PMDC Unit 44 · 9 outcomes

Electrostatics

Unit 8 is still charges: Coulomb, the field as a force-per-charge, field lines, an infinite sheet, potential, and a capacitor charging through a resistor.

Open the Unit 44 quiz Worked numericals

8.1

Coulomb’s law

PMDC: State Coulomb’s law and explain that the force between two point charges is reduced in a medium other than free space.

Two point charges push or pull along the line joining them. Like charges repel. Unlike attract. The force grows with each charge and falls as 1/r².

In a material (dielectric), the medium polarises and weakens the force. Divide by the relative permittivity $ε_{r}$ (or use $ε = ε_{0} ε_{r}$ ).

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F = k q_{1} q_{2} / r^{2} r e p e l

$F = k q_{1} q_{2} / r^{2} k = 1/ (4 π ε_{0}) \approx 9 \times 1 0^{9} N m^{2} C^{- 2}$
$inam e d i u m : F^{'} = F / ε_{r} = k q_{1} q_{2} / (ε_{r} r^{2})$
vacuum: $ε_{r}$ = 1 (largest force)

Hold these

$ε_{r}$ > $1 \Rightarrow w e ak er$ force than in vacuum.
Force is a vector. Opposite charges: attractive, along the join.

8.2

Electric field as a field of force

PMDC: Describe the concept of an electric field as an example of a field of force.

A field of force is a rule that tells you the force on a test object at each point. The electric field E is the force on a small positive test charge, per unit charge. You do not need to put the test charge there for the field to exist.

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E = F / q_{0} = 4 N / C F = 8 N

$E = F / q_{0} (q_{0}$ small, positive)
unit: N C⁻¹ or V m⁻¹

Hold these

E points the way a + charge would be pushed.

8.3

E due to two point charges

PMDC: Calculate the magnitude and direction of the electric field at a point due to two charges of the same or opposite signs.

Each charge makes its own E. Add them as vectors. Same signs: between them the two arrows oppose, so they can cancel. Opposite signs: between them the arrows agree, so they add.

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E_{mi d}

= 2 kq / a² toward −

E (point $c ha r g e) = k$ q / r² (away if q > 0)
$E_{n e t}$ = vector sum of each E

Hold these

Do not add the charges and then use one r — unless they sit together.
Direction first, then size.

8.4

Field lines for two point charges

PMDC: Sketch the electric field lines for two point charges of equal magnitude with the same or opposite signs.

Lines leave + and enter −. They never cross. Density of lines is a picture of |E|. Two pluses: lines leave both and duck away in the middle. A plus and a minus: lines leave + and land on −, crowding between them.

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unlike: + → −

Hold these

Equal + and −: every line that leaves + ends on −.
Two pluses: lines go to infinity, with a quiet point in the middle.

8.5

Infinite charged sheet

PMDC: Describe and draw the electric field due to an infinite conducting plate of positive or negative charge.

An infinite conducting sheet with charge on it has field lines straight out, perpendicular to the plate, same strength everywhere on one side. + sheet: arrows away. − sheet: arrows toward. Unlike a point charge, |E| does not fall with distance for an infinite sheet.

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o u t s i d e E = σ / ε_{0}

infinite conducting sheet: $E = σ / ε_{0}$ (near, one side as used in MDCAT)
direction: away if $σ$ > 0, toward if $σ$ < 0

Hold these

$σ$ is charge per area (C m⁻²).
Parallel arrows, equally spaced — that is the sketch.

8.6

Electric potential

PMDC: Define electric potential at a point in terms of the work done in bringing a unit positive charge from infinity to that point.

Potential V at a point is the work you must do (slowly, no KE left over) to drag a + $1 C$ charge in from infinity. High V is a hill for positive charges. They roll ‘downhill’ toward lower V.

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V = W / q_{0} = k q / r

$V = W_{\infty} \to p o in t / q_{0}$
unit: volt

Hold these

Infinity is the usual zero of V.
V is a scalar.

8.7

The volt

PMDC: Define the unit of potential.

One volt means one joule of work per coulomb. If $V = 5$ V, bringing + $1 C$ from infinity took $5 J$ .

$1 V = 1 J C^{- 1}$
also $1 V$ m⁻¹ = $1 N$ C⁻¹

Hold these

eV is an energy unit: $1 eV = 1.6 \times 1 0^{- 19}$ J. Not a potential.

8.8

Potential due to a point charge

PMDC: Derive an expression for electric potential at a point due to a point charge.

Work against the Coulomb force, from infinity in to distance r. The result falls as 1/r, not 1/r². Several charges: add the V values as numbers (with sign).

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V = k q / r E = k q / r^{2}

$V = k$ q / r (zero at infinity)
$V_{n e t} = Σ k q_{i} / r_{i}$

Hold these

Negative $c ha r g e \Rightarrow n e g a t i v e$ V.
E is the slope of V. Uniform E: $Δ V$ = E d.

8.9

Charging and discharging a capacitor through R

PMDC: Demonstrate charging and discharging of a capacitor through a resistance.

Charge: the capacitor fills through R. Current starts large and dies away. Voltage on C climbs toward the battery voltage. Time scale is $τ$ = R C.

Discharge: battery gone, C empties through R. Voltage and current both fall toward zero, same $τ$ .

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τ = R C = 4 Q / Q_{0} = 1 - e^{- t / τ}

$τ = R C$
$charging : q = q_{0} (1 - e^{- t / R C})$
$discharging : q = q_{0} e^{- t / R C}$

Hold these

After one $τ$ , charge is about 63 % of the way (charging) or 37 % left (discharging).
Large R or C means a slow curve.

Unit 44 formula sheet

Write these until they are cheaper than thinking, then open the quiz.

$F = k q_{1} q_{2} / r^{2} k = 1/ (4 π ε_{0}) \approx 9 \times 1 0^{9} N m^{2} C^{- 2}$
$inam e d i u m : F^{'} = F / ε_{r} = k q_{1} q_{2} / (ε_{r} r^{2})$
vacuum: $ε_{r}$ = 1 (largest force)
$E = F / q_{0} (q_{0}$ small, positive)
unit: N C⁻¹ or V m⁻¹
E (point $c ha r g e) = k$ q / r² (away if q > 0)
$E_{n e t}$ = vector sum of each E
infinite conducting sheet: $E = σ / ε_{0}$ (near, one side as used in MDCAT)
direction: away if $σ$ > 0, toward if $σ$ < 0
$V = W_{\infty} \to p o in t / q_{0}$
unit: volt
$1 V = 1 J C^{- 1}$
also $1 V$ m⁻¹ = $1 N$ C⁻¹
$V = k$ q / r (zero at infinity)
$V_{n e t} = Σ k q_{i} / r_{i}$
$τ = R C$
$charging : q = q_{0} (1 - e^{- t / R C})$
$discharging : q = q_{0} e^{- t / R C}$

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