Dielectrics - Physics 2

1. In a polar dielectric, permanent dipole moments align with an applied electric field, increasing the polarization $\mathbf{P}$. In a nonpolar dielectric, induced dipole moments arise from charge separation within molecules. For both types, the macroscopic relationship $\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P} = \kappa\varepsilon_0\mathbf{E}$ holds. A student argues: *'Since polar dielectrics have larger permanent dipole moments, they always have a larger dielectric constant $\kappa$ than nonpolar dielectrics at room temperature.'* Which of the following best evaluates this claim?
   Answer: The claim is incorrect, because $\kappa$ depends on the density of dipoles and the degree of alignment, not just on the existence of permanent moments. High-symmetry polar molecules may have suppressed net polarization due to geometric cancellation, and some nonpolar materials with highly polarizable electron clouds can have larger $\kappa$ values. Whenever you see a claim about dielectric constants, resist the instinct to think in absolutes — $\kappa$ is a macroscopic quantity that emerges from multiple competing factors, not a single molecular property. The macroscopic relationship $\mathbf{P} = \varepsilon_0(\kappa - 1)\mathbf{E}$ tells you that $\kappa$ reflects how strongly a material polarizes per unit field. That depends on dipole density, degree of alignment, and molecular polarizability — not simply whether permanent dipoles exist. High-symmetry polar molecules (like CCl₄) can have individual bond dipoles that cancel geometrically, yielding negligible net polarization. Meanwhile, nonpolar materials with large, loosely bound electron clouds (like silicon or certain aromatic compounds) can have enormous induced polarizability, producing $\kappa$ values that rival or exceed many polar dielectrics. Answer C correctly captures all of this nuance — it's the right choice. Answer A fails because it treats "permanent dipole exists → larger $\kappa$" as a universal law, ignoring geometric cancellation and the role of electron cloud polarizability in nonpolar materials. This is a classic overgeneralization trap. Answer B gets the temperature logic backwards. At low temperatures, thermal agitation is reduced, which actually helps polar molecules stay aligned — it doesn't help nonpolar materials "catch up." The premise of B is physically inverted. Answer D contains a grain of truth (rotation does matter in liquids), but it's wrong to say crystalline constraints eliminate the permanent-dipole contribution entirely — partial alignment and other mechanisms still contribute to $\kappa$ in polar solids. Your study tip: when evaluating "always" or "never" claims about material properties, immediately look for counterexamples rooted in molecular geometry or alternative physical mechanisms — exams love testing these edge cases.
2. What is the new voltage across the capacitor that does not contain the dielectric?
   Answer: $\dfrac{V}{2}$, because the total charge is conserved and, in a series circuit with isolated plates, each capacitor retains the same charge $Q = C_0 V/2$; dividing by $C_0$ gives $V/2$ for the unmodified capacitor. When capacitors are connected in series and then isolated from the battery, the key constraint shifts from voltage to charge. Charge cannot flow onto or off the isolated inner plates, so the total charge on the series combination is fixed. Before disconnection, the series capacitance is $C_{series} = \dfrac{C_0}{2}$, so the stored charge is $Q = \dfrac{C_0}{2} \cdot V = \dfrac{C_0 V}{2}$. This charge is locked in — inserting the dielectric cannot change it. Since both capacitors share the same series charge, each still carries $Q = \dfrac{C_0 V}{2}$. For the unmodified capacitor (still capacitance $C_0$), the voltage is simply $V_{unmod} = \dfrac{Q}{C_0} = \dfrac{V}{2}$. That confirms B is correct. Choice A incorrectly applies the voltage-divider ratio using the new series capacitance divided by the original element capacitance — this formula doesn't apply here. Voltage dividers work when voltage is fixed (battery connected), not when charge is fixed. Choice C gets the charge-conservation logic partially right but makes an arithmetic error. The dielectric doubles one capacitor to $2C_0$, and the total voltage now splits as $\frac{1}{3}V$ across the modified and $\frac{2}{3}V$ across the unmodified — yet C assigns those backwards, giving $V/3$ to the wrong capacitor. Choice D invents a false symmetry argument. Inserting a dielectric affects only one capacitor; the other is physically unchanged. Study tip: Whenever a battery is disconnected before something changes, freeze the charge — that's your conserved quantity. Ask "what's locked in?" before solving.
3. A capacitor is constructed with two different dielectric slabs of equal thickness $d/2$ and equal area $A/2$ each, placed side by side (i.e., each slab spans the full plate separation $d$ but only half the plate area), with dielectric constants $\kappa_1$ and $\kappa_2$. Which expression gives the total capacitance, and which circuit analogy correctly describes this geometry?
   Answer: $C = \dfrac{\varepsilon_0 A(\kappa_1 + \kappa_2)}{2d}$; the two sections act as capacitors in parallel, since both share the same voltage (connected to the same plates) and their charges add. When a capacitor has dielectric slabs placed side by side (each occupying half the area but the full gap), the key question is: do they share the same voltage, or the same charge? Since both regions are bounded by the same two conducting plates, they share identical voltage $V$ across them. This is the defining condition for a parallel combination. For each half, you have a standard capacitor with area $A/2$ and gap $d$: $C_1 = \frac{\kappa_1 \varepsilon_0 (A/2)}{d}, \quad C_2 = \frac{\kappa_2 \varepsilon_0 (A/2)}{d}$ In parallel, capacitances add directly: $C = C_1 + C_2 = \frac{\varepsilon_0 A}{2d}(\kappa_1 + \kappa_2)$ This confirms D is correct — parallel combination, charges add, voltages are equal. A is wrong on two counts: it gives a harmonic (series-like) formula and incorrectly claims "charges add harmonically." That formula would apply to stacked dielectrics, not side-by-side ones. B describes a series combination and uses the harmonic mean of $\kappa_1$ and $\kappa_2$. Series applies when the electric field passes sequentially through both materials — that's the stacked geometry, not this one. C gets the correct formula but misidentifies the circuit analogy as series. The math is right but the physical reasoning is backward — this is a classic trap designed to test whether you truly understand why the formula works. Study tip: Always ask yourself first — do the dielectric regions share the same voltage or the same charge? Same voltage → parallel. Same charge → series. The geometry tells you everything.
4. A parallel-plate capacitor is connected to a constant-voltage source $V_0$. A dielectric slab of dielectric constant $\kappa > 1$ is slowly inserted between the plates while the voltage source remains connected. Which of the following statements about the free charge on the plates and the work done by the voltage source is correct?
   Answer: The free charge on the plates increases by a factor of $\kappa$, and the work done by the voltage source is twice the increase in stored energy, with the excess energy accounted for by the mechanical work done on the dielectric slab. When a capacitor stays connected to a voltage source while you insert a dielectric, the voltage is locked at $V_0$, but the capacitance changes — and that drives everything else. Here's the core logic: inserting a dielectric increases capacitance by a factor of $\kappa$, since $C = \kappa C_0$. Because $Q = CV$ and $V_0$ is fixed, the free charge on the plates must increase by the same factor: $Q = \kappa C_0 V_0$. This immediately rules out C (charge doesn't stay constant — capacitance changed, so charge must change to maintain the same voltage) and D (charge increases, not decreases; bound charges from the dielectric don't neutralize free charges — they actually cause the battery to supply more free charge to maintain $V_0$). Now for the energy accounting. The new stored energy is $U = \frac{1}{2}\kappa C_0 V_0^2$, an increase of $\Delta U = \frac{1}{2}(\kappa - 1)C_0 V_0^2$. The battery supplies charge $\Delta Q = (\kappa - 1)C_0 V_0$, doing work $W_{battery} = V_0 \cdot \Delta Q = (\kappa - 1)C_0 V_0^2 = 2\Delta U$. The battery does twice the work stored — the extra half goes into the mechanical work done on the slab (the dielectric is pulled in, so the system does negative mechanical work on whoever holds it, or equivalently the field does positive work on the dielectric). This confirms B is correct. A is wrong because it claims the battery's work equals the stored energy increase — it actually supplies twice that amount. Study tip: Whenever a capacitor problem involves a battery staying connected, remember the battery does work $W = V \cdot \Delta Q$, which always equals twice the energy stored from that extra charge. Energy conservation then tells you where the other half went.
5. After the dielectric is inserted (with the battery disconnected), which of the following correctly describes what happens to the electric field between the plates and the energy stored in the capacitor?
   Answer: The electric field decreases by a factor of 3, and the stored energy decreases by a factor of 3, because the charge on the plates remains fixed while the capacitance triples. Whenever a capacitor question involves a dielectric, your first move should be identifying whether the battery is connected or disconnected — this single detail controls everything. If the battery is disconnected, charge is conserved ($Q = \text{const}$); if it stays connected, voltage is conserved. Here, the battery is disconnected before the dielectric is inserted, so the charge $Q$ on the plates cannot change. Inserting a dielectric with $\kappa = 3$ triples the capacitance: $C' = \kappa C_0 = 3C_0$. Since $Q$ is fixed but $C$ triples, the voltage drops by a factor of 3: $V' = Q/C' = V_0/3$. The electric field, which equals $V/d$, therefore also drops by a factor of 3. The stored energy $U = Q^2/(2C)$ then decreases by the same factor of 3, giving $U' = U_0/3$. This confirms answer A is correct. B is wrong because the electric field does change — it's tied to voltage, which drops when capacitance increases at constant charge. There is no "polarization energy" added; in fact, energy is released (it goes into the mechanical work of pulling the dielectric in). C describes the scenario where the battery remains connected (constant voltage), not the disconnected case. With constant voltage, energy would actually increase by $\kappa$, but the field would stay the same — C incorrectly mixes both scenarios. D is wrong on both counts: the field does change, and the idea that the dielectric reduces "effective plate separation" is not a real mechanism here. Study tip: Memorize these two cases as a pair — battery connected means constant $V$; battery disconnected means constant $Q$. Every dielectric insertion problem hinges on that distinction.
6. What is the new voltage across the capacitor after the partial insertion?
   Answer: $\dfrac{2V_0}{5}$, because the partial insertion creates a parallel combination with total capacitance $C_{\kappa} + C_{\text{vac}} = 2C_0 + \frac{C_0}{2} = \frac{5}{2}C_0$, and with fixed charge $Q = C_0 V_0$, the new voltage is $Q/C_{\text{new}} = \frac{2V_0}{5}$. Whenever a dielectric is partially inserted into a capacitor, the key is recognizing the correct circuit model. The two halves — one with dielectric, one without — share the same voltage and act as capacitors in parallel, not in series. Here's the reasoning for choice C. Before disconnection, the charge stored is $Q = C_0 V_0$. After the battery is disconnected, this charge is fixed. The partial insertion creates two parallel capacitors: the dielectric half has capacitance $C_\kappa = \kappa \cdot \frac{C_0}{2} = 4 \cdot \frac{C_0}{2} = 2C_0$, and the vacuum half has $C_{\text{vac}} = \frac{C_0}{2}$. Their total is $C_{\text{new}} = 2C_0 + \frac{C_0}{2} = \frac{5C_0}{2}$. With fixed charge, the new voltage is $V = \frac{Q}{C_{\text{new}}} = \frac{C_0 V_0}{\frac{5C_0}{2}} = \frac{2V_0}{5}$. That confirms C. Choice A incorrectly averages the electric fields across regions, treating the geometry as a series-like problem. The voltage isn't a weighted average of field reductions — it comes from $Q/C_{\text{new}}$. Choice B wrongly applies $\kappa = 4$ to the entire gap, as if the dielectric were fully inserted. Only half the area is covered, so only half the capacitance is scaled by $\kappa$. Choice D claims the vacuum half "contributes nothing," ignoring that it still stores charge and adds $\frac{C_0}{2}$ to the total capacitance. Study tip: When a dielectric covers part of the plate area, always split the capacitor into two parallel capacitors. Parallel means same voltage, additive capacitance — never skip the vacuum half.
7. A student claims: *'Inserting a dielectric always increases the energy stored in a capacitor, because the dielectric constant $\kappa > 1$ increases the capacitance, and larger capacitance means more stored energy.'* Under which condition is this claim correct?
   Answer: When the capacitor plates are connected to a constant-voltage source during insertion, so the voltage remains fixed and $U = \frac{1}{2}CV^2$ increases with $C$. When analyzing dielectric insertion problems, the key is identifying the boundary condition: is voltage fixed (battery connected) or charge fixed (battery disconnected)? The student's claim hinges entirely on which formula applies. With a battery connected, voltage $V$ stays constant. Since $U = \frac{1}{2}CV^2$, inserting a dielectric increases $C$ by a factor of $\kappa$, which directly increases stored energy by the same factor. This makes A correct — the claim holds precisely when voltage is held fixed by an external source. B is actually backwards. When the capacitor is isolated, charge $Q$ is fixed, so you must use $U = \frac{Q^2}{2C}$. Because $C$ appears in the denominator, increasing $C$ decreases stored energy — the opposite of the student's claim. The dielectric is actually pulled in by the electric field, and the system does work on the dielectric, reducing stored energy. C is wrong because the outcome is not universal — it depends entirely on boundary conditions, as shown above. Dielectric polarization doesn't simply "add" energy; the energy bookkeeping changes depending on whether a source is present. D conflates a thermodynamic process condition (quasi-static) with an energy outcome. Inserting slowly affects how energy is transferred, not whether energy stored increases. Even a quasi-static insertion with the battery disconnected still reduces stored energy. Study tip: Always identify the constraint first — constant $V$ or constant $Q$ — then choose the correct energy formula. These two cases give opposite results for energy when a dielectric is inserted, making this a favorite exam trap.

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