Option 1: Electromagnetic Induction and Faraday's Laws
Step 1: Everyday picture. Push a bar magnet towards a coil connected to a galvanometer and the needle deflects; pull the magnet away and it deflects the other way; hold the magnet still and there is no deflection. The coil generates electricity only while the magnetic field threading it is changing. This effect is called electromagnetic induction.
Step 2: The quantity that matters. What the coil responds to is the magnetic flux \(\Phi = BA\cos\theta\). It can be changed in three ways: by altering the field \(B\), the area \(A\), or the orientation angle \(\theta\), and equivalently by moving the magnet and coil relative to each other.
Step 3: Faraday's first law (existence). An emf is induced in a circuit whenever the flux linked with it changes, and this emf exists only for the duration of the change. A faster change gives a bigger effect; no change means no emf.
Step 4: Faraday's second law (magnitude). The induced emf equals the time rate of change of flux linkage. For \(N\) turns,
\[e = -N\frac{d\Phi}{dt}\]
and over a finite time interval the average emf is \(e = -N\,\dfrac{\Delta\Phi}{\Delta t}\).
Step 5: Sign and current. The minus sign is Lenz's law: the induced emf opposes the flux change that causes it, as required by energy conservation. In a circuit of resistance \(R\) the induced current is
\[I = \frac{e}{R} = -\frac{N}{R}\frac{d\Phi}{dt}\]
\[\boxed{e = -N\dfrac{d\Phi}{dt}}\]
Option 2: Parallel Plate Capacitor Partially Filled with Dielectric
Step 1: Setup. Two plates of area \(A\) are separated by a distance \(d\) and carry charges \(+Q\) and \(-Q\). A dielectric slab of dielectric constant \(K\) and thickness \(t\) (with \(t < d\)) is slipped between them; the leftover gap \((d-t)\) stays as air.
Step 2: Two regions, two fields. The surface charge density is \(\sigma = Q/A\). In the air part the field is \(E_0 = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)\). Inside the slab the bound (polarisation) charges weaken this field to \(E = E_0/K\).
Step 3: Add the voltage drops. The potential difference is the field multiplied by the length of each region and then summed:
\[V = E_0(d-t) + \frac{E_0}{K}\,t = E_0\left[(d-t) + \frac{t}{K}\right]\]
Step 4: Insert \(E_0\).
\[V = \frac{Q}{\varepsilon_0 A}\left[(d-t) + \frac{t}{K}\right]\]
Step 5: Capacitance and limits. Using \(C = Q/V\),
\[C = \frac{\varepsilon_0 A}{(d-t) + \dfrac{t}{K}}\]
For \(t = 0\) this reduces to the air capacitor \(C = \varepsilon_0 A/d\), and for \(t = d\) it becomes \(C = K\varepsilon_0 A/d\), exactly as expected. Since \(K > 1\), inserting the slab always increases the capacitance.
\[\boxed{C = \dfrac{\varepsilon_0 A}{(d-t) + \dfrac{t}{K}}}\]