Class 12 Physics Derivations, Chapter-wise (CBSE 2026-27)

1. By symmetry, \(\mathbf E\) is perpendicular to the sheet, the same on both sides, and does not depend on \(y\) or \(z\).
2. Only the two end faces 1 and 2 contribute to the flux. On the curved (side) surface \(\mathbf E\) is parallel to the surface, so its flux is zero.
3. On faces 1 and 2, \(\mathbf E\) is along the outward normal, so each contributes \(EA\): $$\Phi = EA + EA = 2EA$$
4. Charge enclosed \(= \sigma A\). Gauss's law, \(\Phi = q_{\text{enc}}/\varepsilon_0\): $$2EA = \frac{\sigma A}{\varepsilon_0}$$

\(\hat{\mathbf n}\) points away from the sheet (towards it if \(\sigma < 0\)). The field does not depend on the distance \(x\) from the sheet.

Common mistake: writing \(\sigma/\varepsilon_0\). The thin sheet has field lines leaving from both faces, so the flux is \(2EA\), not \(EA\). (\(\sigma/\varepsilon_0\) is the field just outside a charged conductor.)

1. P is the same distance \(\sqrt{r^2 + a^2}\) from both charges, so the two fields are equal in magnitude: $$E_{+q} = E_{-q} = \frac{q}{4\pi\varepsilon_0}\,\frac{1}{r^2 + a^2}$$
2. Their components normal to the dipole axis cancel. The components along the axis add, and point opposite to \(\hat{\mathbf p}\): $$\mathbf E = -(E_{+q} + E_{-q})\cos\theta\;\hat{\mathbf p}$$
3. From the figure, \(\cos\theta = \dfrac{a}{\sqrt{r^2 + a^2}}\), so $$\mathbf E = -\frac{2qa}{4\pi\varepsilon_0 (r^2 + a^2)^{3/2}}\,\hat{\mathbf p}$$
4. With \(\mathbf p = q \times 2a\,\hat{\mathbf p}\), and for a far point \(r \gg a\):

Common mistake: dropping the minus sign. On the equatorial plane the field is opposite to \(\mathbf p\); on the axis it is along \(\mathbf p\) (derivation 1.5).

(i) Outside the shell, \(r > R\).

1. By spherical symmetry, \(\mathbf E\) is radial and has the same magnitude everywhere on the Gaussian sphere. Flux \(= E \times 4\pi r^2\).
2. Charge enclosed is the whole shell: \(q = \sigma \times 4\pi R^2\). Gauss's law: $$E \times 4\pi r^2 = \frac{\sigma}{\varepsilon_0}\,4\pi R^2 \;\Rightarrow\; E = \frac{\sigma R^2}{\varepsilon_0 r^2} = \frac{q}{4\pi\varepsilon_0 r^2}$$

(ii) Inside the shell, \(r < R\). The flux is again \(E \times 4\pi r^2\), but the Gaussian sphere encloses no charge: \(E \times 4\pi r^2 = 0\).

Outside, the shell acts as if all its charge were at the centre. The zero field inside follows from the \(1/r^2\) form of Coulomb's law.

Common mistake: using \(R\) instead of \(r\) in the flux. The Gaussian sphere passes through P, so its area is \(4\pi r^2\).

1. By symmetry, \(\mathbf E\) is radial (in the plane normal to the wire) and its magnitude depends only on \(r\).
2. Flux through the two flat ends is zero, because \(\mathbf E\) is parallel to them. On the curved surface \(\mathbf E\) is normal to the surface and constant: $$\Phi = E \times 2\pi r l$$
3. Charge enclosed \(= \lambda l\). Gauss's law: $$E \times 2\pi r l = \frac{\lambda l}{\varepsilon_0}$$

Directed outward if \(\lambda > 0\). Only the enclosed charge \(\lambda l\) enters Gauss's law, but the field is due to the whole wire; the wire being infinite is what makes \(\mathbf E\) normal to the curved surface.

Common mistake: writing \(1/r^2\). A line charge gives \(E \propto 1/r\); a point charge gives \(1/r^2\); a plane sheet gives no \(r\) dependence.

1. $$\mathbf E_{-q} = -\frac{q}{4\pi\varepsilon_0 (r + a)^2}\,\hat{\mathbf p}, \qquad \mathbf E_{+q} = \frac{q}{4\pi\varepsilon_0 (r - a)^2}\,\hat{\mathbf p}$$
2. Adding: $$\mathbf E = \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{(r-a)^2} - \frac{1}{(r+a)^2}\right]\hat{\mathbf p} = \frac{q}{4\pi\varepsilon_0}\,\frac{4ar}{(r^2 - a^2)^2}\,\hat{\mathbf p}$$
3. For \(r \gg a\), and with \(p = 2qa\):

At the same large distance, the axial field is twice the equatorial field and points the other way.

1. Force \(q\mathbf E\) on \(q\) and \(-q\mathbf E\) on \(-q\). Since \(\mathbf E\) is uniform, the net force is zero.
2. The forces act at different points, so they form a couple. The perpendicular distance between them (arm of the couple) is \(2a\sin\theta\): $$\tau = qE \times 2a\sin\theta = 2qaE\sin\theta = pE\sin\theta$$
3. The direction is normal to the plane of \(\mathbf p\) and \(\mathbf E\), the same as that of \(\mathbf p \times \mathbf E\):

The torque tends to align \(\mathbf p\) with \(\mathbf E\); it is zero when they are parallel. In a non-uniform field there is also a net force.

1. Potential obeys superposition, so with \(r_1, r_2\) the distances of P from \(q\) and \(-q\): $$V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{r_1} - \frac{q}{r_2}\right)$$
2. By geometry: $$r_1^2 = r^2 + a^2 - 2ar\cos\theta, \qquad r_2^2 = r^2 + a^2 + 2ar\cos\theta$$
3. For \(r \gg a\), keep terms only to first order in \(a/r\): $$r_1^2 \cong r^2\left(1 - \frac{2a\cos\theta}{r}\right), \qquad r_2^2 \cong r^2\left(1 + \frac{2a\cos\theta}{r}\right)$$
4. Binomial theorem, to first order: $$\frac{1}{r_1} \cong \frac{1}{r}\left(1 + \frac{a}{r}\cos\theta\right), \qquad \frac{1}{r_2} \cong \frac{1}{r}\left(1 - \frac{a}{r}\cos\theta\right)$$
5. Substituting, with \(p = 2qa\): $$V = \frac{q}{4\pi\varepsilon_0}\,\frac{2a\cos\theta}{r^2} = \frac{p\cos\theta}{4\pi\varepsilon_0 r^2}$$

On the axis (\(\theta = 0\) or \(\pi\)): \(V = \pm\dfrac{p}{4\pi\varepsilon_0 r^2}\). On the equatorial line (\(\theta = \pi/2\)): \(V = 0\).

Common mistake: writing \(1/r\). A dipole's potential falls off as \(1/r^2\) (and its field as \(1/r^3\)); a single charge's potential falls as \(1/r\).

1. Plate 1 has surface charge density \(\sigma = Q/A\), plate 2 has \(-\sigma\). Each produces a field \(\sigma/2\varepsilon_0\) (derivation 1.1).
2. Outside the plates (regions I and II) the two fields cancel: $$E = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$$
3. Between the plates they add, pointing from the positive to the negative plate: $$E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$$
4. The field is uniform, so the potential difference is: $$V = Ed = \frac{1}{\varepsilon_0}\,\frac{Qd}{A}$$

Common mistake: using \(\sigma/2\varepsilon_0\) for the field between the plates. Inside, the fields of both plates add up to \(\sigma/\varepsilon_0\).

1. Without the dielectric: \(E_0 = \sigma/\varepsilon_0\), \(V_0 = E_0 d\), so \(C_0 = \varepsilon_0 A/d\).
2. With the dielectric, the net surface charge density is \(\pm(\sigma - \sigma_p)\): $$E = \frac{\sigma - \sigma_p}{\varepsilon_0}, \qquad V = Ed = \frac{\sigma - \sigma_p}{\varepsilon_0}\,d$$
3. For a linear dielectric, \(\sigma_p\) is proportional to \(\sigma\), so we write \(\sigma - \sigma_p = \sigma/K\) with \(K > 1\): $$V = \frac{\sigma d}{\varepsilon_0 K} = \frac{Qd}{A\varepsilon_0 K}$$

\(K = C/C_0\) is the dielectric constant: the factor by which filling the gap with the dielectric increases the capacitance. \(\varepsilon = \varepsilon_0 K\) is the permittivity of the medium.

Common mistake: using this for a slab that only partly fills the gap. Then the field is \(E_0\) in the air part and \(E_0/K\) in the slab, and you add the two potential drops (NCERT Example 2.8).

1. An external torque \(\tau_{\text{ext}}\) just balances \(\mathbf p \times \mathbf E\) and turns the dipole slowly (no angular acceleration) from \(\theta_0\) to \(\theta_1\). Work done: $$W = \int_{\theta_0}^{\theta_1} \tau_{\text{ext}}(\theta)\,d\theta = \int_{\theta_0}^{\theta_1} pE\sin\theta\,d\theta$$ $$W = pE(\cos\theta_0 - \cos\theta_1)$$
2. This work is stored as potential energy. Choosing \(U = 0\) at \(\theta_0 = \pi/2\): $$U(\theta) = pE\left(\cos\frac{\pi}{2} - \cos\theta\right)$$

Minimum \(U = -pE\) at \(\theta = 0\) (\(\mathbf p\) along \(\mathbf E\), stable equilibrium). Maximum \(U = +pE\) at \(\theta = \pi\) (\(\mathbf p\) opposite to \(\mathbf E\), unstable equilibrium).

Common mistake: saying \(U\) is maximum at \(\theta = 90^\circ\). That is where the torque is maximum; \(U\) is zero there.

1. Bringing \(q_1\) from infinity needs no work: there is no field yet.
2. Bringing \(q_2\) to \(\mathbf r_2\) costs \(q_2\) times the potential of \(q_1\) there. For two charges, $$U = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r_{12}}$$
3. Bringing \(q_3\) costs \(q_3 V_{1,2}(\mathbf r_3)\): $$q_3 V_{1,2}(\mathbf r_3) = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}}\right)$$
4. Adding the work in each step:

2023 question: \(q\), \(2q\), \(-3q\) on a triangle of side \(a\): \(U = \dfrac{1}{4\pi\varepsilon_0 a}\left(2q^2 - 3q^2 - 6q^2\right) = -\dfrac{7q^2}{4\pi\varepsilon_0 a}\).

Common mistake: counting each pair twice. There is exactly one term per pair of charges.

1. Without a field, electrons move randomly; their average velocity is zero: \(\frac{1}{N}\sum \mathbf v_i = 0\).
2. With a field \(\mathbf E\), each electron (charge \(-e\)) accelerates at \(\mathbf a = -e\mathbf E/m\). If \(t_i\) is the time since its last collision and \(\mathbf v_i\) its velocity just after it: $$\mathbf V_i = \mathbf v_i - \frac{e\mathbf E}{m}\,t_i$$
3. Average over all electrons. \((\mathbf v_i)_{\text{avg}} = 0\) and the average of \(t_i\) is the relaxation time \(\tau\): $$\mathbf v_d = (\mathbf v_i)_{\text{avg}} - \frac{e\mathbf E}{m}(t_i)_{\text{avg}} = -\frac{e\mathbf E}{m}\,\tau$$
4. With \(n\) free electrons per unit volume, the number crossing area \(A\) in time \(\Delta t\) is \(n A |v_d|\Delta t\). The charge they carry is \(I\,\Delta t\): $$I\,\Delta t = neA|v_d|\,\Delta t$$

The drift velocity is constant in time even though each electron is accelerated, because collisions keep resetting it. Typical \(v_d\) is about 1 mm/s.

Common mistake: losing the minus sign. Electrons drift opposite to \(\mathbf E\), but the current is along \(\mathbf E\).

1. At balance no current flows through G (\(I_g = 0\)). Kirchhoff's junction rule at D and B then gives $$I_1 = I_3, \qquad I_2 = I_4$$
2. Loop rule for loop ADBA (the G arm has no current, so no drop across it): $$-I_1 R_1 + 0 + I_2 R_2 = 0 \;\Rightarrow\; \frac{I_1}{I_2} = \frac{R_2}{R_1}$$
3. Loop rule for loop CBDC, using \(I_3 = I_1\), \(I_4 = I_2\): $$I_2 R_4 + 0 - I_1 R_3 = 0 \;\Rightarrow\; \frac{I_1}{I_2} = \frac{R_4}{R_3}$$

So an unknown \(R_4\) can be found as \(R_4 = R_3\,R_2/R_1\) by adjusting \(R_3\) until G shows no deflection. The meter bridge uses this principle.

Common mistake: pairing the wrong arms. The condition is "ratio of the two arms on one side of G = ratio of the two arms on the other side"; check it against your own labels.

1. Charge conservation at B₁: \(I = I_1 + I_2\).
2. Both cells have the same terminal voltage \(V = V(B_1) - V(B_2)\): $$V = \varepsilon_1 - I_1 r_1, \qquad V = \varepsilon_2 - I_2 r_2$$
3. Combining: $$I = \frac{\varepsilon_1 - V}{r_1} + \frac{\varepsilon_2 - V}{r_2} = \left(\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}\right) - V\left(\frac{1}{r_1} + \frac{1}{r_2}\right)$$
4. Solving for \(V\): $$V = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} - I\,\frac{r_1 r_2}{r_1 + r_2}$$
5. For one equivalent cell, \(V = \varepsilon_{eq} - I r_{eq}\). Comparing:

Simpler form: \(\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}\) and \(\frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}\).

Common mistake: adding the emfs. In parallel the emfs don't add; the equivalent emf is a weighted average of the two.

1. Biot–Savart law for an element \(d\mathbf l\): $$dB = \frac{\mu_0}{4\pi}\,\frac{I\,|d\mathbf l\times\mathbf r|}{r^3}$$
2. Here \(r^2 = x^2 + R^2\), and every element \(d\mathbf l\) is perpendicular to \(\mathbf r\), so \(|d\mathbf l\times\mathbf r| = r\,dl\): $$dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl}{x^2 + R^2}$$
3. The components \(dB_\perp\) from diametrically opposite elements cancel. Only \(dB_x = dB\cos\theta\) survives, with $$\cos\theta = \frac{R}{(x^2 + R^2)^{1/2}} \;\Rightarrow\; dB_x = \frac{\mu_0 I\,dl}{4\pi}\,\frac{R}{(x^2 + R^2)^{3/2}}$$
4. Summing \(dl\) over the loop gives \(2\pi R\):

For a coil of \(N\) closely wound turns, multiply by \(N\): \(B = \dfrac{\mu_0 N I R^2}{2(x^2 + R^2)^{3/2}}\). Direction by the right-hand thumb rule: curl the fingers along the current; the thumb gives \(\mathbf B\).

Common mistake: adding the full \(dB\) of every element. The perpendicular parts cancel in pairs; only \(dB\cos\theta\) adds up.

1. Conductor a produces the same field at every point of b. From Ampere's law (derivation 4.5): $$B_a = \frac{\mu_0 I_a}{2\pi d}$$
2. A length \(L\) of b carrying \(I_b\) feels a force \(F = I_b L B_a\) (since \(\mathbf B_a \perp\) b), directed towards a: $$F_{ba} = I_b L B_a = \frac{\mu_0 I_a I_b}{2\pi d}\,L$$
3. Similarly, a feels an equal and opposite force: \(\mathbf F_{ba} = -\mathbf F_{ab}\). Force per unit length:

Nature: parallel currents attract; antiparallel currents repel. Ampere: the current which, in each of two very long parallel wires 1 m apart in vacuum, produces a force of \(2 \times 10^{-7}\) N per metre of length.

Common mistake: "like currents repel, like charges repel". It's the opposite for currents: same-direction currents attract.

1. Forces on arms BC and DA are equal, opposite and collinear (along the loop's axis), so they cancel: no force, no torque.
2. Forces on arms AB and CD (length \(b\)) are equal and opposite: \(F_1 = F_2 = IbB\). The net force is zero, but they are not collinear, so they form a couple.
3. The arm of the couple is \(a\sin\theta\) (each force acts at \(\frac{a}{2}\sin\theta\) from the axis): $$\tau = F_1\frac{a}{2}\sin\theta + F_2\frac{a}{2}\sin\theta = IabB\sin\theta = IAB\sin\theta$$
4. Define the magnetic moment \(\mathbf m = I\mathbf A\), with \(\mathbf A\) along the normal by the right-hand thumb rule. Then:

When the plane of the loop is along \(\mathbf B\) (\(\theta = 90^\circ\)), \(\tau = IAB\) is maximum. It is zero when \(\mathbf m\) is parallel (stable) or antiparallel (unstable) to \(\mathbf B\). Compare \(\boldsymbol\tau = \mathbf p \times \mathbf E\) (derivation 1.6).

Common mistake: taking \(\theta\) as the angle between the plane of the loop and \(\mathbf B\). In \(\tau = mB\sin\theta\), \(\theta\) is measured from the normal (\(\mathbf m\)).

1. An electron (charge \(-e\), mass \(m_e\)) moves with speed \(v\) in a circle of radius \(r\). Its period is \(T = 2\pi r/v\), so the equivalent current is $$I = \frac{e}{T} = \frac{ev}{2\pi r}$$
2. Magnetic moment of this current loop (area \(\pi r^2\)): $$\mu_e = I\,\pi r^2 = \frac{ev}{2\pi r}\,\pi r^2 = \frac{evr}{2}$$
3. The orbital angular momentum is \(L = m_e v r\), so \(vr = L/m_e\): $$\mu_e = \frac{e}{2m_e}\,L$$
4. For the negative electron, the current is opposite to its motion, so \(\boldsymbol\mu_e\) points opposite to \(\mathbf L\):

Common mistake: dropping the minus sign in the vector form. It is there because the electron's charge is negative.

1. Ampere's circuital law: \(\oint \mathbf B\cdot d\mathbf l = \mu_0 I\), where \(I\) is the current enclosed by the loop.
2. By symmetry, on a circle of radius \(r\) around the wire, \(\mathbf B\) is tangential and constant in magnitude, so \(\oint \mathbf B\cdot d\mathbf l = B \times 2\pi r\).
3. The loop encloses the whole current \(I\): $$B \times 2\pi r = \mu_0 I$$

Why Ampere's law here? The wire has cylindrical symmetry, so we can pick a loop on which \(\mathbf B\) is tangential and constant. The integral then reduces to \(B \times 2\pi r\), with no integration over current elements as Biot–Savart would need.

Common mistake: drawing \(\mathbf B\) pointing away from the wire. The field lines are circles around the wire (right-hand rule: thumb along the current, fingers curl along \(\mathbf B\)).

1. The magnetic force \(q\mathbf v\times\mathbf B\) is perpendicular to \(\mathbf v\): it does no work and changes only the direction of motion. With \(\mathbf v \perp \mathbf B\) it acts as the centripetal force: $$\frac{mv^2}{r} = qvB \;\Rightarrow\; r = \frac{mv}{qB}$$
2. Angular frequency \(\omega = v/r\): $$\omega = 2\pi\nu = \frac{qB}{m}, \qquad T = \frac{2\pi m}{qB}$$ independent of the speed.
3. Helix. If \(\mathbf v\) has a component \(v_\parallel\) along \(\mathbf B\), that component is unaffected, while the perpendicular component \(v_\perp\) gives circular motion (radius \(mv_\perp/qB\)). The combination is a helix. Distance moved along \(\mathbf B\) in one revolution (pitch):

Common mistake: using the full speed \(v\) for the helix radius. Only \(v_\perp\) goes into the radius; only \(v_\parallel\) goes into the pitch.

(a) Co-axial solenoids.

1. Current \(I_2\) in S₂ gives a uniform field \(B = \mu_0 n_2 I_2\) inside it. The flux linkage with S₁ (\(n_1 l\) turns, area \(\pi r_1^2\)) is $$N_1\Phi_1 = (n_1 l)(\pi r_1^2)(\mu_0 n_2 I_2)$$
2. By definition, \(N_1\Phi_1 = M_{12} I_2\), so \(M_{12} = \mu_0 n_1 n_2 \pi r_1^2 l\).
3. Reverse case: current \(I_1\) in S₁ sets up a field \(\mu_0 n_1 I_1\) only inside S₁, so the flux through S₂ also uses area \(\pi r_1^2\): \(N_2\Phi_2 = (n_2 l)(\pi r_1^2)(\mu_0 n_1 I_1)\), giving the same value.

(b) Concentric coplanar loops, radii \(r_1 \ll r_2\). Current \(I_2\) in the outer loop gives \(B_2 = \dfrac{\mu_0 I_2}{2r_2}\) at its centre (derivation 4.1). The small loop sits in this nearly uniform field: $$\Phi_1 = \pi r_1^2 B_2 = \frac{\mu_0 \pi r_1^2}{2r_2}\,I_2 \;\Rightarrow\; M = \frac{\mu_0 \pi r_1^2}{2r_2}$$

With a core of relative permeability \(\mu_r\), \(M = \mu_r\mu_0 n_1 n_2\pi r_1^2 l\). \(M_{12} = M_{21}\) holds in general, which is useful when one of the two fluxes is hard to calculate.

Common mistake: using the outer radius \(r_2\) in (a). The flux that links both coils passes through the inner solenoid's area, \(\pi r_1^2\).

1. Each element \(dr\) of the rod at distance \(r\) from O moves with speed \(v = \omega r\), perpendicular to both \(\mathbf B\) and the rod.
2. Motional emf across that element (derivation 6.5): $$d\varepsilon = Bv\,dr = B\omega r\,dr$$
3. The elements are in series, so their emfs add along the rod: $$\varepsilon = \int_0^L B\omega r\,dr = B\omega\frac{L^2}{2}$$

Alternative: the rod sweeps area \(\frac{1}{2}L^2\theta\), so \(\varepsilon = B\frac{d}{dt}\left(\frac{1}{2}L^2\theta\right) = \frac{1}{2}BL^2\omega\). For a wheel with many spokes, the emf is the same as for one spoke, because the spokes are in parallel.

Common mistake: using \(\varepsilon = BLv\) with the tip speed \(\omega L\). Different parts move at different speeds; the average speed \(\omega L/2\) is what gives the factor \(\frac{1}{2}\).

1. Flux through PQRS: \(\Phi_B = Blx\). With \(dx/dt = -v\): $$\varepsilon = -\frac{d\Phi_B}{dt} = -Bl\frac{dx}{dt} = Blv$$
2. Lorentz-force view: each charge \(q\) in the rod feels force \(qvB\) along it. Work from P to Q is \(W = qvBl\), so \(\varepsilon = W/q = Blv\).
3. Energy: with total resistance \(r\), the current is \(I = \dfrac{Blv}{r}\). The field pushes back on the current-carrying rod with force \(F = IlB = \dfrac{B^2 l^2 v}{r}\), opposing the motion (Lenz's law).
4. To keep \(v\) constant, an external agent must supply power: $$P = Fv = \frac{B^2 l^2 v^2}{r}$$
5. Joule heating in the circuit: \(P_J = I^2 r = \dfrac{B^2 l^2 v^2}{r}\), the same. The mechanical work becomes electrical energy and then heat.

Common mistake: treating the emf as "free energy". The retarding force means work must be done on the rod; that work is exactly the electrical energy produced, which is why Lenz's law follows from energy conservation.

1. In series, the same current \(i = i_m\sin(\omega t + \phi)\) flows through R, L and C. Voltage amplitudes: $$v_{Rm} = i_m R, \qquad v_{Lm} = i_m X_L, \qquad v_{Cm} = i_m X_C$$
2. \(\mathbf V_R\) is along \(\mathbf I\); \(\mathbf V_L\) and \(\mathbf V_C\) are perpendicular to it and opposite to each other, so they combine to a single phasor of length \(|v_{Cm} - v_{Lm}|\).
3. Kirchhoff's loop rule in phasor form, \(\mathbf V_L + \mathbf V_R + \mathbf V_C = \mathbf V\). \(\mathbf V\) is the hypotenuse of a right triangle: $$v_m^2 = v_{Rm}^2 + (v_{Cm} - v_{Lm})^2$$ $$v_m^2 = i_m^2\left[R^2 + (X_C - X_L)^2\right]$$
4. Writing \(i_m = v_m/Z\):

\(X_C > X_L\): current leads (capacitive). \(X_L > X_C\): current lags (inductive). RC circuit: \(Z = \sqrt{R^2 + X_C^2}\). Minimum impedance \(Z = R\) at resonance (derivation 7.3).

Common mistake: adding the voltage amplitudes directly, \(v_R + v_L + v_C\). They are out of phase, so they add as vectors (NCERT Example 7.6).

1. With charge \(q\) on the capacitor, \(v = q/C\). Kirchhoff's loop rule: $$v_m\sin\omega t = \frac{q}{C} \;\Rightarrow\; q = Cv_m\sin\omega t$$
2. Current: $$i = \frac{dq}{dt} = \omega C v_m\cos\omega t$$ $$i = i_m\sin\left(\omega t + \frac{\pi}{2}\right), \quad i_m = \omega C v_m$$
3. Writing \(i_m = \dfrac{v_m}{1/\omega C}\), the term \(1/\omega C\) plays the role of resistance:

Power: \(p = iv = \frac{i_m v_m}{2}\sin 2\omega t\), whose average over a cycle is zero. A pure capacitor draws no average power.

Common mistake: mixing up lead and lag. Capacitor: current leads. Inductor: current lags (derivation 7.4).

1. Ampere's law \(\oint \mathbf B\cdot d\mathbf l = \mu_0 i(t)\) on the flat disc gives \(B(2\pi r) = \mu_0 i\). On the pot-shaped surface, no current passes through it, so it gives \(B = 0\). A contradiction: the law is missing a term.
2. What does cross S is the electric field between the plates, \(E = \dfrac{Q}{\varepsilon_0 A}\). The flux through S: $$\Phi_E = EA = \frac{Q}{\varepsilon_0}$$
3. As the capacitor charges, \(i = dQ/dt\): $$\frac{d\Phi_E}{dt} = \frac{1}{\varepsilon_0}\frac{dQ}{dt} \;\Rightarrow\; \varepsilon_0\frac{d\Phi_E}{dt} = i$$
4. So \(\varepsilon_0\,d\Phi_E/dt\) is the missing term, the displacement current \(i_d\). Outside the plates \(i_c = i\), \(i_d = 0\); between them \(i_c = 0\), \(i_d = i\). The total \(i_c + i_d\) is the same through every surface, so the current is continuous round the circuit.

For a conductor across a constant voltage, \(\mathbf E\) doesn't change, so \(i_d = 0\) (2026). Kirchhoff's junction rule fails at a single plate for the conduction current alone, but holds once \(i_d\) is included (2025).

Common mistake: calling \(i_d\) a flow of charge. No charge moves between the plates; \(i_d\) is a changing electric field that produces a magnetic field just like a current.

1. Parallel rays from a distant object make angle \(\alpha\) with the axis. The objective forms a real, inverted image A′B′ of height \(h\) at its second focal point, i.e. at distance \(f_o\).
2. In normal adjustment A′B′ also lies at the first focal point of the eyepiece, so the final image is at infinity and the rays leave at angle \(\beta\).
3. The object subtends \(\alpha\) at the objective (it is so far away that this equals the angle at the eye): $$\alpha \approx \tan\alpha = \frac{h}{f_o}$$
4. The final image subtends at the eye: $$\beta \approx \tan\beta = \frac{h}{f_e}$$
5. Magnifying power is \(m = \beta/\alpha\):

Common mistake: using \(1 + D/f_e\) here. That is the microscope near-point case. In normal adjustment the telescope's answer has no \(D\) in it.

1. Objective. It forms a real, inverted, magnified image A′B′ near the eyepiece's focal point. From the similar triangles at the objective's second focal point, $$\tan\beta = \frac{h}{f_o} = \frac{h'}{L}\;\Rightarrow\; m_o = \frac{h'}{h} = \frac{L}{f_o}$$
2. Eyepiece acts as a simple microscope (derivation 9.12) with A′B′ as its object: $$\text{final image at near point: } m_e = 1 + \frac{D}{f_e}$$ $$\text{final image at infinity: } m_e = \frac{D}{f_e}$$
3. Total magnification is the product, \(m = m_o\, m_e\):

Another common way to write \(m_o\) is \(v_o/|u_o|\) (image and object distances for the objective). It gives the same result when the first image is near the eyepiece's focus.

Common mistake: mixing up the two cases. Check whether the question says "least distance of distinct vision" (near point, \(1 + D/f_e\)) or "at infinity" (\(D/f_e\)).

1. Right-angled triangles A′B′F and MPF are similar (for paraxial rays MP is a straight line perpendicular to the axis), and PM = AB: $$\frac{B'A'}{BA} = \frac{B'F}{FP} \quad (9.4)$$
2. Since \(\angle APB = \angle A'PB'\), triangles A′B′P and ABP are similar: $$\frac{B'A'}{BA} = \frac{B'P}{BP} \quad (9.5)$$
3. Comparing, and using B′F = B′P − FP: $$\frac{B'P - FP}{FP} = \frac{B'P}{BP}$$
4. Sign convention (distances measured against incident light are negative): B′P = −v, FP = −f, BP = −u: $$\frac{-v + f}{-f} = \frac{-v}{-u} \;\Rightarrow\; \frac{v}{f} = 1 + \frac{v}{u}$$
5. Dividing by \(v\):

Magnification. From triangles A′B′P and ABP, \(\frac{B'A'}{BA} = \frac{B'P}{BP}\). With signs, \(\frac{-h'}{h} = \frac{-v}{-u}\), so \(m = \frac{h'}{h} = -\frac{v}{u}\).

Common mistake: putting signs into the similar-triangle ratios. Write the ratios with magnitudes first and apply the sign convention only in step 4.

1. In quadrilateral AQNR, the angles at Q and R are right angles, so $$\angle A + \angle QNR = 180^\circ$$
2. In triangle QNR: $$r_1 + r_2 + \angle QNR = 180^\circ$$
3. Comparing the two: $$r_1 + r_2 = A \quad (9.34)$$
4. Total deviation is the sum of the deviations at the two faces: $$\delta = (i - r_1) + (e - r_2) = i + e - (r_1 + r_2)$$

Common mistake: using \(r = A/2\) or \(i = e\) here. Those hold only at minimum deviation (derivation 9.5). \(\delta = i + e - A\) holds for any angle of incidence.

1. For small angles (paraxial rays), with \(\tan\theta \approx \theta\): $$\angle NOM = \frac{MN}{OM},\quad \angle NCM = \frac{MN}{MC}$$ $$\angle NIM = \frac{MN}{MI}$$
2. \(i\) is the exterior angle of \(\triangle NOC\): $$i = \angle NOM + \angle NCM = \frac{MN}{OM} + \frac{MN}{MC} \quad (9.13)$$
3. Similarly, from \(\triangle NCI\): $$r = \angle NCM - \angle NIM = \frac{MN}{MC} - \frac{MN}{MI} \quad (9.14)$$
4. Snell's law \(n_1 \sin i = n_2 \sin r\) becomes \(n_1 i = n_2 r\) for small angles. Substituting: $$\frac{n_1}{OM} + \frac{n_2}{MI} = \frac{n_2 - n_1}{MC} \quad (9.15)$$
5. Sign convention: OM = −u, MI = +v, MC = +R:

Common mistake: skipping the small-angle step. The derivation only works because \(\tan\theta \approx \theta\) and \(\sin\theta \approx \theta\) for paraxial rays, so say so.

1. Apply Eq. 9.15 to the first surface ABC (centre C₁): $$\frac{n_1}{OB} + \frac{n_2}{BI_1} = \frac{n_2 - n_1}{BC_1} \quad (9.17)$$
2. For the second surface ADC (centre C₂), light goes from \(n_2\) back to \(n_1\), and I₁ is a virtual object: $$-\frac{n_2}{DI_1} + \frac{n_1}{DI} = \frac{n_2 - n_1}{DC_2} \quad (9.18)$$
3. Thin lens: \(BI_1 = DI_1\). Adding (9.17) and (9.18): $$\frac{n_1}{OB} + \frac{n_1}{DI} = (n_2 - n_1)\left(\frac{1}{BC_1} + \frac{1}{DC_2}\right) \quad (9.19)$$
4. Object at infinity: \(OB \to \infty\) and \(DI = f\): $$\frac{n_1}{f} = (n_2 - n_1)\left(\frac{1}{BC_1} + \frac{1}{DC_2}\right) \quad (9.20)$$
5. Sign convention: \(BC_1 = +R_1\), \(DC_2 = -R_2\). Dividing by \(n_1\) and writing \(n_{21} = n_2/n_1\):

Also true for a concave lens: there \(R_1\) is negative and \(R_2\) positive, so \(f\) comes out negative.

Common mistake: writing \(+\frac{1}{R_2}\). The minus sign comes from \(DC_2 = -R_2\); for a double-convex lens \(R_2\) is itself negative, so both terms end up adding.

1. Lens A (focal length \(f_1\)) forms I₁: $$\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1} \quad (9.27)$$
2. I₁ is the object for lens B (focal length \(f_2\)): $$\frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2} \quad (9.28)$$
3. Adding: $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \quad (9.29)$$
4. For the equivalent single lens of focal length \(f\), \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\). Comparing:

For any number of lenses in contact, \(P = P_1 + P_2 + P_3 + \dots\) (algebraic sum) and \(m = m_1 m_2 m_3 \dots\)

Common mistake: adding focal lengths. Powers add; focal lengths don't. Keep the sign of each lens (concave = negative).

1. The ray enters at Q at angle \(i\), refracts at angle \(r\), and leaves the bottom face at R. At R it refracts back into air at angle \(i\), so the emergent ray is parallel to the incident ray.
2. Length of the path inside the slab of thickness \(t\): $$QR = \frac{t}{\cos r}$$
3. The angle between QR and the undeviated (dashed) path is \(i - r\). The lateral shift \(d\) is the perpendicular from R onto that path: $$d = QR\,\sin(i - r)$$

When is it minimum? At normal incidence (\(i = 0\), so \(r = 0\)) the shift is zero. It grows with the thickness \(t\) and the angle of incidence \(i\).

Common mistake: using \(t\) as the path length. The path inside is \(t/\cos r\), which is longer than \(t\).

1. With angle of incidence \(\theta\): \(\angle MCP = \theta\) and \(\angle MFP = 2\theta\).
2. $$\tan\theta = \frac{MD}{CD}, \qquad \tan 2\theta = \frac{MD}{FD} \quad (9.1)$$
3. For paraxial rays \(\tan\theta \approx \theta\) and \(\tan 2\theta \approx 2\theta\), so $$\frac{MD}{FD} = 2\,\frac{MD}{CD} \;\Rightarrow\; FD = \frac{CD}{2} \quad (9.2)$$
4. For small \(\theta\), D is very close to P, so \(FD = f\) and \(CD = R\):

1. Plane wavefront AB meets the interface at A. Let \(\tau\) be the time for B to reach C: \(BC = v_1\tau\).
2. In the same time, the secondary wavelet from A spreads into medium 2 as a sphere of radius \(AE = v_2\tau\). The tangent plane CE from C to this sphere is the refracted wavefront.
3. From right triangles ABC and AEC (angle of incidence \(i\) at A in \(\triangle ABC\), angle of refraction \(r\) at C in \(\triangle AEC\)): $$\sin i = \frac{BC}{AC} = \frac{v_1\tau}{AC}, \qquad \sin r = \frac{AE}{AC} = \frac{v_2\tau}{AC}$$
4. Dividing, and using \(n_1 = c/v_1\), \(n_2 = c/v_2\):

If \(r < i\) then \(v_2 < v_1\): light slows down in a denser medium. (The corpuscular model predicted the opposite; experiment sides with waves.) The frequency stays the same, so \(v_1/\lambda_1 = v_2/\lambda_2\).

Common mistake: drawing the refracted wavefront from A. It is the tangent drawn from C to the wavelet centred at A.

1. Force balance (Coulomb force = centripetal force): $$\frac{1}{4\pi\varepsilon_0}\,\frac{e^2}{r^2} = \frac{mv^2}{r} \;\Rightarrow\; mv^2 = \frac{e^2}{4\pi\varepsilon_0 r}$$
2. Bohr's second postulate (quantised angular momentum): $$mvr = \frac{nh}{2\pi} \;\Rightarrow\; v = \frac{nh}{2\pi m r}$$
3. Substituting \(v\) into step 1: $$m\,\frac{n^2h^2}{4\pi^2 m^2 r^2} = \frac{e^2}{4\pi\varepsilon_0 r} \;\Rightarrow\; r = \frac{n^2 h^2\varepsilon_0}{\pi m e^2}$$

For \(n = 1\): Bohr radius \(a_0 = \dfrac{h^2\varepsilon_0}{\pi m e^2} = 5.29\times10^{-11}\) m. Since \(r_n = n^2 a_0\), the gaps \(r_{n+1} - r_n = (2n+1)a_0\) grow with \(n\): the orbits are not equally spaced. Also, from step 2, \(v_n \propto 1/n\).

Common mistake: stating \(r \propto n\). The angular-momentum condition gives \(v \propto 1/n\), and together with the force balance this makes \(r \propto n^2\).