Electric fields

All five subtopics here are printed under "A Level subject content" in the 9702 syllabus, and every objective carries the tier "A Level". None of it is AS. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper, and the whole 9702 bank on this site comes from Paper 1, so no practice is tagged to this topic.

Syllabus points

18.1 Electric fields and field lines

• Understand that an electric field is an example of a field of force and define electric field as force per unit positive charge.
• Recall and use F = qE for the force on a charge in an electric field.
• Represent an electric field by means of field lines.

18.2 Uniform electric fields

• Recall and use E = ΔV/Δd to calculate the field strength of the uniform field between charged parallel plates.
• Describe the effect of a uniform electric field on the motion of charged particles.

18.3 Electric force between point charges

• Understand that, for a point outside a spherical conductor, the charge on the sphere may be considered to be a point charge at its centre.
• Recall and use Coulomb's law F = Q₁Q₂/(4πε₀r²) for the force between two point charges in free space.

18.4 Electric field of a point charge

• Recall and use E = Q/(4πε₀r²) for the electric field strength due to a point charge in free space.

18.5 Electric potential

• Define electric potential at a point as the work done per unit positive charge in bringing a small test charge from infinity to the point.
• Recall and use the fact that the electric field at a point is equal to the negative of potential gradient at that point.
• Use V = Q/(4πε₀r) for the electric potential in the field due to a point charge.
• Understand how the concept of electric potential leads to the electric potential energy of two point charges and use E_P = Qq/(4πε₀r).

The parallel with gravitation, and where it breaks

Almost every equation here has a twin in topic 13. Learning them as pairs is far more efficient than learning them twice.

Two differences, and both matter:

• Gravitational forces are only attractive, because there is no negative mass. Electric forces can be attractive or repulsive, so the electric equations carry no built-in minus sign and the sign of the answer comes from the signs of the charges. A negative value of F means attraction; a positive value means repulsion.
• Gravitational potential is always negative, tending up to zero at infinity. Electric potential is positive near a positive charge and negative near a negative charge, and in both cases it tends to zero at infinity.

The constant 1/(4πε₀) is 8.99 × 10⁹ N m² C⁻², where ε₀, the permittivity of free space, is 8.85 × 10⁻¹² F m⁻¹.

Electric field strength

Electric field strength at a point is the force per unit positive charge acting on a small stationary test charge placed at that point:

E = F / q

with unit N C⁻¹, or equivalently V m⁻¹. It is a vector, and its direction is the direction of the force on a positive charge.

Rearranged, the force on a charge in a field is

F = qE

and note that a negative charge experiences a force in the opposite direction to the field.

The words positive and small are both examined. Positive, because the direction of the field is defined by convention using a positive test charge. Small, because a large test charge would disturb the very field being measured.

Field lines

Field lines show the direction of the force on a positive charge, and their spacing shows the strength.

• Around an isolated positive point charge, radial lines pointing outwards; around a negative one, radial lines pointing inwards.
• Between two opposite charges, lines run from the positive to the negative, curving between them.
• Between two like charges, the lines repel each other and there is a neutral point on the line joining them where the resultant field is zero.
• Between parallel charged plates, the lines are parallel, equally spaced and perpendicular to the plates, running from the positive plate to the negative. The field is uniform except for a slight curving outwards at the edges.
• Field lines always meet a conductor's surface at right angles, and there is no field inside a hollow conductor.

Field lines never cross, because the field at a point has one direction.

Uniform fields between parallel plates

For two parallel plates separated by a distance d with a potential difference V between them, the field between them is uniform and

E = ΔV / Δd

Worked example. Two plates 4.0 mm apart have a potential difference of 250 V.

E = 250 / 0.0040 = 62500

so E is 6.3 × 10⁴ V m⁻¹, and it points from the positive plate towards the negative.

The force on an electron in that field:

F = 1.60 × 10⁻¹⁹ × 62500 = 1.0 × 10⁻¹⁴

so about 1.0 × 10⁻¹⁴ N, directed towards the positive plate, since the electron is negative.

Notice that the field is the same everywhere between the plates, so the force on the charge does not change as it moves across, which is the property that makes the next section work.

Charged particles moving in a uniform field

A charged particle entering a uniform field at right angles to it behaves exactly like a projectile in a gravitational field, and for the same reason: a constant force at right angles to the initial velocity.

• The component of velocity along the original direction is unchanged, because there is no force that way.
• The component along the field accelerates uniformly, with a = qE/m = qV/(md).
• The path is therefore a parabola while the particle is between the plates, and a straight line once it leaves.

Worked example. An electron enters midway between the plates above, travelling parallel to them at 2.0 × 10⁷ m s⁻¹. The plates are 60 mm long. How far is it deflected?

Time between the plates:

t = 0.060 / (2.0 × 10⁷) = 3.0 × 10⁻⁹

Acceleration:

a = 1.0 × 10⁻¹⁴ / (9.11 × 10⁻³¹) = 1.098 × 10¹⁶

Deflection, from s = ½at² with no initial velocity across the gap:

s = 0.5 × 1.098 × 10¹⁶ × (3.0 × 10⁻⁹)² = 0.0494

so about 4.9 cm, which is far more than the 2 mm of clearance available, so this electron in fact hits the plate. Checking an answer against the geometry like that is worth doing, because a deflection larger than the gap tells you the particle never gets through.

A particle entering along the field is simply accelerated or decelerated in a straight line, and the energy method is quicker there: the work done is qV, so

½mv² = qV

Coulomb's law

The force between two point charges in free space:

F = Q₁Q₂ / (4πε₀r²)

This is an inverse square law, exactly like Newton's law of gravitation. Doubling the separation quarters the force.

Two points about applying it:

• r is the separation of the centres. For a point outside a spherical conductor, the charge on the sphere may be treated as a point charge at its centre, which is what makes the law usable for charged spheres.
• Substitute the signs of the charges. Two like charges give a positive F, meaning repulsion; unlike charges give a negative F, meaning attraction.

Worked example. Two small spheres carry charges of +3.0 nC and -5.0 nC, with centres 20 mm apart.

F = 8.99 × 10⁹ × 3.0 × 10⁻⁹ × 5.0 × 10⁻⁹ / (0.020)² = 3.371 × 10⁻⁴

so the force is 3.4 × 10⁻⁴ N, and it is attractive because the charges are unlike.

Field of a point charge

Dividing Coulomb's law by the test charge gives the field strength due to a point charge Q:

E = Q / (4πε₀r²)

Worked example. Find the field strength 50 mm from a point charge of +8.0 nC.

E = 8.99 × 10⁹ × 8.0 × 10⁻⁹ / (0.050)² = 28768

so E is 2.9 × 10⁴ V m⁻¹, directed away from the charge.

Where two charges both contribute, the resultant field is the vector sum of the individual fields, so the directions must be considered and not just the magnitudes. This is the main difference from the potential calculations below.

Electric potential

Electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point. It is a scalar, measured in volts, where 1 V is 1 J C⁻¹.

For the field of a point charge:

V = Q / (4πε₀r)

Because V is a scalar, potentials due to several charges are added as ordinary numbers with their signs, which is much easier than adding fields. A point midway between equal positive and negative charges has zero potential but a large field, and a point midway between two equal positive charges has zero field but a large potential. Those two situations are a favourite question precisely because they force you to keep the scalar and vector natures apart.

Worked example. Find the potential 50 mm from the +8.0 nC charge above.

V = 8.99 × 10⁹ × 8.0 × 10⁻⁹ / 0.050 = 1438.4

so V is about 1.4 × 10³ V.

Note the relationship between that answer and the field strength found earlier: 1438.4 / 0.050 = 28768, which is E. That is no coincidence, and it is the next objective.

Field as the potential gradient

The electric field at a point is the negative of the potential gradient at that point:

E = -ΔV / Δr

The minus sign says the field points from high potential to low potential, which is the direction a positive charge would be pushed, and it is the reason the uniform-field equation E = ΔV/Δd works with the magnitudes alone once the direction is known.

Practically, this means that on a graph of V against r, the gradient at a point gives -E, and on a graph of E against r, the area under the curve between two radii gives the change in potential. Both readings are examined.

Since V goes as 1/r and E as 1/r², their graphs have different shapes: the potential curve falls away more gently. Sketching one and assuming the other is the same shape is a common error.

Electric potential energy

The potential energy of a charge q at a point where the potential is V is

E_P = qV

so for two point charges:

E_P = Qq / (4πε₀r)

The sign carries meaning:

• For like charges, E_P is positive and increases as they are brought closer, because work must be done against repulsion. Release them and they fly apart.
• For unlike charges, E_P is negative and becomes more negative as they approach, exactly as in the gravitational case. Energy must be supplied to separate them.

Worked example. How much work is needed to bring an alpha particle, charge +2e, from a long distance away to within 2.0 × 10⁻¹⁴ m of a gold nucleus, charge +79e? Take e as 1.60 × 10⁻¹⁹ C.

Charge on the alpha particle:

2 × 1.60 × 10⁻¹⁹ = 3.20 × 10⁻¹⁹

Charge on the nucleus:

79 × 1.60 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷

E_P = 8.99 × 10⁹ × 3.20 × 10⁻¹⁹ × 1.264 × 10⁻¹⁷ / (2.0 × 10⁻¹⁴) = 1.818 × 10⁻¹²

so about 1.8 × 10⁻¹² J, which is roughly 11 MeV. Because energy is conserved, that is also the kinetic energy an alpha particle must have to reach that distance, which is how the closest approach in the Rutherford scattering experiment is calculated, and how the size of a nucleus was first estimated.

Common mistakes

• Defining electric field strength as force per unit charge without the word positive, so the direction is undefined.
• Forgetting that the force on a negative charge is opposite to the field direction.
• Measuring r from the surface of a charged sphere rather than from its centre.
• Adding electric fields as though they were scalars. Fields are vectors; potentials are scalars.
• Confusing a point of zero field with a point of zero potential. They are different places.
• Assuming V and E fall off with distance in the same way. V goes as 1/r, E as 1/r².
• Using E = ΔV/Δd for the field of a point charge, where the field is not uniform.
• Leaving the plate separation in millimetres in E = ΔV/Δd.
• Putting a minus sign into the electric equations by analogy with gravitation. The sign here comes from the charges.
• Saying an electron is deflected towards the negative plate.
• Treating a charged particle's path as parabolic after it has left the plates, where there is no field and the path is straight.
• Confusing ε₀ with the constant 1/(4πε₀). One is 8.85 × 10⁻¹², the other 8.99 × 10⁹.