Flashcards · 9702 · A Level Topic 18

Electric fields (A Level) flashcards

23 cards on A Level Topic 18, subtopics 18.1 to 18.5. Pick a subtopic or study them all.

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Electric field: A region of space in which a charge experiences a force due to other charges. A field of force.
Which way do electric field lines point?: In the direction of the force on a small positive test charge: away from positive charges, towards negative charges. Closer lines mean a stronger field.
Field-line pattern around an isolated point charge: Radial lines: outward from a positive charge, inward to a negative charge, spreading apart with distance.
Field-line pattern between two charged parallel plates: Parallel, equally spaced lines from the positive to the negative plate: a uniform field (with some curving at the edges).

Field strength between charged parallel plates: E = V ÷ d, where V is the p.d. between the plates and d their separation. Unit V m⁻¹ (= N C⁻¹). It holds only for a uniform field.
Plates 5.0 mm apart have a p.d. of 200 V. Find the field strength.: E = V ÷ d = 200 ÷ 0.0050 = 4.0 × 10⁴ V m⁻¹.
A charged particle enters a uniform electric field at right angles to it. Describe its path.: A parabola, like a projectile: constant velocity across the field, constant acceleration a = qE ÷ m along it.
An electron is accelerated from rest through 500 V. Find its speed.: eV = ½mv², so v = √(2eV ÷ m) = √(2 × 1.60 × 10⁻¹⁹ × 500 ÷ 9.11 × 10⁻³¹) = 1.33 × 10⁷ m s⁻¹.
The electronvolt: The energy gained by an electron accelerated through a p.d. of 1 V: 1 eV = 1.60 × 10⁻¹⁹ J.

Coulomb’s law: F = Qq ÷ (4πε₀r²): the force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation.
Value of 1/(4πε₀), and the unit of ε₀: 1/(4πε₀) ≈ 8.99 × 10⁹ N m² C⁻². ε₀ = 8.85 × 10⁻¹² F m⁻¹.
Two +2.0 μC point charges are 0.10 m apart. Find the force between them.: F = 8.99 × 10⁹ × (2.0 × 10⁻⁶)² ÷ 0.10² = 3.6 N, repulsive.
Electric vs gravitational force between point objects: Both are inverse square laws. Gravity is always attractive; the electric force is attractive or repulsive, depending on the signs of the charges.

Electric field strength: The force per unit positive charge on a small test charge at that point: E = F ÷ q. Unit N C⁻¹. A vector.
Field strength of a point charge Q: E = Q ÷ (4πε₀r²), with r measured from the charge.
Find the field strength 0.20 m from a point charge of +5.0 × 10⁻⁶ C.: E = 8.99 × 10⁹ × 5.0 × 10⁻⁶ ÷ 0.20² = 1.12 × 10⁶ N C⁻¹, directed radially outward.
How is an isolated charged sphere treated for points outside it?: As a point charge at its centre, so r is measured from the centre, not the surface.
Force on a charge q in a field of strength E: F = Eq, in the direction of E for a positive charge and opposite to E for a negative charge.

Electric potential: The work done per unit positive charge in bringing a small test charge from infinity to that point. V = Q ÷ (4πε₀r). Unit V (J C⁻¹). A scalar.
How does the sign of electric potential differ from gravitational potential?: Gravitational potential is always negative (gravity only attracts). Electric potential is positive near a positive charge and negative near a negative charge.
Electric potential energy of a pair of point charges: Ep = Qq ÷ (4πε₀r), the work done bringing one charge from infinity to distance r from the other.
Relationship between field strength and potential: E = −dV/dr: the field strength is minus the potential gradient. The field points towards decreasing potential.
Equipotentials: Lines or surfaces of constant potential. They are perpendicular to field lines, and no work is done moving a charge along one.