Electric and magnetic fields

**Two** — positive and negative. **Like charges repel** (push apart); **unlike charges attract** (pull together). Unit: the coulomb (C).

The force between two point charges is $F = k\dfrac{q_{1}q_{2}}{r^{2}}$ — proportional to each charge and to the inverse square of the distance r between them.

**k = 8.99 × 10⁹ N m² C⁻²** — given in the data booklet. It sets the strength of the electric force.

It **halves** — F is proportional to each charge, so halving q_{1} (or q_{2}) halves F.

It is divided by **2² = 4** — F is proportional to 1/r² (the inverse-square law).

**Electrons** (tiny negative particles). Gaining electrons makes an object negative; losing them makes it positive.

**Friction** (rubbing), **contact** (touching a charged object), and **induction** (bringing a charge near and grounding — no contact).

The **opposite** sign to the charge brought near — and it never needs contact.

Charge is never created or destroyed, only **moved**. If one object gains −q, another is left with +q, so the total is unchanged.

**Attractive** — they have opposite signs, so they pull together.

It is **inversely proportional to r²** (inverse-square): double r → quarter F; triple r → one-ninth F.

**−Q/2** each: the total (−Q) is shared equally.

**k = 1/(4πε₀)**; in a medium use its permittivity ε, which gives a smaller force.

So the electrons that flowed away cannot flow back; the sphere keeps the opposite charge to the rod.

The **force per unit charge** on a small positive test charge: $E = \dfrac{F}{q}$. It is a **vector**. Unit: **N C⁻¹**.

**N C⁻¹** (newtons per coulomb).

$E = \dfrac{kQ}{r^{2}}$ — Coulomb constant k × charge Q ÷ distance² (derived from Coulomb's law with E = F ÷ q).

**Outward** — away from the charge (a positive test charge is pushed away).

**Inward** — toward the charge (a positive test charge is pulled in).

E falls to a **quarter** — the field is inverse-square ($E \propto 1/r^{2}$).

**Superposition** — add the field from each charge **as a vector** (same direction → add sizes; opposite → subtract).

At the **midpoint** — the two equal fields point in opposite directions and cancel (the null point).

Rearrange $E = \dfrac{F}{q}$ to $F = qE$ — multiply the charge by the field strength.

A **vector** — it has size and direction (the direction a +test charge is pushed).

$F = qE = (2.0\times10^{-9})(5.0\times10^{4}) = 1.0\times10^{-4}$ N, along the field.

**Zero.** Outside it is the same as a point charge at the centre: E = kQ/r².

Closer lines mean a **stronger** field; they never cross.

On the line joining them, **nearer the smaller charge**, where their fields are equal and opposite.

A field with the **same strength and direction everywhere** — drawn as **evenly-spaced, parallel** lines. You get one in the gap between two parallel charged plates.

**Evenly-spaced parallel lines** running from the **+ plate** to the **− plate** (the direction a positive charge is pushed).

$E = \dfrac{V}{d}$ — voltage across the plates ÷ the gap between them. Given in the data booklet. Unit: V m⁻¹.

**Volts per metre (V m⁻¹)**, which is the same as **N C⁻¹** (newtons per coulomb).

E **doubles** — field strength is inversely proportional to the separation d (E = V ÷ d).

$F = qE$ (rearranged from the data-booklet definition $E = \dfrac{F}{q}$). Bigger charge or stronger field → bigger force.

$W = qV$ (in joules): the data booklet's V = W/q rearranged. It is the energy the charge gains, and for a charge from rest its kinetic energy.

The energy a charge of **e** (1.6 × 10⁻¹⁹ C) gains moving through **1 V**: 1 eV = 1.6 × 10⁻¹⁹ J. A charge e through V volts gains V eV.

Multiply by 1.6 × 10⁻¹⁹: 250 × 1.6 × 10⁻¹⁹ = 4.0 × 10⁻¹⁷ J.

E = V ÷ d = 600 ÷ 0.020 = 3.0 × 10⁴ V m⁻¹.

From the **+ plate to the − plate** — the direction a **positive** charge would be pushed.

They **bulge outwards** and spread apart: the field is weaker there (edge effects).

**qE = mg**, so q = mg/E = mgd/V.

Every charge measured was a **whole-number multiple of e**: charge is **quantised**.

The region around a magnet **or a current** where a magnetic force is felt. We picture it with **field lines** — closer lines mean a stronger field.

**Concentric circles** centred on the wire. Use the **right-hand grip rule**: thumb along the current I, fingers curl the way the circles point.

From the **N pole to the S pole** (outside the magnet). Unlike poles (N–S) attract; like poles (N–N) repel.

They **attract** (parallel currents come together).

They **repel** (anti-parallel currents push apart).

$\dfrac{F}{L} = \mu_{0}\dfrac{I_{1}I_{2}}{2\pi r}$ — given in the data booklet.

The **permeability of free space**, a constant equal to 4π × 10⁻⁷ T m A⁻¹.

It is **inversely proportional** to r: F/L ∝ 1/r. Doubling r halves F/L.

It **doubles** — F/L is proportional to each current (F/L ∝ I_{1} I_{2}).

Each wire sits in the **magnetic field** created by the other, so each feels a force. By Newton's third law the forces are equal and opposite.

It flips between attraction and repulsion (the currents become anti-parallel, or parallel, instead).

Attraction — same-direction (parallel) currents attract.

**Uniform**: parallel, evenly spaced lines. Outside it looks like the field of a **bar magnet**.

Right hand: fingers curl with the current, **thumb points to the N end** (where the field lines leave).

Along the field line, with its **N end pointing in the field direction**.

The **potential energy per unit positive charge** at a point: $V = kQ/r$. Unit: **volt** (J C⁻¹). It is a **scalar**.

It falls off as **1/r** — double r and V halves. (The field E falls off as 1/r².)

**Yes** — near a negative charge V is negative. Unlike gravity (always negative), charge has two signs.

At **infinity**. V is the work per coulomb to bring a small +charge from infinity to the point.

$E_p = kQq/r$ (a scalar for the **pair**, in joules, with the sign of Qq).

**Like** charges (Qq > 0) ⇒ **positive** E_p; **unlike** charges (Qq < 0) ⇒ **negative** E_p (bound).

$W = q\,\Delta V$, where $\Delta V = V_{final} - V_{initial}$.

Because **potential depends only on position**, so W = qΔV uses only the two endpoints.

$V = kQ/r = (8.99\times10^9)(2.0\times10^{-9})/0.30 = 60$ V.

**Potential** V is a **scalar** (just add them); **field** E is a **vector** (add by components).

The **field** did the work — energy is **released**. A positive W means **you** supplied energy.

Compute each $V = kQ/r$ **with its sign**, then **add** them as numbers (scalars).

$E = -\Delta V_e/\Delta r$: the field is **minus the potential gradient**; it points towards **lower** potential.

**None**: ΔV = 0, so W = qΔV = 0. Field lines cross equipotentials at **90°**.

**Constant**, equal to its value at the surface, kQ/R; so the field inside is **zero**.