4.P — Worked Problems: Electricity and Magnetism

Fourteen problems across Part 4, climbing from standard to genuinely awkward. Every solution shows the arithmetic. Attempt each before opening it.

Problem 1 — Three charges on a triangle

Charges $+2.0\ \mu$C, $+2.0\ \mu$C and $-3.0\ \mu$C sit at the corners of an equilateral triangle of side 0.30 m. Find the net force on the negative charge.

Solution

The negative charge is attracted to both positive ones. By symmetry the two forces have equal magnitude, and the resultant points along the line from the negative charge to the midpoint of the other two.

Magnitude of each:

F = \frac{(8.988\times10^{9})(3.0\times10^{-6})(2.0\times10^{-6})}{(0.30)^2} = \frac{5.393\times10^{-2}}{0.090} = 0.599\ \text{N}

The angle between the two force vectors is 60°, since the triangle is equilateral. Adding two equal vectors at 60°:

F_{\text{net}} = 2F\cos(30°) = 2(0.599)(0.8660) = 1.038\ \text{N}

1.04 N, directed towards the midpoint of the two positive charges.

What to notice. The rule 2F\cos(\theta/2) for two equal vectors separated by \theta is worth committing to memory; it saves resolving into components every time. Check the extremes: \theta = 0 gives 2F and \theta = 180° gives 0, both correct.

Problem 2 — The field where nothing pushes

Charges $+9.0\ \mu$C and $-4.0\ \mu$C are 1.0 m apart. Find every point on the line through them where the electric field is zero.

Solution

Set up coordinates with the +9 at x = 0 and the -4 at x = 1.0 m.

Between them the two fields both point towards the negative charge, so they add and can never cancel. No solution there.

To the left of the positive charge the fields point in opposite directions, but you are always closer to the larger charge, so the positive one always wins. No solution.

To the right of the negative charge the fields oppose, and here you are closer to the smaller charge, so a balance is possible. Let the point be at x = 1 + d.

\frac{k(9\times10^{-6})}{(1+d)^2} = \frac{k(4\times10^{-6})}{d^2}

\frac{9}{(1+d)^2} = \frac{4}{d^2}

Cross-multiply and take square roots — both sides are positive so this is safe:

3d = 2(1+d) \quad\Longrightarrow\quad 3d = 2 + 2d \quad\Longrightarrow\quad d = 2.0\ \text{m}

The field is zero at x = 3.0 m, two metres beyond the negative charge.

What to notice. The null point is always on the far side of the smaller charge, and never between charges of opposite sign. For two charges of the same sign the situation reverses: the null point lies between them, nearer the smaller one. Knowing which region to search saves solving a quadratic that has a spurious root.

Problem 3 — Gauss's law on a nested shell

A solid conducting sphere of radius 5.0 cm carries $+6.0\ \mu$C. It sits inside a concentric conducting shell of inner radius 10 cm and outer radius 12 cm, which carries a net charge of $-2.0\ \mu$C. Find the field at r = 3, 7, 11 and 20 cm, and the charge on each surface.

Solution

At r = 3 cm — inside the solid conductor. E = 0, because a conductor in equilibrium has no interior field.

At r = 7 cm — between the sphere and the shell. Enclosed charge is $+6.0\ \mu$C:

E = \frac{kQ}{r^2} = \frac{(8.988\times10^{9})(6.0\times10^{-6})}{(0.070)^2} = \frac{5.393\times10^{4}}{4.90\times10^{-3}} = 1.10\times10^{7}\ \text{V/m}

outward.

At r = 11 cm — inside the metal of the shell. E = 0 again.

That tells us the charge on the shell's inner surface. A Gaussian sphere at 11 cm has zero flux, so zero enclosed charge, so:

+6.0 + q_{\text{inner}} = 0 \quad\Longrightarrow\quad q_{\text{inner}} = -6.0\ \mu\text{C}

The shell's total is $-2.0\ \mu$C, so:

q_{\text{outer}} = -2.0 - (-6.0) = +4.0\ \mu\text{C}

At r = 20 cm — outside everything. Enclosed charge is $6.0 - 2.0 = +4.0\ \mu$C:

E = \frac{(8.988\times10^{9})(4.0\times10^{-6})}{(0.20)^2} = \frac{3.595\times10^{4}}{0.040} = 8.99\times10^{5}\ \text{V/m}

outward.

What to notice. The shell's outer surface ends up positive even though the shell itself carries net negative charge. The inner charge forces $-6\ \mu$C onto the inner surface, and whatever is left over has nowhere to go but outside. This is the general rule: the outer surface of any conductor carries whatever the enclosed charge plus the conductor's own charge adds up to, and it is why a Faraday cage does not shield the outside from charge placed inside.

Problem 4 — Capacitor with a slab

A parallel-plate capacitor has plates of area 200 cm² separated by 2.0 mm and is charged to 100 V, then disconnected from the supply. A slab of dielectric with \kappa = 4.0 and thickness 1.0 mm is inserted, centred in the gap. Find the new voltage and the energy before and after.

Solution

Initial capacitance:

C_0 = \frac{\varepsilon_0A}{d} = \frac{(8.854\times10^{-12})(0.0200)}{2.0\times10^{-3}} = 8.854\times10^{-11}\ \text{F} = 88.5\ \text{pF}

Charge, which stays fixed since the supply is disconnected:

Q = C_0V = (8.854\times10^{-11})(100) = 8.854\times10^{-9}\ \text{C}

The field in each region. With Q fixed, the field in the air gaps is unchanged:

E_0 = \frac{V_0}{d} = \frac{100}{2.0\times10^{-3}} = 5.0\times10^{4}\ \text{V/m}

Inside the dielectric it is reduced by \kappa:

E_{\text{diel}} = \frac{5.0\times10^{4}}{4.0} = 1.25\times10^{4}\ \text{V/m}

New voltage — add the potential drops across the 1.0 mm of air and the 1.0 mm of dielectric:

V = (5.0\times10^{4})(1.0\times10^{-3}) + (1.25\times10^{4})(1.0\times10^{-3}) = 50 + 12.5 = 62.5\ \text{V}

Energy before:

U_0 = \frac{1}{2}C_0V_0^2 = \frac{1}{2}(8.854\times10^{-11})(100)^2 = 4.43\times10^{-7}\ \text{J}

Energy after, using U = \frac{1}{2}QV since Q is what stayed constant:

U = \frac{1}{2}(8.854\times10^{-9})(62.5) = 2.77\times10^{-7}\ \text{J}

What to notice. The energy fell by 1.66\times10^{-7} J. Where did it go? Into the work done by the capacitor on the slab: the fringing field at the edge of the plates pulls the dielectric in, and if you let go of it, it accelerates inwards. If you hold it back you can extract that energy. This is the operating principle of an electrostatic actuator, and it is also why a dielectric is pulled in rather than pushed out — the system lowers its energy by admitting the slab.

Problem 5 — Two capacitors, one charged

A 6.0 μF capacitor is charged to 200 V and then connected across an uncharged 3.0 μF capacitor. Find the final voltage, and the energy before and after.

Solution

Charge is conserved — it redistributes but does not vanish:

Q_{\text{total}} = (6.0\times10^{-6})(200) = 1.20\times10^{-3}\ \text{C}

In parallel the two share a common voltage, and capacitances add:

C_{\text{total}} = 6.0 + 3.0 = 9.0\ \mu\text{F}

V = \frac{Q}{C} = \frac{1.20\times10^{-3}}{9.0\times10^{-6}} = 133.3\ \text{V}

Energy before:

U_i = \frac{1}{2}(6.0\times10^{-6})(200)^2 = 0.120\ \text{J}

Energy after:

U_f = \frac{1}{2}(9.0\times10^{-6})(133.3)^2 = \frac{1}{2}(9.0\times10^{-6})(17{,}769) = 0.0800\ \text{J}

One third of the energy has vanished.

What to notice. This is a famous puzzle, and the answer is that the lost energy is real and unavoidable. Charge flowed from one capacitor to the other through the connecting wires, and any real wire has resistance, so I^2R turned a third of the energy into heat. Make the resistance smaller and the current is larger and the time is shorter, and the total heat comes out the same — you cannot escape it by using better wire. With genuinely zero resistance the circuit becomes an oscillator (an LC circuit, Volume III Chapter 1.5) and the energy is radiated away as electromagnetic waves instead. Either way, exactly one third is lost whenever charge is shared between an equal and a half-sized capacitor.

Problem 6 — Drift velocity in a thin film

A gold film 100 nm thick and 2.0 μm wide carries 1.0 mA. Gold has 5.90\times10^{28} free electrons per m³. Find the drift velocity and compare with the copper wire from Chapter 4.4.

Solution

A = (100\times10^{-9})(2.0\times10^{-6}) = 2.0\times10^{-13}\ \text{m}^2

v_d = \frac{I}{nAe} = \frac{1.0\times10^{-3}}{(5.90\times10^{28})(2.0\times10^{-13})(1.602\times10^{-19})}

Denominator: (5.90\times10^{28})(2.0\times10^{-13}) = 1.18\times10^{16}, times 1.602\times10^{-19} gives 1.890\times10^{-3}.

v_d = \frac{1.0\times10^{-3}}{1.890\times10^{-3}} = 0.529\ \text{m/s}

Compared with 3.7\times10^{-4} m/s in the household wire — 1400 times faster.

What to notice. The current is 5000 times smaller and the drift velocity is 1400 times larger, entirely because the cross-section is 5\times10^{6} times smaller. Current density J = I/A is the quantity that actually matters for a conductor's behaviour, and here it is 5\times10^{9} A/m² against 5\times10^{6} in the wire. Metal films in chips run at current densities that would instantly vaporise a wire, and they survive only because they are in thermal contact with a silicon substrate that carries the heat away. At even higher densities the electron flow physically pushes metal atoms along — electromigration — which slowly opens circuits and is a genuine wear-out mechanism in processors.

Problem 7 — The heater that is too long

A 2.0 kW heater is designed for 230 V using nichrome wire of resistivity 1.1\times10^{-6} Ω·m and diameter 0.50 mm. Find the length required. Then find what power it delivers if the supply drops to 210 V.

Solution

Required resistance:

R = \frac{V^2}{P} = \frac{230^2}{2000} = \frac{52900}{2000} = 26.45\ \Omega

Cross-section:

A = \pi r^2 = \pi(0.25\times10^{-3})^2 = \pi(6.25\times10^{-8}) = 1.963\times10^{-7}\ \text{m}^2

Length, from R = \rho L/A:

L = \frac{RA}{\rho} = \frac{(26.45)(1.963\times10^{-7})}{1.1\times10^{-6}} = \frac{5.193\times10^{-6}}{1.1\times10^{-6}} = 4.72\ \text{m}

At 210 V, with R unchanged:

P = \frac{210^2}{26.45} = \frac{44100}{26.45} = 1667\ \text{W}

What to notice. A 9 % voltage drop causes a 17 % power drop, because power goes as V^2. This is why brownouts noticeably weaken heating appliances and why a long extension lead makes a kettle slow — the lead's resistance eats part of the voltage. Note also the 4.7 m of wire: that is why heating elements are always coiled, and why the coil must be supported on ceramic so the turns cannot touch and short out a section, which would reduce R and raise the power in the remaining wire until it melted.

Problem 8 — Velocity selector

A beam of ions of varying speeds passes through a region with a magnetic field of 0.080 T and an electric field perpendicular to it. Find the electric field needed to let ions of 3.0\times10^{5} m/s pass straight through, and show the answer does not depend on charge or mass.

Solution

For the ion to go straight, the electric and magnetic forces must cancel:

qE = qvB

The charge cancels:

E = vB = (3.0\times10^{5})(0.080) = 2.4\times10^{4}\ \text{V/m}

Nothing about q or m survives. Only the speed matters.

What to notice. This is the front end of every mass spectrometer, and it exists because the magnetic radius formula r = mv/qB from Chapter 4.5 mixes mass and speed together. Without a velocity selector, a heavy slow ion and a light fast one land in the same place and cannot be told apart. Select a single speed first and the subsequent bend measures m/q cleanly. J. J. Thomson used exactly this crossed-field arrangement in 1897 to measure the electron's charge-to-mass ratio, which is how the electron was discovered.

Problem 9 — Force between two wires

Two long parallel wires 5.0 cm apart carry 20 A and 30 A in the same direction. Find the force per metre and its direction. Then find where between them the magnetic field is zero.

Solution

\frac{F}{L} = \frac{\mu_0I_1I_2}{2\pi d} = \frac{(4\pi\times10^{-7})(20)(30)}{2\pi(0.050)} = \frac{(2\times10^{-7})(600)}{0.050}

Using \mu_0/2\pi = 2\times10^{-7}:

\frac{F}{L} = \frac{1.2\times10^{-4}}{0.050} = 2.4\times10^{-3}\ \text{N/m}

Attractive, since the currents are parallel.

The null point. Between the wires the two fields point in opposite directions. Let the point be x from the 20 A wire:

\frac{\mu_0(20)}{2\pi x} = \frac{\mu_0(30)}{2\pi(0.050-x)}

20(0.050-x) = 30x \quad\Longrightarrow\quad 1.0 = 50x \quad\Longrightarrow\quad x = 0.020\ \text{m}

2.0 cm from the 20 A wire, which is 3.0 cm from the 30 A wire — the distances are in the ratio of the currents, as the 1/r law requires.

What to notice. Compare with Problem 2. For the electric field of two like charges the null point is also between them, and it sits where the ratio of the squares of the distances matches the ratio of the charges, so it is nearer the smaller charge but not proportionally. Here, because the magnetic field of a wire falls as 1/r rather than 1/r^2, the null point sits at exactly the ratio of the currents. Different falloff, different geometry.

Problem 10 — Designing a solenoid

An electromagnet must produce 0.050 T over a 30 cm length using wire that can safely carry 5.0 A. How many turns are needed, and what length of wire if the coil radius is 2.0 cm? If the wire is copper of diameter 1.0 mm, what power does it dissipate?

Solution

Turns per metre, from B = \mu_0nI:

n = \frac{B}{\mu_0I} = \frac{0.050}{(4\pi\times10^{-7})(5.0)} = \frac{0.050}{6.283\times10^{-6}} = 7958\ \text{turns/m}

Over 0.30 m:

N = 7958\times0.30 = 2387\ \text{turns}

Wire length — each turn is a circle of radius 2.0 cm:

L_{\text{wire}} = N(2\pi r) = 2387\times2\pi(0.020) = 2387\times0.1257 = 300\ \text{m}

Resistance, with A = \pi(0.5\times10^{-3})^2 = 7.854\times10^{-7} m² and \rho_{\text{Cu}} = 1.68\times10^{-8}:

R = \frac{(1.68\times10^{-8})(300)}{7.854\times10^{-7}} = \frac{5.04\times10^{-6}}{7.854\times10^{-7}} = 6.42\ \Omega

Power:

P = I^2R = (25)(6.42) = 160\ \text{W}

What to notice. 160 watts of heat to make a field one thirtieth as strong as an MRI's, from a coil the size of a mug. That is why every serious electromagnet either uses an iron core, which multiplies the field by a factor of hundreds for free (Chapter 4.8), or superconducting wire, which has R = 0 and therefore P = 0 once the current is running. Note also that 2387 turns of 1 mm wire needs a winding depth of several layers over a 30 cm former, which the calculation above quietly assumed does not change the radius — a real design has to iterate.

Problem 11 — The falling magnet

A square loop of wire, side 10 cm and resistance 0.20 Ω, falls out of a region of uniform 0.50 T field at 2.0 m/s, with the field perpendicular to the loop. Find the induced EMF, the current, the force opposing the fall, and the power dissipated. Check that the power balances.

Solution

While the loop is leaving, only the bottom edge is outside the field, so the area still in the field shrinks at Lv per second.

EMF:

\mathcal{E} = BLv = (0.50)(0.10)(2.0) = 0.10\ \text{V}

Current:

I = \frac{\mathcal{E}}{R} = \frac{0.10}{0.20} = 0.50\ \text{A}

Force. The top edge is still in the field and carries this current, so it feels F = BIL:

F = (0.50)(0.50)(0.10) = 0.025\ \text{N}

directed upwards, opposing the fall, as Lenz's law requires.

Power dissipated:

P = I^2R = (0.25)(0.20) = 0.050\ \text{W}

Mechanical power delivered against the retarding force:

P = Fv = (0.025)(2.0) = 0.050\ \text{W}\ \checkmark

What to notice. The two match exactly, which they must — the loop's gravitational potential energy is being converted into electrical energy and then into heat, with nothing left over. If the loop is heavy enough that this force is small compared with its weight it keeps accelerating; if it is light enough, it reaches a terminal velocity where BILv equals mg. Setting B^2L^2v/R = mg gives v_{\text{term}} = mgR/B^2L^2, and that formula is the whole theory of eddy current braking.

Problem 12 — Transformer with real losses

A transformer steps 11 kV down to 240 V to supply a 50 kW load. Find the turns ratio and both currents. If the transformer is 97 % efficient, how much heat does it produce, and what does that cost over a year at 15 p per kWh?

Solution

Turns ratio:

\frac{N_s}{N_p} = \frac{240}{11000} = \frac{1}{45.8}

Secondary current:

I_s = \frac{P}{V_s} = \frac{50000}{240} = 208\ \text{A}

Primary current, allowing for the 97 % efficiency so that the input power is higher:

P_{\text{in}} = \frac{50000}{0.97} = 51{,}546\ \text{W}

I_p = \frac{51546}{11000} = 4.69\ \text{A}

Heat produced:

P_{\text{loss}} = 51546 - 50000 = 1546\ \text{W}

Annual cost, running continuously:

E = 1.546\ \text{kW}\times8760\ \text{h} = 13{,}543\ \text{kWh}

\text{Cost} = 13543\times0.15 = £2032

What to notice. Two thousand pounds a year to run a device that does nothing but change a voltage, and it is 97 % efficient. That number is why distribution transformer efficiency is regulated in most countries and why utilities will pay a large premium up front for a design one percentage point better — over a thirty-year life the saving dwarfs the purchase price. The 1546 W splits roughly into copper losses (I^2R in the windings, which rise with load) and iron losses (hysteresis and eddy currents in the core, which are constant whenever it is energised). That is why a transformer left connected with nothing plugged in still gets warm.

Problem 13 — Radiation pressure on a solar sail

A solar sail of area 1200 m² and mass 5.0 kg, perfectly reflecting, is at Earth's distance from the Sun where the intensity is 1361 W/m². Find its acceleration, and how long to reach 1 km/s.

Solution

Radiation pressure on a perfect reflector is twice that on an absorber, because the light's momentum is reversed:

P_{\text{rad}} = \frac{2S}{c} = \frac{2(1361)}{3.00\times10^{8}} = 9.07\times10^{-6}\ \text{Pa}

Force:

F = P_{\text{rad}}A = (9.07\times10^{-6})(1200) = 1.09\times10^{-2}\ \text{N}

Acceleration:

a = \frac{F}{m} = \frac{1.09\times10^{-2}}{5.0} = 2.18\times10^{-3}\ \text{m/s}^2

Time to 1 km/s:

t = \frac{v}{a} = \frac{1000}{2.18\times10^{-3}} = 4.59\times10^{5}\ \text{s} = 5.3\ \text{days}

What to notice. Eleven millinewtons — about the weight of a paperclip — from a sail the size of a tennis court. And that feeble force reaches a kilometre per second in under a week, because it never stops and carries no fuel. Over a year it would reach 69 km/s, far beyond anything chemical rockets achieve, which is the entire argument for solar sailing. The catch is the 1/r^2 falloff: the same sail at Jupiter's distance gets 27 times less. Sails are good for going inward or for slow patient missions outward, and useless for anything urgent.

Problem 14 — The capacitor and the displacement current

A parallel-plate capacitor with circular plates of radius 5.0 cm is charged by a steady current of 0.30 A. Find the rate of change of the electric field between the plates, and the magnetic field at a radius of 3.0 cm inside the gap.

Solution

Rate of change of E. From E = Q/\varepsilon_0A:

\frac{dE}{dt} = \frac{1}{\varepsilon_0A}\frac{dQ}{dt} = \frac{I}{\varepsilon_0A}

A = \pi(0.050)^2 = 7.854\times10^{-3}\ \text{m}^2

\frac{dE}{dt} = \frac{0.30}{(8.854\times10^{-12})(7.854\times10^{-3})} = \frac{0.30}{6.954\times10^{-14}} = 4.31\times10^{12}\ \text{V/m per second}

Magnetic field at r = 3.0 cm. Use the corrected Ampère law from Chapter 4.7. Inside the gap there is no real current, only displacement current, and a circular loop of radius r encloses the fraction of it passing through area \pi r^2:

I_{\text{disp,enc}} = I\frac{\pi r^2}{\pi R^2} = I\frac{r^2}{R^2} = 0.30\times\frac{(0.030)^2}{(0.050)^2} = 0.30\times0.36 = 0.108\ \text{A}

B(2\pi r) = \mu_0(0.108)

B = \frac{(4\pi\times10^{-7})(0.108)}{2\pi(0.030)} = \frac{(2\times10^{-7})(0.108)}{0.030} = 7.2\times10^{-7}\ \text{T}

What to notice. 0.72 microtesla — about seventy times weaker than the Earth's field, in a gap where no charge is flowing at all. It is real and it has been measured, and it is exactly what you would get at 3 cm from a 0.108 A wire. This is Maxwell's added term producing a field where the uncorrected Ampère law predicts none, and it is the term that makes electromagnetic waves possible. Note also the r^2/R^2: inside the gap B grows linearly with radius, exactly as it does inside a uniform current-carrying wire, because in both cases the enclosed current grows as the area.