Q.A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius $R = 6.0\ \text{cm}$ has a capacitance $C = 100\ \text{pF}$. The capacitor is connected to a $230\ \text{V}$ ac supply with a (angular) frequency of $300\ \text{rad s}^{-1}$.

Concept and Intuition

Between the plates of a capacitor there is no conduction current, yet a changing electric field there produces a displacement current that seamlessly continues the circuit's current — this was Maxwell's key insight completing Ampere's law. For a parallel-plate capacitor, since $E=V/d$, a changing voltage across the plates directly means a changing $E$-field, and the displacement current works out to have exactly the same value as the conduction (charging) current feeding the plates, $I_d = C\,\dfrac{dV}{dt}$.

Step-by-Step Solution

1. Displacement current: $I_d=\varepsilon_0\dfrac{d\Phi_E}{dt}=\varepsilon_0 A\dfrac{dE}{dt}=\dfrac{\varepsilon_0 A}{d}\dfrac{dV}{dt}$.
2. Treat the source voltage of 400 V as being established (capacitor charged from 0 to 400 V) over the given interval $\Delta t=10^{-6}$ s, so $\dfrac{dV}{dt}\approx\dfrac{400}{10^{-6}}$ V/s.
3. Data: $A=60\ \text{cm}^2=60\times10^{-4}\ \text{m}^2=6\times10^{-3}\ \text{m}^2$, $d=2\ \text{mm}=2\times10^{-3}\ \text{m}$, $\varepsilon_0=8.85\times10^{-12}\ \text{F/m}$.
4. $I_d=\dfrac{8.85\times10^{-12}\times6\times10^{-3}\times400}{2\times10^{-3}\times10^{-6}}$.
5. Numerator: $8.85\times10^{-12}\times6\times10^{-3}=5.31\times10^{-14}$; $\times400=2.124\times10^{-11}$.
6. Denominator: $2\times10^{-3}\times10^{-6}=2\times10^{-9}$.
7. $I_d=\dfrac{2.124\times10^{-11}}{2\times10^{-9}}=1.062\times10^{-2}$ A.

Common Mistakes

• Forgetting to convert plate area from cm² to m² (a factor-of-$10^4$ error).
• Mixing up mm and m for the plate separation.
• Using $\varepsilon_0 A/d$ (capacitance) but forgetting to divide by the time interval to get $dV/dt$.

ANSWER: B

Concept and Intuition

Maxwell noticed that Ampere's original law, $\oint \vec B\cdot d\vec l=\mu_0 I_{enc}$, gives inconsistent results depending on which surface bounded by the loop you use, when the current is discontinuous — for example, at a charging capacitor's gap, no conduction current crosses the surface between the plates. Maxwell resolved this by proposing a "displacement current" $I_d=\epsilon_0\dfrac{d\Phi_E}{dt}$ due to the changing electric flux, giving the corrected Ampere–Maxwell law:

Gauss's laws (electricity and magnetism) and Faraday's law were left in their original forms; only Ampere's law needed this correction.

Step-by-Step Solution

1. Recall the four Maxwell's equations: Gauss's law (electricity), Gauss's law (magnetism), Faraday's law, Ampere–Maxwell law.
2. Identify which one Maxwell himself modified from its pre-existing (Ampere) form: it is Ampere's circuital law, via the displacement current term.
3. The other three retain their classical forms as first stated by Gauss and Faraday.

Common Mistakes

• Confusing Faraday's law (which Maxwell used as-is) with the one he modified.
• Not recognizing "displacement current" as the specific fix that was added.

ANSWER: D

Concept and Intuition

Maxwell's displacement current arises from a time-varying electric field, exactly analogous to how a real conduction current arises from moving charge. Between the plates of a charging/discharging capacitor, $I_d = \varepsilon_0 \dfrac{d\Phi_E}{dt} = \varepsilon_0 A \dfrac{dE}{dt}$, where $A$ is the plate area. This displacement current is what "completes the circuit" through the gap between the plates in Ampère–Maxwell's law.

Step-by-Step Solution

1. Plate radius $r = 2\,\mathrm{cm} = 0.02\,\mathrm{m}$, so area $A = \pi r^2 = \pi (0.02)^2 = 1.2566\times10^{-3}\,\mathrm{m^2}$.
2. $I_d = \varepsilon_0 A \dfrac{dE}{dt} = (8.85\times10^{-12})(1.2566\times10^{-3})(4.5\times10^7)$.
3. $(8.85\times10^{-12})(1.2566\times10^{-3}) \approx 1.112\times10^{-14}$.
4. $1.112\times10^{-14}\times4.5\times10^7 \approx 5.0\times10^{-7}\,\mathrm{A} = 0.5\,\mu\mathrm{A}$.

Common Mistakes

• Using the diameter instead of the radius in the area formula.
• Forgetting to multiply by $\varepsilon_0$ (treating $A\,dE/dt$ alone as the current).

ANSWER: D

Concept and Intuition

Maxwell's displacement current is defined precisely so that it equals the conduction current charging/discharging the capacitor: $i_d=\dfrac{dq}{dt}$, ensuring current continuity even through the gap between the plates (where no charge physically flows).

Step-by-Step Solution

1. $q=q_0\cos\omega t$.
2. $i_d=\dfrac{dq}{dt}=-q_0\omega\sin\omega t$.

Common Mistakes

• Missing the negative sign or the factor of $\omega$ that comes from differentiating the cosine.
• Confusing displacement current with the charge itself, forgetting to differentiate.

ANSWER: C

Concept and Intuition

Maxwell noticed that Ampere's law, $\oint \vec{B}\cdot d\vec{l} = \mu_0 I$, is inconsistent when applied to a capacitor being charged (no real current flows between the plates, yet a changing electric field exists there). He fixed this by adding a term $\epsilon_0\dfrac{d\phi_E}{dt}$, calling it the displacement current $I_d$, so that the corrected law reads $\oint \vec{B}\cdot d\vec{l} = \mu_0(I + I_d)$. Since $I_d$ is added to an ordinary current $I$ inside the same equation, it must have the same dimensions as current — that is baked into how Maxwell constructed the term.

Step-by-Step Solution

1. Maxwell's displacement current is defined as $I_d = \epsilon_0\dfrac{d\phi_E}{dt}$, where $\phi_E$ is the electric flux.
2. The generalized Ampere's law is $\oint \vec{B}\cdot d\vec{l} = \mu_0(I_{conduction} + I_d)$.
3. For this equation to be dimensionally consistent, $I_d$ must carry the same dimensions as $I_{conduction}$, i.e. dimensions of electric current.
4. Therefore $\epsilon_0\dfrac{d\phi_E}{dt}$ has dimensions of current.

Common Mistakes

• Confusing this with $\dfrac{d\phi_E}{dt}$ alone (without the $\epsilon_0$ factor), which instead has dimensions related to potential/EMF-type quantities, not current.
• Trying to derive the dimensions from scratch via $\epsilon_0$'s and flux's separate SI units instead of recalling the physical definition, which is faster and less error-prone.

ANSWER: B

Concept and Intuition

Original Ampere's Law related magnetic fields only to conduction currents. Maxwell noticed an inconsistency in circuits with capacitors (where no conduction current flows between the plates) and introduced the displacement current $I_d = \varepsilon_0 \dfrac{d\Phi_E}{dt}$, showing that a changing electric flux itself acts as a source of magnetic field. This symmetric partner to Faraday's law (changing $B$ produces $E$) completes Maxwell's equations.

Step-by-Step Solution

1. Faraday's Law: changing $B$ field induces an EMF/electric field — not what's described here.
2. Biot–Savart Law: relates magnetic field to steady conduction current, not to a changing electric field.
3. Lenz's Law: only gives the direction of induced EMF, unrelated to this statement.
4. Modified (Ampere–Maxwell) Law: explicitly states a changing electric field produces a magnetic field via the displacement current term — matches the statement exactly.

Common Mistakes

• Confusing this with Faraday's law, which is the reverse effect (changing $B$ → $E$).
• Mixing up Biot-Savart's (steady current) with Maxwell's displacement-current addition.

ANSWER: C