.