Physics (Theory)

Use \(E_x = -\dfrac{dV}{dx}\).
\(\dfrac{dV}{dx} = -50\) V/m, so \(E_x = -(-50) = +50\) V/m = 50 N/C.
The positive sign means the field points along +x (towards decreasing potential).
Common mistake: forgetting the minus sign and choosing −x.

Joined by a wire, both spheres reach the same potential: \(\dfrac{kQ_1}{r_1} = \dfrac{kQ_2}{r_2}\), so \(Q_1/Q_2 = r_1/r_2\).
Surface field \(E = \dfrac{kQ}{r^2}\), so \(\dfrac{E_A}{E_B} = \dfrac{Q_1}{Q_2}\cdot\dfrac{r_2^2}{r_1^2} = \dfrac{r_1}{r_2}\cdot\dfrac{r_2^2}{r_1^2} = \dfrac{r_2}{r_1}\).
So the smaller sphere has the stronger surface field (this is why charge collects at sharp points).

Apply Ampere's circuital law on a circle of radius r = a/2 inside the wire.
Current enclosed: \(I_{enc} = I\dfrac{\pi r^2}{\pi a^2} = \dfrac{I}{4}\).
\(B(2\pi r) = \mu_0 I_{enc}\) ⇒ \(B = \dfrac{\mu_0 (I/4)}{2\pi (a/2)} = \dfrac{\mu_0 I}{4\pi a}\).
Check: inside the wire \(B = \dfrac{\mu_0 I r}{2\pi a^2}\), which also gives μ₀I/(4πa) at r = a/2.

Points of constant path difference S₂P − S₁P lie on hyperbolas (in 3-D, hyperboloids).
When D ≫ d, the part of these hyperbolas that falls on the screen near the centre is almost straight, so the fringes look like straight lines parallel to the slits.

Speed in the medium: \(v = \dfrac{c}{\sqrt{\varepsilon_r \mu_r}} = \dfrac{c}{\sqrt{(3/2)(8/3)}} = \dfrac{c}{\sqrt 4} = \dfrac{c}{2}\).
Frequency is fixed by the source and does not change on entering a new medium.
\(\lambda = v/\nu\), so the wavelength becomes half.

V_R = V_L = V_C = 10 V ⇒ the circuit is at resonance, source voltage V = V_R = 10 V, and R = X_L (same current through both).
After shorting C: \(Z = \sqrt{R^2 + X_L^2} = \sqrt2\,R\), so \(I = \dfrac{10}{\sqrt2 R}\).
\(V_L = I X_L = \dfrac{10}{\sqrt2 R}\cdot R = \dfrac{10}{\sqrt2} = 5\sqrt2\) V ≈ 7.07 V.

The diagnostic tool is X-ray imaging. X-rays have wavelengths from about 1 nm down to 10⁻³ nm.
400 nm–1 nm is ultraviolet, 1 mm–700 nm is infrared and 0.1 m–1 mm is microwaves.

At closest approach all KE becomes electrostatic PE: \(\tfrac12 m v^2 = \dfrac{k(2e)(Ze)}{d}\), so \(d \propto \dfrac{1}{v^2}\).
Halving v makes v² one-fourth, so d becomes 4 times: d' = 4d.

The original lens is like two identical plano-concave lenses in contact: \(\dfrac1F = \dfrac1f + \dfrac1f = \dfrac2f\).
So f = 2F = 2 × 10 = 20 cm (each piece is diverging, f = −20 cm).
By lens maker's formula: one curved surface instead of two halves the power.

Sustained interference needs waves of the same frequency with a constant phase difference.
(i), (iii) and (iv) all have angular frequency ω: cos ωt = sin(ωt + π/2), and (iv) has a fixed phase π/3.
(ii) has frequency 2ω, so it cannot interfere steadily with the others.

Resistances: \(R_1 = V^2/P_1\), \(R_2 = V^2/P_2\).
Series: \(R = V^2\left(\dfrac{1}{P_1}+\dfrac{1}{P_2}\right) = \dfrac{V^2 (P_1+P_2)}{P_1P_2}\).
Power at V/2: \(P = \dfrac{(V/2)^2}{R} = \dfrac{V^2}{4}\cdot\dfrac{P_1P_2}{V^2(P_1+P_2)} = \dfrac{P_1P_2}{4(P_1+P_2)}\).

Diffusion of majority carriers builds the depletion layer and barrier field; this field drives a drift current of minority carriers in the opposite direction.
Equilibrium is reached when the two currents are equal in size, so the net current is zero.

A is false: in diamagnetic substances the magnetic moments of electrons cancel, so their atoms have zero net magnetic moment.
R is false: a current loop always behaves as a magnetic dipole with moment m = IA.

Moving electrons have a de Broglie wavelength λ = h/p. Diffraction is a wave effect, so it is observed only because electrons behave as waves too.
Both statements are true and R correctly explains A.

A is true: this is the mass defect ΔM.
R is false: when nucleons bind, energy (binding energy = ΔMc²) is released, not absorbed. That released energy is why the nucleus is lighter.

Both statements are true.
But quantisation of energy comes from Bohr's postulate that angular momentum is quantised (mvr = nh/2π), not from the electrostatic force providing centripetal force (which holds even in a classical orbit). So R is not the explanation.

Einstein's equation: \(\dfrac{hc}{\lambda} = \phi_0 + eV_0\), with stopping potential V₀ = 0.80 V.
Photon energy: \(\dfrac{hc}{\lambda} = \dfrac{6.63\times10^{-34}\times3\times10^8}{200\times10^{-9}} = 9.945\times10^{-19}\) J = \(\dfrac{9.945\times10^{-19}}{1.6\times10^{-19}}\) ≈ 6.2 eV.
\(\phi_0 = 6.2 - 0.80 = 5.4\) eV.

Minima of a single slit: \(a\sin\theta = n\lambda\), position \(y = \dfrac{n\lambda D}{a}\).
For coincidence: \(n_1(400) = n_2(600)\). Smallest integers: n₁ = 3, n₂ = 2 (both give 1200 nm).
\(y = \dfrac{2\times600\times10^{-9}\times1.5}{1\times10^{-3}} = 1.8\times10^{-3}\) m = 1.8 mm.
Check: \(3\times400\times10^{-9}\times1.5/10^{-3}\) = 1.8 mm as well.

Bright fringe: \(y = \dfrac{n\lambda D}{d}\).
Coincidence: \(n_1(440) = n_2(660)\) ⇒ smallest n₁ = 3, n₂ = 2 (both 1320 nm).
\(y = \dfrac{2\times660\times10^{-9}\times1.5}{0.6\times10^{-3}} = 3.3\times10^{-3}\) m = 3.3 mm.

Maximum torque τ = NIAB, so with N, I, B equal, τ ∝ A.
Square: perimeter of one turn = L/N ⇒ side a = L/(4N), area \(A_1 = \dfrac{L^2}{16N^2}\).
Circle: \(2\pi r = L/N\) ⇒ r = L/(2πN), area \(A_2 = \pi r^2 = \dfrac{L^2}{4\pi N^2}\).
\(\dfrac{\tau_1}{\tau_2} = \dfrac{A_1}{A_2} = \dfrac{L^2/16N^2}{L^2/4\pi N^2} = \dfrac{\pi}{4}\).

Order: drift speeds are tiny, about 10⁻⁴ m/s (of the order of mm per second or less).
Derivation: let n = free electrons per unit volume, A = cross-section, v_d = drift speed.
In time Δt electrons move a distance v_dΔt, so the electrons in volume A v_dΔt cross a section.
Charge crossing: \(Q = neAv_d\,\Delta t\) (electrons move opposite to E, so current is along E).
\(I = \dfrac{Q}{\Delta t} = neAv_d\).

Graph: x-axis r (fm), y-axis potential energy (MeV). The curve is very high (positive) at small r, falls steeply, crosses zero, reaches a negative minimum (about −100 MeV) at r₀ ≈ 0.8 fm, then rises and approaches zero by about 2–3 fm.
Conclusions (any two):
1. PE is minimum at about r₀ = 0.8 fm, the stable separation of nucleons.
2. For r > r₀ the slope is positive, so the nuclear force is attractive.
3. For r < r₀ the force is strongly repulsive, which keeps nucleons from collapsing.
4. The force is short-range: it becomes negligible beyond a few fermi.

(a) Take a cylindrical Gaussian surface of end-cap area A that pierces the sheet, with caps on both sides at equal distance.
By symmetry E is normal to the sheet, so the curved surface gives zero flux; each cap gives EA.
Total flux = 2EA. Charge enclosed = σA.
Gauss's law: \(2EA = \dfrac{\sigma A}{\varepsilon_0}\) ⇒ \(E = \dfrac{\sigma}{2\varepsilon_0}\), directed away from the sheet if σ > 0. It does not depend on distance.
(b) Each sheet gives σ/(2ε₀).
(i) Between the sheets the two fields point in opposite directions: E = σ/2ε₀ − σ/2ε₀ = 0.
(ii) Outside, they point the same way: E = σ/2ε₀ + σ/2ε₀ = σ/ε₀, directed away from the sheets.

(a) Bridge ABCD: P in AB, Q in BC, R in AD, S in DC; galvanometer G between B and D; cell between A and C. Let I₁ flow in P, I₂ in R.
At balance I_g = 0, so P and Q carry the same current I₁; R and S carry I₂, and V_B = V_D.
Loop ABDA: \(-I_1 P + I_2 R = 0\) ⇒ \(I_1P = I_2R\).
Loop BCDB: \(-I_1 Q + I_2 S = 0\) ⇒ \(I_1Q = I_2S\).
Dividing: \(\dfrac{P}{Q} = \dfrac{R}{S}\).
(b) Between M and N: arms M–O (R), O–N (R), M–P (R), P–N (R), with O–P (R) as the bridge.
R/R = R/R, so the bridge is balanced and O–P carries no current.
R_MN = (R + R) ∥ (R + R) = 2R ∥ 2R = R.
R_AB = 2R + R + 3R = 6R.

After disconnecting, charge Q stays constant.
(a) With slab C = Kε₀A/d; without slab C' = ε₀A/d = C/K, so capacitance decreases K times.
(b) \(U = \dfrac{Q^2}{2C}\); with C' = C/K, \(U' = \dfrac{Q^2}{2C/K} = KU\). Energy increases (the extra energy comes from the work done in pulling the slab out against attraction).
(c) V = Q/C ⇒ V' = Q/C' = KV, so the potential difference increases K times.

Field between plates E = V/d with d = 2 cm; acceleration \(a = \dfrac{eE}{m} = \dfrac{eV}{md}\) (vertical).
Time in plates: \(t = \dfrac{x}{u}\), x = 3 cm.
The beam starts midway, so it must deflect y = 1 cm: \(y = \dfrac12 a t^2 = \dfrac12\dfrac{eV}{md}\dfrac{x^2}{u^2}\).
\(V = \dfrac{2ymd u^2}{e x^2} = \dfrac{2(0.01)(9.1\times10^{-31})(0.02)(3\times10^7)^2}{(1.6\times10^{-19})(0.03)^2}\).
Numerator = 3.276 × 10⁻¹⁹, denominator = 1.44 × 10⁻²², V ≈ 2275 V.
Common mistake: using y = 2 cm (the full gap) instead of half the gap.

(a) With A positive, D₂ (anode at A) is forward biased and conducts; D₁ (cathode at A) is reverse biased and acts as an open switch.
(b) Equivalent circuit: A is joined directly to R (ideal D₂ = a plain wire); D₁ is removed. From R, current divides: one path R → 3 kΩ → B; the other path R → 1 kΩ → P → 2 kΩ → B.
(c) Peak: V_R − V_B = 12 V.
The 3 kΩ resistor is directly across 12 V: V_RB = 12 V.
The 1 kΩ and 2 kΩ in series (3 kΩ) are also across 12 V: current = 12/3000 = 4 mA, so V_RP = 4 mA × 1 kΩ = 4 V and V_PB = 4 mA × 2 kΩ = 8 V.

1. Diffusion: there are many holes in the p-side and many electrons in the n-side. Due to this concentration difference, holes diffuse p → n and electrons diffuse n → p, giving a diffusion current from p to n. They leave behind immobile ions: positive donor ions on the n-side and negative acceptor ions on the p-side, forming the depletion region.
2. Drift: the ion layers set up an electric field from the n-side to the p-side (barrier potential). This field pushes minority carriers across the junction (electrons p → n, holes n → p), giving a drift current opposite to the diffusion current.
Diffusion current falls and drift current rises until they become equal; then equilibrium is reached and net current is zero.

(a) Diagram: parallel rays from a distant object fall on the objective (large f₀) and form a real, inverted image A′B′ at its focus. This point is also the focus of the eyepiece (small fₑ), so rays leave the eyepiece parallel and the final image is at infinity. Tube length = f₀ + fₑ.
Angular magnification \(m = \dfrac{\beta}{\alpha} = \dfrac{f_0}{f_e}\).
(b) Any two:
1. A mirror has no chromatic aberration.
2. A parabolic mirror removes spherical aberration.
3. A mirror can be supported over its whole back, so a large aperture is lighter and easier to mount than a large lens.
4. A large mirror is cheaper to make and gives a brighter image (more light gathered).

(a) 1. Light must travel from an optically denser to a rarer medium.
2. Angle of incidence must be greater than the critical angle.
(b) A → B (Snell): \(n\sin\theta = \dfrac{3n}{4}\sin\theta_1\) ⇒ \(\sin\theta_1 = \dfrac43\sin\theta\).
Interfaces are parallel, so θ₁ is the angle of incidence at B → C:
\(\dfrac{3n}{4}\sin\theta_1 = \dfrac{2n}{3}\sin r\) ⇒ \(\sin r = \dfrac98\sin\theta_1 = \dfrac98\cdot\dfrac43\sin\theta = \dfrac32\sin\theta\).
(Shortcut: \(n\sin\theta = \dfrac{2n}{3}\sin r\) directly.)
For sin θ = 2/3: sin r = 1, r = 90°, the ray just grazes the B–C surface.
For sin θ > 2/3, sin r would exceed 1, so the ray is totally reflected at B–C. Hence it never enters C for sin θ ≥ 2/3.

(a) Let the dipole lie along the axis with p = q(2a) from −q to +q, and P be on the perpendicular bisector at distance r.
Distance of P from each charge = \(\sqrt{r^2 + a^2}\), so \(E_{+q} = E_{-q} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2 + a^2}\).
Components perpendicular to the dipole axis cancel; components parallel to the axis add, both pointing from +q towards −q (opposite to p).
With \(\cos\theta = \dfrac{a}{\sqrt{r^2+a^2}}\): \(E = 2E_{+q}\cos\theta = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2qa}{(r^2 + a^2)^{3/2}}\).
Vector form: \(\vec E = -\dfrac{\vec p}{4\pi\varepsilon_0 (r^2+a^2)^{3/2}}\).
For r ≫ a: \(\vec E = -\dfrac{\vec p}{4\pi\varepsilon_0 r^3}\).
(b) Force: \(\vec F = q\vec E + (-q)\vec E = q(2\hat i) - q(2\hat i) = 0\) (uniform field).
Torque: p points from −q to +q, i.e. along the x-axis (parallel or antiparallel to E). \(\vec\tau = \vec p\times\vec E\) = 0 since the angle is 0 or π.
So both force and torque are zero.

(a) Let the common terminal voltage be V and currents I₁, I₂ leave the positive terminals; external current I = I₁ + I₂.
\(V = E_1 - I_1 r_1\) ⇒ \(I_1 = \dfrac{E_1 - V}{r_1}\); similarly \(I_2 = \dfrac{E_2 - V}{r_2}\).
\(I = \left(\dfrac{E_1}{r_1} + \dfrac{E_2}{r_2}\right) - V\left(\dfrac1{r_1} + \dfrac1{r_2}\right)\).
Rearranging: \(V = \dfrac{E_1 r_2 + E_2 r_1}{r_1 + r_2} - I\dfrac{r_1 r_2}{r_1 + r_2}\).
Compare with V = E_eq − I r_eq: \(E_{eq} = \dfrac{E_1 r_2 + E_2 r_1}{r_1 + r_2}\), \(r_{eq} = \dfrac{r_1 r_2}{r_1 + r_2}\).
(b) \(E_{eq} = \dfrac{E\cdot R + 3E\cdot R}{2R} = 2E\); \(r_{eq} = \dfrac{R\cdot R}{2R} = \dfrac R2\).
\(I = \dfrac{E_{eq}}{2R + r_{eq}} = \dfrac{2E}{2R + R/2} = \dfrac{4E}{5R}\).

(a) Thin lens of index n₂ in a medium n₁, surfaces of radii R₁ and R₂.
First surface (object O at distance u, image I₁ at v₁): \(\dfrac{n_2}{v_1} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R_1}\).
I₁ acts as the object for the second surface (lens is thin): \(\dfrac{n_1}{v} - \dfrac{n_2}{v_1} = \dfrac{n_1 - n_2}{R_2}\).
Adding: \(n_1\left(\dfrac1v - \dfrac1u\right) = (n_2 - n_1)\left(\dfrac1{R_1} - \dfrac1{R_2}\right)\).
For u = ∞, v = f: \(\dfrac1f = \left(\dfrac{n_2}{n_1} - 1\right)\left(\dfrac1{R_1} - \dfrac1{R_2}\right)\), and 1/v − 1/u = 1/f.
(b) L₁: u = −80, f = 40: \(\dfrac1v = \dfrac1{40} - \dfrac1{80} = \dfrac1{80}\) ⇒ v₁ = 80 cm right of L₁.
L₂ is 120 cm away, so this image is 40 cm in front of L₂ (u = −40) = at its focus ⇒ rays leave L₂ parallel (v₂ = ∞).
L₃ receives parallel rays ⇒ image at its focus, 40 cm right of L₃.
Distance from object = 80 + 120 + 20 + 40 = 260 cm.

(a) Diagram: object AB between F and C. A ray from B parallel to the axis reflects through F; a ray from B through F reflects parallel to the axis (or a ray to the pole P reflects symmetrically). They meet at B′, giving a real, inverted, magnified image A′B′ beyond C.
Triangles A′B′P and ABP are similar: \(\dfrac{A'B'}{AB} = \dfrac{PA'}{PA}\).
Ray from B parallel to the axis hits the mirror at M (near P, so MP ≈ AB); triangles A′B′F and PMF are similar: \(\dfrac{A'B'}{AB} = \dfrac{FA'}{FP} = \dfrac{PA' - PF}{PF}\).
So \(\dfrac{PA'}{PA} = \dfrac{PA' - PF}{PF}\). Sign convention: PA = −u, PA′ = −v, PF = −f:
\(\dfrac{-v}{-u} = \dfrac{-v + f}{-f}\) ⇒ \(vf = uv - uf\) ⇒ dividing by uvf: \(\dfrac1f = \dfrac1v + \dfrac1u\).
(b) Virtual, erect image: m = +2 = −v/u, u = −10 cm ⇒ v = +20 cm (behind the mirror).
\(\dfrac1f = \dfrac1{20} + \dfrac1{-10} = -\dfrac1{20}\) ⇒ f = −20 cm. Focal length is 20 cm.

(a) The induced emf in a circuit equals the negative rate of change of magnetic flux through it: \(\varepsilon = -\dfrac{d\Phi_B}{dt}\) (for N turns, \(-N\dfrac{d\Phi}{dt}\)).
(b) Field inside: \(B = \mu_0 n I = \dfrac{\mu_0 N I}{l}\).
Flux linkage: \(N\Phi = N(BA) = \dfrac{\mu_0 N^2 A I}{l}\).
Since NΦ = LI: \(L = \dfrac{\mu_0 N^2 A}{l}\).
(c) \(\omega = 2\pi\times\dfrac{60}{60} = 2\pi\) rad/s.
\(\varepsilon = \tfrac12 B\omega l^2 = \tfrac12(4.0\times10^{-3})(2\pi)(0.5)^2 = \pi\times10^{-3}\) V ≈ 3.14 mV.

(a) Diagram: laminated soft-iron core; primary coil with few turns on one limb, secondary with many more turns (Ns > Np) on the other limb (or wound over the primary).
Principle: mutual induction – an alternating current in the primary produces changing flux in the core, which links the secondary and induces an emf in it.
If Φ is the flux per turn: \(\varepsilon_s = -N_s\dfrac{d\Phi}{dt}\), back emf \(\varepsilon_p = -N_p\dfrac{d\Phi}{dt}\).
For an ideal transformer (zero resistance, no flux leakage) v_p = ε_p and v_s = ε_s, so \(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\).
No losses: \(V_pI_p = V_sI_s\) ⇒ \(\dfrac{V_s}{V_p} = \dfrac{I_p}{I_s}\).
(b) (i) \(I_p = \dfrac{P}{V_p} = \dfrac{5000}{200} = 25\) A.
(ii) \(V_s = V_p\dfrac{N_s}{N_p} = 200\times5 = 1000\) V.