Electric fields flashcards

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Electric field
A region in which a charged particle experiences a force.
Electric field strength E
The force per unit positive charge on a small test charge placed at that point. Unit N C⁻¹ = V m⁻¹; a vector.
Electric potential V at a point
The work done per unit charge in bringing a small positive test charge from infinity to that point. Unit V = J C⁻¹; a scalar.
Equipotential
A line or surface joining points at the same potential; no work is done moving a charge along it.
Uniform electric field
A field with the same strength and direction at every point, e.g. between the central parts of parallel plates.
Potential gradient
The rate of change of potential with distance, ΔV/Δr; the field strength is minus the potential gradient.
Field strength from force and charge
E = F/Q (Yes)
Coulomb's law
F = Q₁Q₂/4πε₀r² (1/4πε₀ = 8.99 × 10⁹ N m² C⁻²) (Yes)
Field strength of a point charge
E = Q/4πε₀r² (radial, inverse square) (Yes)
Potential of a point charge
V = Q/4πε₀r (zero at infinity, 1/r) (Yes)
Uniform field between parallel plates
E = V/d (Yes)
Field strength and potential gradient
E = −ΔV/Δr: the field points the way the potential falls. (No — learn it)
Speed of a charge accelerated from rest through a p.d. V
½mv² = QV, so v = √(2QV/m). (No — derive it)
Field lines of an isolated positive point charge
Straight radial lines, evenly spread, arrows pointing outwards; equipotentials are concentric circles crossing them at 90°.
Field between two parallel plates
Parallel, equally spaced lines perpendicular to the plates from + to −; equipotentials parallel to the plates and equally spaced; lines bulge at the edges.
Graphs of E and V against r for a positive point charge
Both fall with distance: V ∝ 1/r, E ∝ 1/r² (steeper). E at any r = −(gradient of the V–r graph).
Two like charges, field lines between them
Lines bend away from each other, leaving a neutral point (E = 0) closer to the smaller charge.
Path of an electron fired between charged parallel plates
A parabola curving towards the positive plate, then a straight line after leaving the field.
Why are field lines perpendicular to equipotentials?
No work is done moving along an equipotential, so the force (and field) has no component along it.
Why do equipotentials around a point charge spread out for equal steps of V?
V ∝ 1/r, so the same change in V needs a larger distance further out, where the field is weaker.
Why is the path of a charge across a uniform field a parabola?
Constant velocity perpendicular to the field, constant acceleration QE/m along it — just like a horizontally launched projectile.
How can a charged oil drop be held stationary between horizontal plates?
Adjust V so the upward electric force QV/d equals the weight mg; a negative drop needs the upper plate positive.
What did balancing many charged drops show?
Every charge was a whole-number multiple of 1.60 × 10⁻¹⁹ C: charge is quantised.
Why is gravity ignored for electrons in electric fields?
The electric force eE is many orders of magnitude larger than the weight mg.
A +2.5 nC charge feels a force of 6.0 × 10⁻⁵ N. Field strength?
E = F/Q = 6.0 × 10⁻⁵ ÷ 2.5 × 10⁻⁹ = 2.4 × 10⁴ N C⁻¹.
+45 nC and −30 nC, centres 12 cm apart. Force?
F = 8.99 × 10⁹ × 45 × 10⁻⁹ × 30 × 10⁻⁹ ÷ 0.12² = 8.4 × 10⁻⁴ N, attractive.
+16 nC and +4.0 nC are 0.60 m apart. Where is the neutral point?
16/x² = 4.0/(0.60 − x)² → 4/x = 2/(0.60 − x) → x = 0.40 m from the 16 nC charge.
Potential 0.25 m from a +2.0 nC charge?
V = 8.99 × 10⁹ × 2.0 × 10⁻⁹ ÷ 0.25 = 72 V.
3.0 kV across plates 1.5 cm apart. Field strength?
E = V/d = 3000 ÷ 0.015 = 2.0 × 10⁵ V m⁻¹.
A 3.2 × 10⁻¹⁵ kg drop balances at 245 V between plates 5.0 mm apart. Charge?
Q = mgd/V = 3.2 × 10⁻¹⁵ × 9.81 × 5.0 × 10⁻³ ÷ 245 = 6.4 × 10⁻¹⁹ C = 4e.
Electron accelerated from rest through 500 V. Speed?
eV = 8.0 × 10⁻¹⁷ J = ½mv² → v = 1.3 × 10⁷ m s⁻¹.
A 2.5 MeV proton heads straight for a nucleus of charge +50e. Closest approach?
Eₖ = (50e)e/4πε₀r → r = 8.99 × 10⁹ × 50 × 1.60 × 10⁻¹⁹ ÷ 2.5 × 10⁶ = 2.9 × 10⁻¹⁴ m.
1/r or 1/r²?
Force and field strength: 1/r². Potential and potential energy: 1/r. Check which one the question asks for.
Adding fields versus adding potentials
Fields are vectors: add with directions (subtract if opposite). Potentials are scalars: add with the sign of each charge.
Direction of the force on an electron in a field
Opposite to the field direction; electrons are pushed towards higher potential.
Units before substituting in field questions
r and d in metres, charges in coulombs (nC = 10⁻⁹, μC = 10⁻⁶), p.d. in volts.
Deflection between plates: which suvat?
Time from the constant velocity parallel to the plates (t = L/v); then y = ½at² with a = QE/m across them.

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