Q.Explain what would happen if in the capacitor given in Exercise 2.8, a $3\ \text{mm}$ thick mica sheet (of dielectric constant = 6) were inserted between the plates,

Concept: dielectric insertion. The $3\,\text{mm}$ mica ($K=6$) exactly fills the $3\,\text{mm}$ gap, so $C$ rises to $KC_0=6C_0$. What else changes depends on the supply.

With $C_0=\dfrac{\varepsilon_0 A}{d}=17.7\,\text{pF}$ and $Q_0=C_0V_0=1.77\,\text{nC}$ (from Exercise 2.8: $A=6\times10^{-3}\,\text{m}^2$, $V_0=100\,\text{V}$):

(a) Supply connected — $V$ fixed at $100\,\text{V}$: $C=6C_0=106.2\,\text{pF}$, charge rises to $Q=CV=10.62\,\text{nC}$ ($6\times$), energy $\tfrac12CV^2$ rises $6\times$. The battery supplies the extra charge. …

The core idea

A dielectric filling the gap multiplies the capacitance by $K$:

What else changes depends on what is held fixed. A connected battery fixes the voltage; a disconnected capacitor fixes the charge.

Setup (from Exercise 2.8)

Plate area $A=6\times10^{-3}\,\text{m}^2$, separation $d=3\,\text{mm}=3\times10^{-3}\,\text{m}$, charged to $V_0=100\,\text{V}$. The mica is exactly $3\,\text{mm}$ thick, so it fills the gap completely — no air layer, no series combination.

Part (a) — Supply remains connected (V fixed)

The battery holds $V=V_0=100\,\text{V}$.

• Capacitance: $C=106.2\,\text{pF}$.
• Charge: $Q=CV=(106.2\times10^{-12})(100)=1.062\times10^{-8}\,\text{C}=10.62\,\text{nC}$ — six times larger; the battery pumps in the extra charge.
• Energy: $U=\tfrac12CV^2$ rises by the factor $K=6$.
• Field: since $V$ is fixed and the mica fills the gap, $E=\dfrac{V}{d}=\dfrac{100}{3\times10^{-3}}=3.33\times10^{4}\,\text{V/m}$ — unchanged from before insertion.

Part (b) — Supply disconnected (Q fixed)

The charge is trapped at $Q=Q_0=1.77\,\text{nC}$.

• Capacitance: $C=106.2\,\text{pF}$.
• Voltage: $V=\dfrac{Q}{C}=\dfrac{1.77\times10^{-9}}{106.2\times10^{-12}}=16.67\,\text{V}$ — the original $100\,\text{V}$ divided by $K=6$.
• Energy: $U=\dfrac{Q^2}{2C}$ falls to $\tfrac16$ of its original value. …

Method: Effect of Inserting a Dielectric — Battery Connected vs Disconnected

This method finds how capacitance, voltage, charge, energy, and field change when a dielectric slab is inserted, depending on whether the battery stays connected.

Steps

Step 1: Decide what is held fixed

This is the single branching decision the whole problem hinges on: if the battery stays connected, it maintains a fixed voltage $V$ across the plates (charge is free to flow in/out). If the battery is disconnected first, the charge $Q$ on the isolated plates is trapped and cannot change.

Step 2: Find the new capacitance (same in both cases)

If the dielectric (constant $K$) fills the gap completely, the capacitance scales by $K$:

This step doesn't depend on the battery — geometry and the medium alone set $C$.

Step 3a: If voltage is fixed (battery connected)

Use $Q_{\text{new}} = C_{\text{new}}V$ (charge rises by factor $K$, drawn in from the battery) and $U_{\text{new}} = \tfrac12 C_{\text{new}}V^2$ (energy rises by factor $K$). The field $E = V/d$ is unchanged, since neither $V$ nor $d$ changed. …