Effect of Dielectric on Capacitance 🔌

1. Quick recap: bare capacitor

For two large parallel plates (area \(A\)) separated by a gap \(d\) and holding charges \(\pm Q\):

• Surface charge density: \(\sigma = Q/A\).
• Uniform electric field between the plates: \(E_0 = \sigma/\varepsilon_0\). :contentReference[oaicite:0]{index=0}
• Potential difference: \(V_0 = E_0 d\). :contentReference[oaicite:1]{index=1}
• Capacitance: \(C_0 = \dfrac{\varepsilon_0 A}{d}\). :contentReference[oaicite:2]{index=2}

Because \(\varepsilon_0\) is tiny, \(1\;\text{farad}\) demands an enormous plate area—about 30 km × 30 km for a 1 cm gap! :contentReference[oaicite:3]{index=3}

2. Drop a dielectric in 🎯

Slide a solid dielectric between the plates and the field polarises the material. Equivalent “bound” charge sheets appear on the dielectric faces with densities \(\pm\sigma_p\). The plates now effectively hold \(\pm(\sigma-\sigma_p)\). :contentReference[oaicite:4]{index=4}

2.1 New electric field

\(E = \dfrac{\sigma-\sigma_p}{\varepsilon_0}\). :contentReference[oaicite:5]{index=5}

2.2 Relating \(\sigma_p\) to \(\sigma\)

For a linear dielectric \(\sigma_p\) scales with \(\sigma\). Write \[ \sigma-\sigma_p=\dfrac{\sigma}{K}, \] where \(K\) (> 1) is the dielectric constant. :contentReference[oaicite:6]{index=6}

2.3 New potential and capacitance 🚀

• Potential: \(V = \dfrac{Qd}{K\varepsilon_0A}\). :contentReference[oaicite:7]{index=7}
• Capacitance: \(C = \dfrac{\varepsilon_0 K A}{d} = K C_0\). :contentReference[oaicite:8]{index=8}

2.4 Permittivity jargon 🧠

• Absolute permittivity: \(\varepsilon = K\varepsilon_0\). :contentReference[oaicite:9]{index=9}
• Dielectric constant: \(K=\dfrac{\varepsilon}{\varepsilon_0}\). :contentReference[oaicite:10]{index=10}
• Key fact: The dielectric multiplies the capacitance by \(K\) no matter the capacitor’s shape. :contentReference[oaicite:11]{index=11}

3. Worked example 📚 (slab ¾ d thick)

A slab with dielectric constant \(K\) and thickness \((3/4)d\) sits snugly between the plates (area remains \(A\)). Keep the original free charge \(Q_0\).

1. Original field: \(E_0=V_0/d\).
2. Field inside the slab: \(E=E_0/K\).
3. Total potential difference: \[ V = E\!\left(\tfrac{3}{4}d\right) + E_0\!\left(\tfrac{1}{4}d\right) = V_0\frac{K+3}{4K}. \]
4. Capacitance jump: \[ C = \frac{Q_0}{V} = C_0 \frac{4K}{K+3}. \]

Notice how a partial dielectric boosts capacitance but less dramatically than a full insert. :contentReference[oaicite:12]{index=12}

4. Why this matters for NEET 🎯

• Capacitance multiplier: a complete dielectric increases \(C\) by \(K\).
• Field & voltage shrink: inside a dielectric \(E\) and \(V\) drop by \(1/K\).
• Permittivity link: \(\varepsilon = K\varepsilon_0\).
• Problem pattern: partial slabs (different thicknesses) change \(C\) through series-like reasoning (\(C=C_0\frac{4K}{K+3}\) for a ¾ d slab).
• Bound charge idea: induced surface charge (\(\sigma_p\)) explains why the net field weakens.

Keep exploring—each dielectric twist makes capacitors even cooler! 😊