4.4 - Electric and Magnetic Fields

An electric field is a region in which a charged particle experiences an electric force. The field exists because of source charges; a test charge reveals the field but does not create the field being described.

A particle's motion depends on the resultant of electric force and any other forces. Between horizontal plates, for example, a charged particle can accelerate vertically while retaining horizontal velocity, producing a curved path. A stationary charged particle initially accelerates in the force direction.

Electric field strength at a point is $\mathbf{E}=\mathbf{F}/Q$ for a small positive test charge $Q$. It is a vector with unit $\mathrm{N\,C^{-1}}$, equivalent to $\mathrm{V\,m^{-1}}$.

For a particle of charge $q$, the electric force is $\mathbf{F}=q\mathbf{E}$. Use the magnitude $|q|E$ for size; a positive charge is forced along $\mathbf{E}$ and a negative charge in the opposite direction. Combine this force vector with weight or other forces before applying $F=ma$.

A charge of $4.0\times10^{-12}\,\mathrm{C}$ in a field of $8.0\times10^5\,\mathrm{N\,C^{-1}}$ experiences force $F=EQ=3.2\times10^{-6}\,\mathrm{N}$. Whether it rises depends on whether this upward force exceeds its weight.

For two point charges separated by centre-to-centre distance $r$ in free space, the force magnitude is $F=|Q_1Q_2|/(4\pi\varepsilon_0r^2)$. The force lies along the line joining the charges.

The two charges exert equal-magnitude, opposite forces on one another. Doubling one charge doubles $F$; doubling separation reduces $F$ to one quarter. Convert nanocoulombs and centimetres to coulombs and metres before substitution. With several source charges, calculate each force vector and add components.

A point charge $Q$ produces electric field magnitude $E=|Q|/(4\pi\varepsilon_0r^2)$ at distance $r$. The direction is radially outward for positive $Q$ and radially inward for negative $Q$.

At a point between equal positive and negative charges, both field contributions point from the positive charge towards the negative charge, so their magnitudes add. Between two equal positive charges, the midpoint contributions oppose and cancel.

Electric potential is potential energy per unit positive charge; electric field describes how rapidly potential falls with position. Along one dimension, $E=-\mathrm{d}V/\mathrm{d}x$, so the field points from higher potential towards lower potential.

Moving a positive charge along the field lowers its potential energy; moving it against the field requires external work. For a radial field with $V(\infty)=0$, the potential at radius $r$ equals the area under the $E$-against-$r$ curve from $r$ to infinity, with sign set by the source charge.

Between large parallel plates, away from the edges, the electric field is approximately uniform: $E=V/d$, where $V$ is the potential difference and $d$ is the perpendicular plate separation.

For $V=800\,\mathrm{V}$ and $d=0.050\,\mathrm{m}$, $E=1.6\times10^4\,\mathrm{V\,m^{-1}}$. An electron has force magnitude $eE=2.6\times10^{-15}\,\mathrm{N}$ opposite to the field direction.

With zero potential chosen at infinity, a point charge produces $V=Q/(4\pi\varepsilon_0r)$. The sign of $V$ follows the source charge, while $r$ is a positive distance.

Potentials from several charges add algebraically because potential is scalar: calculate each $Q_i/(4\pi\varepsilon_0r_i)$ and sum the signed values. A charge $q$ at potential $V$ has potential energy $U=qV$; a change obeys $\Delta U=q\Delta V$.

For a positive particle approaching a positive nucleus, $V$ and $U$ rise. If it momentarily stops at closest approach, its lost kinetic energy equals the increase in electric potential energy, allowing $r$ to be found from $U=Qq/(4\pi\varepsilon_0r)$.

A positive point charge has straight radial lines directed outward and concentric spherical equipotentials; a negative charge reverses the arrows. A uniform field has parallel, equally spaced field lines and parallel equipotentials perpendicular to them.

Line density is a drawing convention for relative field strength, not a count of physical strands. Curved field lines show the local acceleration direction, but a moving charge with sideways velocity need not follow the line.

Capacitance is $C=Q/V$: the charge stored on either plate per unit potential difference across the capacitor. Its unit is the farad, $1\,\mathrm{F}=1\,\mathrm{C\,V^{-1}}$.

For an ideal fixed capacitor, $C$ is set by its construction; increasing $V$ increases $Q$ in proportion rather than changing $C$. During charging, current transfers charge, so $Q=\int I\,\mathrm{d}t$; the area under an current-time graph gives the delivered charge.

A $220\,\mu\mathrm{F}$ capacitor at $12.0\,\mathrm{V}$ stores $Q=CV=(220\times10^{-6})(12.0)=2.64\times10^{-3}\,\mathrm{C}$. Convert microfarads before calculating.

Charging requires work because the potential difference rises as charge accumulates. For a fixed capacitor, a graph of $V$ against $Q$ is a straight line from the origin to $(Q,V)$, so the stored energy is its triangular area: $W=\tfrac12QV$.

The last two forms follow by substituting $Q=CV$. For $C=47\,\mu\mathrm{F}$ and $V=400\,\mathrm{V}$, $W=\tfrac12(47\times10^{-6})(400)^2=3.76\,\mathrm{J}$.

The time constant is $\tau=RC$. After one time constant, a discharging value is $e^{-1}\approx0.37$ of its initial value; a charging capacitor has reached $1-e^{-1}\approx0.63$ of its final value. A larger $R$ or $C$ stretches the curve horizontally.

During charging, increasing capacitor p.d. leaves decreasing p.d. across the resistor, so current falls. When the capacitor reaches the supply p.d., resistor p.d. and current are zero. During discharge, the capacitor itself drives the current.

Connect a capacitor and resistor in series with a d.c. supply and a switch that selects charging or discharging. Connect an oscilloscope or voltage sensor/data logger in parallel with the capacitor so it records $V_C$ without becoming part of the series current path.

Use a high-input-resistance sensor, record the actual resistor value, and sample much faster than the time constant. Discharge the capacitor safely before rewiring and observe polarity for an electrolytic capacitor.

For discharge through resistance $R$, $Q=Q_0e^{-t/(RC)}$, $V=V_0e^{-t/(RC)}$ and current magnitude $I=I_0e^{-t/(RC)}$. Each quantity falls by the same fractional factor over equal time intervals.

A fitted gradient $m$ gives $RC=-1/m$ and then $C=-1/(mR)$ if $R$ is known. Alternatively, substitute one paired value and time into the exponential equation, retaining consistent units.

The angle $\theta$ is between the magnetic field and the normal to the coil plane. Flux is maximum when the field is perpendicular to the plane and zero when it lies in the plane. For identical turns, linkage multiplies the one-turn flux by $N$.

A 50-turn coil of area $1.2\times10^{-3}\,\mathrm{m^2}$ perpendicular to $B=0.018\,\mathrm{T}$ has $N\Phi=50(0.018)(1.2\times10^{-3})=1.1\times10^{-3}\,\mathrm{Wb}$.

A charge moving through magnetic flux density $B$ experiences force magnitude $F=B|q|v\sin\theta$, where $\theta$ is the angle between velocity and field. The force is zero for parallel motion and maximum at $90^\circ$.

For a positive charge, Fleming's left-hand rule uses first finger for field, second finger for conventional current/positive-charge motion, and thumb for force. Reverse the force direction for a negative charge. The force is perpendicular to both $\mathbf{v}$ and $\mathbf{B}$.

Because the magnetic force is perpendicular to velocity, it does no work and changes direction rather than speed. With $\mathbf{v}\perp\mathbf{B}$ in a uniform field, it can supply centripetal force and produce a circular path. Crossed electric and magnetic forces balance when $qE=Bqv$.

A straight conductor of active length $l$ carrying current $I$ in magnetic flux density $B$ experiences $F=BIl\sin\theta$. Here $\theta$ is the angle between conventional current and the field, and $l$ is only the conductor length inside the field.

Apply Fleming's left-hand rule: first finger field from north to south, second finger conventional current, thumb force. Opposite sides of a current-carrying coil can experience opposite forces, forming a couple and a turning moment.

An e.m.f. is induced only while magnetic flux linkage through the coil changes. Relative motion between magnet and coil changes field strength and geometry at the turns; faster change gives a larger induced e.m.f.

Before the magnet reaches the coil, linkage is nearly constant and e.m.f. is near zero. Approaching and leaving produce opposite polarities because the linkage first changes one way and then the other. A falling magnet can give a larger, narrower later peak because it is moving faster.

Current in a primary coil creates a magnetic field. When that current changes, the field and the flux linkage through a nearby secondary coil change, inducing an e.m.f. in the secondary.

A secondary current flows only if its circuit is complete. A diode may select one polarity so a connected capacitor charges rather than alternately charging and discharging. Switching steady d.c. on or off produces only transient secondary e.m.f.

The combined Faraday-Lenz law is $\mathcal{E}=-\mathrm{d}(N\Phi)/\mathrm{d}t$. The magnitude is the rate of change of flux linkage; the minus sign states that any induced current produces effects opposing the change that caused it.

If linkage changes from $0.012$ to $0.003\,\mathrm{Wb}$ in $0.020\,\mathrm{s}$, the average e.m.f. magnitude is $|0.003-0.012|/0.020=0.45\,\mathrm{V}$. Direction requires a declared positive normal and loop direction, or a clear Lenz-law statement.