Electromagnetic Waves MCQs with Answers – Part 1 (NEET Physics)

Explanation: A steady conduction current produces a magnetic field, as described by Ampere's original circuital law. However, Maxwell recognised that conduction current is not the only possible source of a magnetic field. An electric field that changes with time can also give rise to a magnetic field. The relevant quantity is the rate of change of electric flux through a surface, not merely the magnitude of the electric field. Maxwell represented this additional contribution through displacement current: \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt}. \] Here, \(\Phi_E\) is the electric flux through the chosen surface and \(\varepsilon_0\) is the permittivity of free space. A nonzero rate of change of electric flux can therefore produce a displacement-current contribution even when no free charges physically cross that surface. For example, the insulating gap of a charging capacitor contains a changing electric field but has no conduction current in the ideal model. The displacement current accounts for the associated magnetic field in that region. Conversely, a steady electric field can have substantial electric flux while producing zero displacement current. Thus, the essential distinction is between the existence of electric flux and its variation with time.

Explanation: Ampere's original circuital law relates magnetic circulation around a closed path to the conduction current passing through a surface bounded by that path: \[ \oint\vec B\cdot d\vec l=\mu_0 i_c. \] The left-hand side is evaluated along a particular closed boundary. Therefore, for a fixed magnetic-field configuration, choosing different surfaces with that same boundary cannot change the value of the circulation. For the surface cutting the connecting wire, a conduction current \(i_c\) passes through the surface. The original law consequently predicts a nonzero magnetic circulation. For the surface passing through the insulating capacitor gap, no conduction charge crosses the surface. Applying the original law without modification would therefore predict zero circulation. These two predictions contradict each other because the boundary and the physical magnetic field have not changed. Maxwell identified the missing contribution as the displacement current associated with changing electric flux: \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt}. \] During ideal capacitor charging, this contribution supplies the missing current term for the gap-crossing surface. The corrected Ampere–Maxwell law therefore gives a consistent magnetic circulation for both surfaces. The failure lies in the incompleteness of the original current term, not in any physical dependence of the magnetic field on an arbitrarily chosen spanning surface.

Explanation: Statement I is correct. During charging, conduction current flows through the external connecting wires and supplies charge to the capacitor plates. The plates acquire equal and opposite charges, producing an electric field in the gap. For an ideal capacitor with a perfectly insulating gap, free charge carriers do not physically travel from one plate to the other through that gap. Nevertheless, the changing electric field is associated with a displacement current. Displacement current is defined by the rate of change of electric flux: \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt}. \] Statement II is incorrect because displacement current has the same SI unit as conduction current, namely ampere. Although electric charge is measured in coulomb, current is a rate of charge transfer or its displacement-current equivalent. For an ideal charging capacitor, the relation becomes \[ i_d=\frac{dQ}{dt}, \] where \(Q\) denotes the magnitude of charge accumulated on either plate. Since \(dQ/dt\) has the unit coulomb per second, it is measured in ampere. Equality of displacement current and conduction current in this arrangement does not mean that their microscopic mechanisms are identical. Therefore, Statement I is correct and Statement II is incorrect.

Explanation: The Assertion is correct because the magnetic circulation is determined along the common closed boundary, not by an arbitrary choice of spanning surface. For the surface cutting the connecting wire, the relevant contribution in the ideal capacitor model is the conduction current. Consequently, the Ampere–Maxwell law gives \[ \oint\vec B\cdot d\vec l=\mu_0 i_c. \] For the surface passing through the insulating gap, the conduction-current contribution is zero. However, the electric field between the plates changes as charge accumulates. Neglecting edge effects, the electric flux through the full plate area is \[ \Phi_E=\frac{Q}{\varepsilon_0}. \] Differentiating this relation gives \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt}=\frac{dQ}{dt}. \] The charging current in the connecting wire is also \(i_c=dQ/dt\), with a consistent choice of current direction. Hence, \(i_d=i_c\) for the two corresponding surfaces. The gap-crossing surface therefore gives the same magnetic circulation, \(\mu_0 i_d=\mu_0 i_c\). The Reason is correct and provides precisely the physical and mathematical explanation required by the Assertion. The currents are not added together as \(2i_c\), because the two specified surfaces contain different current contributions.

Explanation: \(\textbf{Given:}\) The rate of change of electric flux is \[ \frac{d\Phi_E}{dt}=4.0\times10^5\,\text{N m}^2\text{C}^{-1}\text{s}^{-1} \] The permittivity of free space is \[ \varepsilon_0=8.85\times10^{-12}\,\text{F m}^{-1} \] \(\textbf{Displacement-current relation:}\) Maxwell's displacement current is \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt} \] The supplied rate is positive, so the displacement current is positive with respect to the selected surface orientation. \(\textbf{Substitution:}\) \[ i_d=(8.85\times10^{-12})(4.0\times10^5)\,\text{A} \] Separating the numerical factors and powers of ten, \[ i_d=(8.85\times4.0)\times10^{-7}\,\text{A} \] \[ i_d=35.4\times10^{-7}\,\text{A} \] \[ i_d=3.54\times10^{-6}\,\text{A} \] \(\textbf{Unit check:}\) Electric flux has the unit \(\text{N m}^2\text{C}^{-1}\), while \[ \varepsilon_0=\frac{\text{C}^2}{\text{N m}^2} \] Therefore, the product \(\varepsilon_0\,d\Phi_E/dt\) has the unit \(\text{C s}^{-1}=\text{A}\), as required. \(\textbf{Final answer:}\) \[ \boxed{i_d=3.54\,\mu\text{A}} \] The displacement current depends on the rate of change of flux, not on the instantaneous magnitude of flux. Only the supplied flux-changing rate is needed for this calculation.

Explanation: \(\textbf{Key graph principle:}\) The displacement current through a fixed surface is proportional to the slope of its electric-flux versus time graph: \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt} \] The vertical intercept represents the initial electric flux, whereas the slope represents its rate of change. Graph II has a much larger vertical intercept than Graph I, but this does not establish that its displacement current is larger. \(\textbf{Slope of Graph I:}\) \[ \Phi_1=(5+3t)\times10^5 \] Differentiating with respect to time, \[ \frac{d\Phi_1}{dt}=3\times10^5\,\text{N m}^2\text{C}^{-1}\text{s}^{-1} \] Therefore, \[ i_{d1}=\varepsilon_0(3\times10^5) \] \(\textbf{Slope of Graph II:}\) \[ \Phi_2=(100+2t)\times10^5 \] \[ \frac{d\Phi_2}{dt}=2\times10^5\,\text{N m}^2\text{C}^{-1}\text{s}^{-1} \] Hence, \[ i_{d2}=\varepsilon_0(2\times10^5) \] \(\textbf{Comparison:}\) \[ \frac{i_{d1}}{i_{d2}}=\frac{\varepsilon_0(3\times10^5)}{\varepsilon_0(2\times10^5)} \] \[ \frac{i_{d1}}{i_{d2}}=\frac32 \] \(\textbf{Final answer:}\) \[ \boxed{i_{d1}:i_{d2}=3:2} \] Both slopes are positive, so both currents have the positive orientation selected for their respective surfaces. The essential reasoning is that displacement current is determined by flux variation, not by the height of the flux graph.

Explanation: \(\textbf{Given:}\) The surface area is \[ A=2.0\times10^{-2}\,\text{m}^2 \] The angle between the electric field and the surface normal is \[ \theta=60^\circ \] The field-changing rate is \[ \frac{dE}{dt}=4.0\times10^4\,\text{V m}^{-1}\text{s}^{-1} \] \(\textbf{Electric-flux geometry:}\) Electric flux through a plane surface in a uniform field is \[ \Phi_E=EA\cos\theta \] The angle is measured from the surface normal, not from the plane itself. Since both the area and orientation remain constant, \[ \frac{d\Phi_E}{dt}=A\cos\theta\frac{dE}{dt} \] The displacement-current relation becomes \[ i_d=\varepsilon_0 A\cos\theta\frac{dE}{dt} \] \(\textbf{Substitution:}\) \[ i_d=(8.85\times10^{-12})(2.0\times10^{-2})\cos60^\circ(4.0\times10^4) \] Using \(\cos60^\circ=1/2\), \[ i_d=(8.85\times10^{-12})(2.0\times10^{-2})\left(\frac12\right)(4.0\times10^4) \] First, evaluate the geometrical and field-changing factor: \[ A\cos\theta\frac{dE}{dt}=(2.0\times10^{-2})\left(\frac12\right)(4.0\times10^4) \] \[ A\cos\theta\frac{dE}{dt}=4.0\times10^2\,\text{V m s}^{-1} \] Therefore, \[ i_d=(8.85\times10^{-12})(4.0\times10^2)\,\text{A} \] \[ i_d=3.54\times10^{-9}\,\text{A} \] \(\textbf{Final answer:}\) \[ \boxed{i_d=3.54\,\text{nA}} \] Ignoring the inclination would give twice the correct displacement current. The field component normal to the surface determines electric flux, so a field variation along the surface contributes no flux through it.

Explanation: \(\textbf{Interpretation of the vertical axis:}\) The graph does not directly plot electric flux. It plots the product \(\varepsilon_0\Phi_E\), which has the dimension of electric charge. From Maxwell's displacement-current relation, \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt} \] Since \(\varepsilon_0\) is constant, \[ i_d=\frac{d(\varepsilon_0\Phi_E)}{dt} \] Therefore, the slope of the supplied graph directly gives displacement current. \(\textbf{First segment:}\) At \(t=1\,\text{ms}\), the relevant interval is from \(0\) to \(2\,\text{ms}\). \[ i_{d1}=\frac{9-1}{2-0}\,\text{nC ms}^{-1} \] \[ i_{d1}=4\,\text{nC ms}^{-1} \] Since \[ 1\,\text{nC ms}^{-1}=\frac{10^{-9}\,\text{C}}{10^{-3}\,\text{s}}=10^{-6}\,\text{A} \] we obtain \[ i_{d1}=+4\,\mu\text{A} \] \(\textbf{Second segment:}\) At \(t=3\,\text{ms}\), the relevant interval is from \(2\) to \(5\,\text{ms}\). \[ i_{d2}=\frac{3-9}{5-2}\,\text{nC ms}^{-1} \] \[ i_{d2}=\frac{-6}{3}\,\text{nC ms}^{-1} \] \[ i_{d2}=-2\,\mu\text{A} \] \(\textbf{Direction check:}\) The first segment has a positive slope, so its displacement current is positive. The second segment has a negative slope, so its displacement current is negative relative to the same selected normal. The sign reversal is caused by the reversal of the flux-changing rate, even though the plotted flux-related quantity remains positive throughout. \(\textbf{Final answer:}\) \[ \boxed{\big(i_d(1\,\text{ms}),\,i_d(3\,\text{ms})\big)=(+4,\,-2)\,\mu\text{A}} \]

Explanation: \(\textbf{Given charge variation:}\) \[ Q(t)=(2+6t-t^2)\,\mu\text{C} \] The required magnetic circulation is to be determined at \(t=1\,\text{s}\). The selected spanning surface cuts the connecting wire and therefore receives the conduction-current contribution. \(\textbf{Charging current:}\) The instantaneous current entering the positive capacitor plate equals its rate of increase of charge: \[ i_c=\frac{dQ}{dt} \] Differentiating the given expression, \[ i_c=(6-2t)\,\mu\text{C s}^{-1} \] At \(t=1\,\text{s}\), \[ i_c=(6-2\times1)\,\mu\text{A} \] \[ i_c=4\,\mu\text{A} \] \(\textbf{Magnetic circulation:}\) For the specified surface, the ideal Ampere–Maxwell law gives \[ \oint\vec B\cdot d\vec l=\mu_0 i_c \] Substituting, \[ \oint\vec B\cdot d\vec l=(4\pi\times10^{-7})(4\times10^{-6})\,\text{T m} \] \[ \oint\vec B\cdot d\vec l=16\pi\times10^{-13}\,\text{T m} \] \[ \oint\vec B\cdot d\vec l=1.6\pi\times10^{-12}\,\text{T m} \] \(\textbf{Consistency with the capacitor gap:}\) For the ideal capacitor, the displacement current through the full gap satisfies \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt}=\frac{dQ}{dt}=4\,\mu\text{A} \] Thus, a suitable alternative gap-crossing surface with the same boundary also gives the same magnetic circulation. This agreement is the essential surface-independence property of the Ampere–Maxwell law. \(\textbf{Final answer:}\) \[ \boxed{\oint\vec B\cdot d\vec l=1.6\pi\times10^{-12}\,\text{T m}} \] The instantaneous current must be obtained by differentiating the charge expression. Substituting the accumulated charge itself into Ampere's law would be physically and dimensionally incorrect.

Explanation: The electric field between oppositely charged capacitor plates is directed from the positive plate towards the negative plate. During the specified charging stage, the positive upper plate becomes more positively charged while the lower plate becomes more negatively charged. The electric field retains its downward direction, but its magnitude increases. During the subsequent discharging stage, the magnitudes of the plate charges decrease. Because the signs of the charges have not reversed, the electric field still points from the upper plate towards the lower plate. Its magnitude now decreases rather than increases. The displacement-current contribution depends on the time derivative of electric flux: \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt}. \] For the chosen normal directed from the upper plate towards the lower plate, the displacement current is positive during charging and negative during discharging. According to the Ampere–Maxwell law, this reversal of the displacement-current contribution reverses the circulation sense of the associated magnetic field. The magnetic field at corresponding points therefore reverses direction under the same ideal symmetrical arrangement. The electric field itself does not need to reverse for the magnetic field to reverse. This distinction separates the direction of a field from the sign of its rate of change. The correct conclusion is that the electric-field direction remains unchanged while the magnetic-field direction reverses.

Explanation: Statement I is correct for the specified ideal arrangement. A surface cutting only the connecting wire receives a conduction-current contribution equal to \(I\). The displacement-current contribution through that surface is zero under the stated ideal approximation. Statement II is also correct. The insulating gap contains no conduction current, but the changing electric flux produces a displacement current. For the full plate region, \[ \Phi_E=\frac{Q}{\varepsilon_0}. \] Therefore, \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt}=\frac{dQ}{dt}=I. \] Statement III is incorrect because the current contributions must be evaluated through the particular surface chosen. The general Ampere–Maxwell law is \[ \oint\vec B\cdot d\vec l=\mu_0(i_c+i_d). \] For the first specified surface, \[ i_c+i_d=I+0=I. \] For the second specified surface, \[ i_c+i_d=0+I=I. \] Thus, both surfaces produce the same magnetic circulation without any double-counting. Statement IV is incorrect because displacement current does not require free charge carriers to cross an insulating gap. The equality \(i_d=i_c\) expresses consistency of the field description and continuity of the charging process, not identical microscopic transport mechanisms. Consequently, Statements I and II alone are correct.

Explanation: \(\textbf{Magnetic-field symmetry:}\) For circular plates with a uniform changing electric field, cylindrical symmetry permits the use of a concentric circular Amperian loop. The magnetic field is tangential to the loop and has equal magnitude at all points on it. Therefore, the Ampere–Maxwell law gives \[ B(2\pi r)=\mu_0 i_{d,\text{enclosed}} \] \(\textbf{Inside the plate radius:}\) For \(r\lt R\), the loop encloses only part of the total changing electric flux. Since the electric field is uniform over the circular plate area, the enclosed displacement current is proportional to the enclosed area: \[ i_{d,\text{enclosed}}=I\frac{\pi r^2}{\pi R^2} \] \[ i_{d,\text{enclosed}}=I\frac{r^2}{R^2} \] Substituting into the Ampere–Maxwell law, \[ B(2\pi r)=\mu_0 I\frac{r^2}{R^2} \] Hence, \[ B=\frac{\mu_0 Ir}{2\pi R^2} \] For fixed \(I\) and \(R\), this gives \(B\propto r\). \(\textbf{Outside the plate radius:}\) For \(r\gt R\), the loop encloses the entire displacement current \(I\). Thus, \[ B(2\pi r)=\mu_0 I \] \[ B=\frac{\mu_0 I}{2\pi r} \] Therefore, \(B\propto1/r\) outside the plate radius. At \(r=R\), the two expressions give the same field magnitude: \[ B(R)=\frac{\mu_0 I}{2\pi R}. \] The field increases linearly from the axis to the plate edge and then decreases inversely with radial distance. The important distinction is that the enclosed displacement current increases with \(r^2\) inside the plates but stops increasing once the entire active plate area is enclosed. The magnetic-field magnitude does not increase quadratically because the circular path length also increases with \(r\).

Explanation: \(\textbf{Given:}\) The plate radius is \[ R=6\,\text{cm}=0.06\,\text{m} \] The radial position is \[ r=3\,\text{cm}=0.03\,\text{m} \] The total charging current is \[ I=12\,\text{mA}=12\times10^{-3}\,\text{A} \] \(\textbf{Identify the enclosed current:}\) Since \(r\lt R\), the circular Amperian loop lies entirely within the plate radius. The electric field is assumed uniform across the plate area. Consequently, the displacement current enclosed by the smaller loop is proportional to its area: \[ i_{d,\text{enclosed}}=I\frac{r^2}{R^2} \] Substituting, \[ i_{d,\text{enclosed}}=12\times10^{-3}\left(\frac{0.03}{0.06}\right)^2 \] \[ i_{d,\text{enclosed}}=12\times10^{-3}\left(\frac12\right)^2 \] \[ i_{d,\text{enclosed}}=12\times10^{-3}\times\frac14 \] \[ i_{d,\text{enclosed}}=3.0\times10^{-3}\,\text{A} \] Only one-fourth of the total displacement current is enclosed because the loop radius is half the plate radius. \(\textbf{Apply the Ampere–Maxwell law:}\) By cylindrical symmetry, \[ B(2\pi r)=\mu_0 i_{d,\text{enclosed}} \] Therefore, \[ B=\frac{\mu_0 i_{d,\text{enclosed}}}{2\pi r} \] \(\textbf{Substitution:}\) \[ B=\frac{(4\pi\times10^{-7})(3.0\times10^{-3})}{2\pi(0.03)} \] \[ B=\frac{12\pi\times10^{-10}}{0.06\pi} \] \[ B=200\times10^{-10}\,\text{T} \] \[ B=2.0\times10^{-8}\,\text{T} \] \(\textbf{Physical check:}\) The field magnitude must approach zero as \(r\to0\) under the stated uniform-field model. The result agrees with the interior-field relation \[B=\frac{\mu_0 Ir}{2\pi R^2}\] Using the entire charging current \(12\,\text{mA}\) for the smaller loop would overestimate the magnetic field by a factor of four. \(\textbf{Final answer:}\) \[ \boxed{B=2.0\times10^{-8}\,\text{T}} \]

Explanation: \(\textbf{Identify the radial regions:}\) Point \(P\) lies inside the plate radius because \[ r_P=\frac R2\lt R \] Point \(Q\) lies outside the plate radius because \[ r_Q=2R\gt R \] The enclosed displacement current must therefore be evaluated differently for the two points. \(\textbf{Field at point P:}\) For a concentric circular loop with \(r\lt R\), \[ i_{d,\text{enclosed}}=I\frac{r^2}{R^2} \] The magnetic-field magnitude is \[ B=\frac{\mu_0 Ir}{2\pi R^2} \] At \(r=R/2\), \[ B_P=\frac{\mu_0 I(R/2)}{2\pi R^2} \] \[ B_P=\frac{\mu_0 I}{4\pi R} \] \(\textbf{Field at point Q:}\) For \(r\gt R\), the entire displacement current \(I\) is enclosed. Therefore, \[ B=\frac{\mu_0 I}{2\pi r} \] At \(r=2R\), \[ B_Q=\frac{\mu_0 I}{2\pi(2R)} \] \[ B_Q=\frac{\mu_0 I}{4\pi R} \] \(\textbf{Comparison:}\) \[ \frac{B_P}{B_Q}=\frac{\mu_0 I/(4\pi R)}{\mu_0 I/(4\pi R)} \] \[ \frac{B_P}{B_Q}=1 \] \(\textbf{Final answer:}\) \[ \boxed{B_P:B_Q=1:1} \] Although point \(Q\) is four times as far from the axis as point \(P\), the magnetic fields are equal. Inside the plate radius, the magnetic field increases with radial distance because the enclosed displacement current grows with area. Outside the plate radius, the entire current is already enclosed, so the magnetic field decreases with distance. These competing radial dependences produce equal field magnitudes at the two specified points. Using the same \(1/r\) dependence at both points would incorrectly treat the interior loop as though it enclosed the entire charging current.

Explanation: \(\textbf{Identify the fixed conditions:}\) The total charging current \(I\) remains unchanged. The observation points also remain fixed relative to the common axis. Only the plate radius changes from \(R\) to \(2R\). The first observation point is inside the plates in both arrangements, while the second is outside the plates in both arrangements. \(\textbf{Interior magnetic field:}\) For a concentric loop of radius \(r\) inside a uniformly charging circular capacitor, \[ B_{\text{in}}=\frac{\mu_0 Ir}{2\pi R^2} \] The interior observation radius is fixed at \[ r_{\text{in}}=\frac R3 \] For the original capacitor, \[ B_{\text{in}}=\frac{\mu_0 I(R/3)}{2\pi R^2} \] \[ B_{\text{in}}=\frac{\mu_0 I}{6\pi R} \] When the plate radius becomes \(2R\), the same point remains inside the plates. The new field is \[ B'_{\text{in}}=\frac{\mu_0 I(R/3)}{2\pi(2R)^2} \] \[ B'_{\text{in}}=\frac{\mu_0 I(R/3)}{8\pi R^2} \] \[ B'_{\text{in}}=\frac{\mu_0 I}{24\pi R} \] Therefore, \[ \frac{B'_{\text{in}}}{B_{\text{in}}}=\frac{\mu_0 I/(24\pi R)}{\mu_0 I/(6\pi R)} \] \[ \frac{B'_{\text{in}}}{B_{\text{in}}}=\frac14 \] \(\textbf{Exterior magnetic field:}\) The second observation point is situated at \(r_{\text{out}}=3R\). It remains outside even the enlarged plate radius \(2R\). For an exterior circular loop, \[ B_{\text{out}}=\frac{\mu_0 I}{2\pi r_{\text{out}}} \] Before enlargement, \[ B_{\text{out}}=\frac{\mu_0 I}{2\pi(3R)} \] \[ B_{\text{out}}=\frac{\mu_0 I}{6\pi R} \] After enlargement, the observation radius and the total enclosed current are unchanged. Hence, \[ B'_{\text{out}}=\frac{\mu_0 I}{6\pi R} \] Consequently, \[ \frac{B'_{\text{out}}}{B_{\text{out}}}=1 \] \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{B'_{\text{in}}}{B_{\text{in}}},\frac{B'_{\text{out}}}{B_{\text{out}}}\right)=\left(\frac14,1\right)} \] The interior field decreases because the same total displacement current is now distributed uniformly over four times the plate area. The exterior field remains unchanged because its Amperian loop continues to enclose the entire charging current. The decisive reasoning is to identify which radial field relation applies before and after the geometrical change.

Explanation: \(\textbf{Capacitor-current relation:}\) For a capacitor with constant capacitance, \[ Q=CV \] Differentiating with respect to time, \[ \frac{dQ}{dt}=C\frac{dV}{dt} \] The total displacement current through the full plate region equals the rate of change of plate charge: \[ i_d=C\frac{dV}{dt} \] Here, \[ C=4\times10^{-6}\,\text{F} \] \(\textbf{Current enclosed by the smaller loop:}\) The concentric loop has radius \(r=R/2\). With a uniform changing electric field over the plate area, the fraction of total displacement current enclosed is \[ \frac{i_{d,\text{enclosed}}}{i_d}=\frac{\pi r^2}{\pi R^2} \] \[ \frac{i_{d,\text{enclosed}}}{i_d}=\left(\frac{R/2}{R}\right)^2 \] \[ \frac{i_{d,\text{enclosed}}}{i_d}=\frac14 \] Therefore, \[ i_{d,\text{enclosed}}=\frac C4\frac{dV}{dt} \] \(\textbf{Stage 1 — Rising voltage:}\) At \(t=1\,\text{s}\), the graph lies on the first straight-line segment. The voltage slope is \[ \frac{dV}{dt}=\frac{6-0}{2-0}=3\,\text{V s}^{-1} \] The total displacement current is \[ i_d=(4\times10^{-6})(3) \] \[ i_d=12\times10^{-6}\,\text{A}=12\,\mu\text{A} \] The enclosed displacement current is \[ i_{d1}=\frac14(12)=+3\,\mu\text{A} \] \(\textbf{Stage 2 — Constant voltage:}\) At \(t=3\,\text{s}\), the voltage graph is horizontal. Consequently, \[ \frac{dV}{dt}=0 \] Therefore, \[ i_d=C(0)=0 \] The enclosed displacement current is also \[ i_{d2}=0 \] Although the capacitor still has a nonzero voltage and electric field, its electric flux is not changing. Hence, there is no displacement current during this constant-voltage interval. \(\textbf{Stage 3 — Falling voltage:}\) At \(t=6\,\text{s}\), the graph lies on the descending straight-line segment. The voltage slope is \[ \frac{dV}{dt}=\frac{2-6}{7-5} \] \[ \frac{dV}{dt}=-2\,\text{V s}^{-1} \] The total displacement current becomes \[ i_d=(4\times10^{-6})(-2) \] \[ i_d=-8\,\mu\text{A} \] The enclosed displacement current is \[ i_{d3}=\frac14(-8)=-2\,\mu\text{A} \] \(\textbf{Direction and physical check:}\) The positive first current corresponds to increasing plate charge and increasing electric flux. The zero current corresponds to constant electric flux. The negative third current corresponds to decreasing plate charge and decreasing electric flux. The same geometrical area fraction applies in all three intervals because the plate radius and observation-loop radius do not change. Using the full capacitor current instead of the enclosed current would give \(+12,0,-8\,\mu\text{A}\), which is not the quantity requested. \(\textbf{Final answer:}\) \[ \boxed{(i_{d1},i_{d2},i_{d3})=(+3,\,0,\,-2)\,\mu\text{A}} \]

Explanation: \(\textbf{Electric-flux condition:}\) Displacement current is determined by the rate of change of electric flux through the specified surface. For a uniform electric field and a fixed plane surface, \[ \Phi_E=\vec E\cdot\vec A \] The area vector is \[ \vec A=A\hat n \] Here, \[ A=0.10\,\text{m}^2 \] and \[ \hat n=\frac{3\hat i+4\hat j}{5} \] \(\textbf{Resolve the changing field:}\) The electric-field components are \[ E_x=200+6t \] \[ E_y=300-8t \] Their rates of change are \[ \frac{dE_x}{dt}=6\,\text{V m}^{-1}\text{s}^{-1} \] \[ \frac{dE_y}{dt}=-8\,\text{V m}^{-1}\text{s}^{-1} \] Because the surface is stationary, its area vector does not change. Therefore, \[ \frac{d\Phi_E}{dt}=A\frac{d\vec E}{dt}\cdot\hat n \] \(\textbf{Calculate the flux-changing rate:}\) \[ \frac{d\Phi_E}{dt}=0.10\left[6\left(\frac35\right)-8\left(\frac45\right)\right] \] \[ \frac{d\Phi_E}{dt}=0.10\left(\frac{18-32}{5}\right) \] \[ \frac{d\Phi_E}{dt}=0.10(-2.8) \] \[ \frac{d\Phi_E}{dt}=-0.28\,\text{V m s}^{-1} \] \(\textbf{Apply Maxwell's relation:}\) \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt} \] \[ i_d=(8.85\times10^{-12})(-0.28)\,\text{A} \] \[ i_d=-2.478\times10^{-12}\,\text{A} \] \[ i_d\approx-2.48\,\text{pA} \] \(\textbf{Sign check:}\) Although the \(x\)-component of the electric field increases, the \(y\)-component decreases. The contribution of the decreasing \(y\)-component to the selected normal direction is larger than that of the increasing \(x\)-component. Consequently, the net electric flux decreases. The negative sign indicates displacement current opposite to the direction defined by the chosen surface normal. It does not indicate that the displacement-current magnitude is negative. \(\textbf{Final answer:}\) \[ \boxed{i_d=-2.48\,\text{pA}} \]

Explanation: \(\textbf{Relation between voltage and displacement current:}\) For a capacitor of constant capacitance, \[ Q=CV \] The displacement current through the full capacitor gap equals the rate of change of plate charge. Hence, \[ i_d=\frac{dQ}{dt}=C\frac{dV}{dt} \] \(\textbf{Differentiate the applied voltage:}\) \[ V(t)=V_0\sin(\omega t) \] Therefore, \[ \frac{dV}{dt}=V_0\omega\cos(\omega t) \] The instantaneous displacement current is \[ i_d(t)=CV_0\omega\cos(\omega t) \] \(\textbf{Current at one-quarter period:}\) At \(t=T/4\), \[ \omega t=\omega\frac T4=\frac{2\pi}{T}\frac T4=\frac{\pi}{2} \] Thus, \[ i_d\left(\frac T4\right)=CV_0\omega\cos\frac{\pi}{2} \] \[ i_d\left(\frac T4\right)=0 \] At this instant, the capacitor voltage has its positive maximum value. However, the voltage is momentarily neither increasing nor decreasing. Therefore, the displacement current vanishes even though the electric field is maximum in magnitude. \(\textbf{Current at half period:}\) At \(t=T/2\), \[ \omega t=\pi \] Hence, \[ i_d\left(\frac T2\right)=CV_0\omega\cos\pi \] \[ i_d\left(\frac T2\right)=-CV_0\omega \] At this instant, the voltage passes through zero while decreasing. The displacement current therefore has its maximum negative value. \(\textbf{Physical interpretation:}\) The voltage and displacement current are not in phase for an ideal capacitor. Current depends on the rate of voltage variation rather than on voltage magnitude. The current reaches zero at voltage extrema and reaches its greatest magnitude when the sinusoidal voltage crosses zero. The negative current indicates decreasing positive-plate charge. \(\textbf{Final answer:}\) \[ \boxed{\left(i_d\left(\frac T4\right),i_d\left(\frac T2\right)\right)=(0,-CV_0\omega)} \]

Explanation: For an ideal capacitor, the conduction current in the connecting wire equals the rate of accumulation of charge on the corresponding plate. The displacement current through the full insulating gap has the same signed value when consistent orientations are selected. Thus, \[ i_c=i_d=\frac{dQ}{dt} \] Since the charge variation is uniform in each experiment, the derivative can be calculated from \[ i=\frac{\Delta Q}{\Delta t} \] \(\textbf{Experiment P:}\) The charge increases by \(6\,\text{nC}\) in \(2\,\text{ms}\). Therefore, \[ i_P=\frac{+6\times10^{-9}}{2\times10^{-3}} \] \[ i_P=+3\,\mu\text{A} \] This agrees with the measured current, so P is consistent. \(\textbf{Experiment Q:}\) The charge decreases by \(8\,\text{nC}\) in \(4\,\text{ms}\). Therefore, \[ i_Q=\frac{-8\times10^{-9}}{4\times10^{-3}} \] \[ i_Q=-2\,\mu\text{A} \] This agrees with the measured negative current, so Q is consistent. The negative sign represents charge leaving the designated plate. \(\textbf{Experiment R:}\) The charge increases by \(5\,\text{nC}\) in \(1\,\text{ms}\). Hence, \[ i_R=\frac{+5\times10^{-9}}{1\times10^{-3}} \] \[ i_R=+5\,\mu\text{A} \] The measured current is \(-5\,\mu\text{A}\), which has the opposite sign. Thus, R is inconsistent with the stated current convention. \(\textbf{Experiment S:}\) The plate charge remains unchanged throughout the interval. Consequently, \[ i_S=\frac{0}{3\times10^{-3}}=0 \] A continuing wire current of \(+1\,\mu\text{A}\) would cause the plate charge to change in the ideal model. Therefore, S is also inconsistent. The consistency test requires agreement in both current magnitude and sign. Experiments P and Q satisfy this condition, while R and S do not. Hence, only P and Q are physically consistent with the given ideal-capacitor assumptions.

Explanation: \(\textbf{Initial capacitor charge:}\) The capacitance is \[ C=2\,\mu\text{F}=2\times10^{-6}\,\text{F} \] The initial voltage is \[ V_0=1\,\text{V} \] Therefore, \[ Q_0=CV_0 \] \[ Q_0=(2\times10^{-6})(1) \] \[ Q_0=2\,\mu\text{C} \] \(\textbf{Charge transferred during the first interval:}\) The charge transferred equals the signed area under the current–time graph. From \(t=0\) to \(2\,\text{s}\), the graph forms a triangle. Thus, \[ \Delta Q_1=\frac12\times2\times6\,\mu\text{C} \] \[ \Delta Q_1=6\,\mu\text{C} \] \(\textbf{Charge transferred during the second interval:}\) From \(t=2\) to \(6\,\text{s}\), the current changes linearly from \(+6\,\mu\text{A}\) to \(-2\,\mu\text{A}\). The signed area of this straight-line segment can be obtained using average current: \[ \Delta Q_2=\frac{I_{\text{initial}}+I_{\text{final}}}{2}\Delta t \] \[ \Delta Q_2=\frac{6+(-2)}{2}\times4\,\mu\text{C} \] \[ \Delta Q_2=8\,\mu\text{C} \] The signed area is positive because the positive-current contribution exceeds the negative-current contribution over this interval. \(\textbf{Final capacitor voltage:}\) The net accumulated charge is \[ Q_6=Q_0+\Delta Q_1+\Delta Q_2 \] \[ Q_6=2+6+8 \] \[ Q_6=16\,\mu\text{C} \] Using \(Q=CV\), \[ V_6=\frac{Q_6}{C} \] \[ V_6=\frac{16\,\mu\text{C}}{2\,\mu\text{F}} \] \[ V_6=8\,\text{V} \] \(\textbf{Instantaneous current at five seconds:}\) The slope of the second segment is \[ \frac{\Delta I}{\Delta t}=\frac{-2-6}{6-2} \] \[ \frac{\Delta I}{\Delta t}=-2\,\mu\text{A s}^{-1} \] At \(t=5\,\text{s}\), three seconds have elapsed since the beginning of this segment. Therefore, \[ I(5)=6+(-2)(5-2) \] \[ I(5)=6-6=0\,\mu\text{A} \] The instantaneous displacement current through the full capacitor gap equals this wire current. Thus, \[ i_d(5\,\text{s})=0 \] \(\textbf{Physical check:}\) The accumulated charge depends on the area under the current graph, while instantaneous displacement current depends on the graph's current value at the selected time. At \(t=5\,\text{s}\), the current is momentarily zero even though the capacitor has accumulated charge and possesses a nonzero electric field. \(\textbf{Final answer:}\) \[ \boxed{\left(V(6\,\text{s}),i_d(5\,\text{s})\right)=(8\,\text{V},0)} \]