Exercises — Dielectrics — polarization, dielectric constant, effect on capacitance

WHAT: only the medium between plates changed; the geometry $A$ (plate area) and $d$ (gap) are untouched. WHY: the formula $C=\frac{\kappa {\epsilon }_{0}A}{d}=\kappa C_0$ shows $\kappa$ is the only extra factor when geometry is fixed. $C=\kappa C_0=4\times 6=24\mu \text{F}.$ Larger, by exactly $\kappa =4$. A dielectric always raises $C$ (it "cuts the field, cranks the C").

WHAT: the field shrank from $E_0$ to $E$ at fixed charge. WHY: the definition of $\kappa$ is precisely this ratio, $\kappa =E_0/E$ — "how many times weaker the tug got." $\kappa =\frac{E_0}{E}=\frac{5\times 10^5}{1\times 10^5}=5.$

WHAT: we plug straight into the parallel-plate-with-dielectric formula. WHY that formula: parallel plates make a uniform field, so capacitance is purely geometric ($A/d$) times the medium factor $\kappa$. $C=\frac{\kappa {\epsilon }_{0}A}{d}=\frac{3\times 8.85\times 10^{-12}\times 0.02}{1\times 10^{-3}}.$ Numerator $=3\times 8.85\times 10^{-12}\times 0.02=5.31\times 10^{-13}$. Divide by $10^{-3}$: $C=5.31\times 10^{-10}\text{F}=531\text{pF}.$

(i) WHAT/WHY: the field a sheet of free charge would make on its own is $E_0={\sigma }_f/{\epsilon }_0$ (this comes from Gauss's law for parallel plates). $E_0=\frac{1.77\times 10^{-5}}{8.85\times 10^{-12}}=2.0\times 10^6\text{V/m}.$ (ii) WHY divide by $\kappa$: the net field inside is the vacuum field cut by $\kappa$ (charge fixed): $E=\frac{E_0}{\kappa }=\frac{2.0\times 10^6}{2}=1.0\times 10^6\text{V/m}.$ (iii) WHY this bound-charge formula: the leftover field is made by ${\sigma }_f-{\sigma }_b$, so $E=({\sigma }_f-{\sigma }_b)/{\epsilon }_0={\sigma }_f/(\kappa {\epsilon }_0)$. Rearranged, ${\sigma }_b={\sigma }_f(1-1/\kappa )$. ${\sigma }_b=1.77\times 10^{-5}\left(1-\frac{1}{2}\right)=8.85\times 10^{-6}{\text{C/m}}^2.$

WHAT is held fixed: the wire is cut, so no charge can leave — $Q$ is frozen. $Q=C_0{V}_0=(2\times 10^{-6})(200)=4\times 10^{-4}\text{C}=400\mu \text{C}\text{(unchanged)}.$ New capacitance: $C=\kappa C_0=4\times 2=8\mu \text{F}$. New voltage — WHY it drops: with $Q$ fixed, $V=Q/C$ falls as $C$ rises: $V=\frac{Q}{C}=\frac{4\times 10^{-4}}{8\times 10^{-6}}=50\text{V}=\frac{V_0}{\kappa }.$ Energy — WHICH formula: since $Q$ is the fixed quantity, use $U=Q^2/(2C)$. $U_0=\frac{Q^2}{2C_0}=\frac{(4\times 10^{-4}{)}^2}{2(2\times 10^{-6})}=0.04\text{J}=40\text{mJ}.$ $U=\frac{Q^2}{2C}=\frac{(4\times 10^{-4}{)}^2}{2(8\times 10^{-6})}=0.01\text{J}=10\text{mJ}.$ Energy fell by $30$ mJ. That missing energy is the work the field does pulling the slab in — the slab is sucked into the gap.

WHAT is held fixed: the battery clamps $V=200$ V no matter what. New capacitance: $C=\kappa C_0=8\mu \text{F}$. New charge — WHY it rises: $Q=CV$ with $V$ fixed, so $Q$ grows with $C$: $Q=CV=(8\times 10^{-6})(200)=1.6\times 10^{-3}\text{C}=1600\mu \text{C}.$ Original $Q_0=C_{0}V=(2\times 10^{-6})(200)=400\mu \text{C}$, so the battery pushed in $\Delta Q=1600-400=1200\mu \text{C}\text{extra}.$ Energy — WHICH formula: $V$ is fixed, so use $U=\frac{1}{2}C{V}^2$. $U_0=\frac{1}{2}(2\times 10^{-6})(200{)}^2=0.04\text{J}=40\text{mJ}.$ $U=\frac{1}{2}(8\times 10^{-6})(200{)}^2=0.16\text{J}=160\text{mJ}.$ Energy rose by $120$ mJ — pumped in by the battery (see Energy Stored in a Capacitor).

WHAT/WHY series: the same charge $Q$ sits on the plates, and the voltage across the gap is the sum of the drops across each layer — that is the defining signature of capacitors in series (see Capacitance and Capacitors). Layer 1 (vacuum, thickness $d/2$): $C_1=\frac{{\epsilon }_{0}A}{d/2}=\frac{2{\epsilon }_{0}A}{d}$. Layer 2 (dielectric, thickness $d/2$): $C_2=\frac{\kappa {\epsilon }_{0}A}{d/2}=\frac{2\kappa {\epsilon }_{0}A}{d}$. Series combination — WHY reciprocals add: in series the voltages add at fixed $Q$, and $V=Q/C$, so $1/C=1/C_1+1/C_2$. $\frac{1}{C}=\frac{d}{2{\epsilon }_{0}A}+\frac{d}{2\kappa {\epsilon }_{0}A}=\frac{d}{2{\epsilon }_{0}A}\left(1+\frac{1}{\kappa }\right).$ Invert: $C=\frac{2{\epsilon }_{0}A}{d}\cdot \frac{\kappa }{\kappa +1}=\frac{{\epsilon }_{0}A}{d}\cdot \frac{2\kappa }{\kappa +1}.$ So ${\kappa }_{\text{eff}}=\frac{2\kappa }{\kappa +1}=\frac{2\times 3}{3+1}=\frac{6}{4}=\mathrm{1.5.}$ Sanity: ${\kappa }_{\text{eff}}=1.5$ sits between vacuum ($1$) and full fill ($3$) — the half-fill helps, but the vacuum layer is the "weak link" that dominates the series stack.

WHY parallel: both halves see the same voltage $V$ (they share the same top and bottom plates) — same-voltage-different-charge is the signature of parallel. Each half has area $A/2$. Left (dielectric): $C_L=\frac{\kappa {\epsilon }_0(A/2)}{d}=\frac{\kappa {\epsilon }_{0}A}{2d}$. Right (vacuum): $C_R=\frac{{\epsilon }_0(A/2)}{d}=\frac{{\epsilon }_{0}A}{2d}$. WHY they add: in parallel the charges add at the same $V$, so $C=C_L+C_R$. $C=\frac{{\epsilon }_{0}A}{2d}(\kappa +1)=\frac{{\epsilon }_{0}A}{d}\cdot \frac{\kappa +1}{2}.$ So ${\kappa }_{\text{eff}}=\frac{\kappa +1}{2}=\frac{3+1}{2}=2.$ Compare with L4·A: same slab, same amount of material, but the parallel arrangement gives ${\kappa }_{\text{eff}}=2$ vs the series $1.5$. Filling across the field lines side-by-side beats stacking along them — the vacuum gap no longer throttles the whole path.

WHY energy method: the plates exert a fringing force on the slab that is awkward to sum directly; but energy bookkeeping is exact. With the battery gone, $Q$ is fixed, so the stored energy is $U=Q^2/(2C)$, and the mechanical work done on the slab by the field equals the drop in stored energy (energy conservation, no battery involved). $U_{\text{out}}=\frac{Q^2}{2C_0}=\frac{(4\times 10^{-4}{)}^2}{2(2\times 10^{-6})}=0.04\text{J}.$ $U_{\text{in}}=\frac{Q^2}{2C}=\frac{(4\times 10^{-4}{)}^2}{2(8\times 10^{-6})}=0.01\text{J}.$ $W_{\text{on slab}}=U_{\text{out}}-U_{\text{in}}=0.04-0.01=0.03\text{J}=30\text{mJ}.$ Since energy drops as the slab enters, the system "wants" the slab in: the force pulls the slab inward. (Physically the fringing field grabs the induced bound charges and drags them into the strong-field region.)

WHY $V_{\max }=E_{\max }d$: the field inside is uniform, $E=V/d$, so it first reaches the breakdown value $E_{\max }$ when $V=E_{\max }d$. Beyond that a spark jumps. Material A: $V_{\max }=E_{\max }d=(3\times 10^6)(2\times 10^{-3})=6000\text{V}.$ Material B: $V_{\max }=(20\times 10^6)(2\times 10^{-3})=40000\text{V}.$ Comment: the dielectric wins twice — it raises capacitance by $\kappa =5$ and it has a higher breakdown field ($20$ vs $3\text{MV/m}$), so you can apply a far larger voltage before it sparks. That double benefit (more $C$, safer $V$) is exactly why real capacitors are stuffed with dielectric, not left as vacuum.

Step 1 — $\kappa$: by definition $\kappa =1+{\chi }_e=1+3=4$ (see Electric Susceptibility). Step 2 — field $E$: the free sheet alone makes $E_0={\sigma }_f/{\epsilon }_0$; inside it shrinks by $\kappa$: $E=\frac{{\sigma }_f}{\kappa {\epsilon }_0}=\frac{2\times 10^{-5}}{4\times 8.85\times 10^{-12}}=5.65\times 10^5\text{V/m}.$ Step 3 — polarization $P$: $P={\epsilon }_0{\chi }_{e}E$ (the linear-material definition of how much dipole-per-volume the field induces). $P=(8.85\times 10^{-12})(3)(5.65\times 10^5)=1.5\times 10^{-5}{\text{C/m}}^2.$ Step 4 — bound charge: ${\sigma }_b=P=1.5\times 10^{-5}{\text{C/m}}^2$ (the exposed-face charge equals the polarization magnitude). Cross-check: ${\sigma }_f(1-1/\kappa )=2\times 10^{-5}(1-0.25)=1.5\times 10^{-5}$. ✓ Step 5 — the $D$-field: the displacement field is defined so it "sees only free charge": $D={\epsilon }_0\kappa E$ (see Gauss's Law and the D-field). $D=(8.85\times 10^{-12})(4)(5.65\times 10^5)=2.0\times 10^{-5}{\text{C/m}}^2={\sigma }_f.✓$ The beauty: $D$ came out equal to ${\sigma }_f$ exactly, confirming that Gauss's law for $D$ ignores the bound charge and reads off the free plate charge directly.