Current Electricity MCQs with Answers – Part 1 (NEET Physics)

Explanation: Electric current is the rate at which net electric charge crosses a specified cross-section of a conductor. The conventional direction of current is defined as the direction in which positive charge would move. In a metallic conductor, the mobile charge carriers are predominantly free electrons. An electron carries negative charge and therefore experiences an electric force opposite to the applied electric field. Consequently, the average directed motion, or drift velocity, of electrons is towards the left. Because the moving charges are negative, their contribution to conventional current is towards the right. The electrons also possess random thermal velocities, which do not disappear when an electric field is applied. In the absence of an applied field, their random motion produces no net macroscopic current in thermal equilibrium. The applied field produces a small non-zero average drift velocity superimposed on this random motion. Thus, electron drift and conventional current have opposite directions in a metallic conductor.

Explanation: \(\textbf{Given charge transport:}\) Positive ions transport \(2\,\mu\text{C}\) per second towards the right. Negative ions transport \(3\,\mu\text{C}\) per second towards the left. Conventional current follows positive-charge motion but is opposite to negative-charge motion. Therefore, the positive ions produce a rightward conventional current. The leftward-moving negative ions also produce a rightward conventional current. \(\textbf{Current contributions:}\) \[ I_+=\frac{2\,\mu\text{C}}{1\,\text{s}}=2\,\mu\text{A} \] \[ I_-=\frac{3\,\mu\text{C}}{1\,\text{s}}=3\,\mu\text{A} \] Since both contributions are directed towards the right, their magnitudes must be added. \[ I_{\text{net}}=I_++I_- \] \[ I_{\text{net}}=2+3=5\,\mu\text{A} \] The opposite directions of ionic motion do not imply opposing conventional currents, because the ions have opposite charge signs. If the negative-ion contribution were subtracted merely because those ions move leftward, the sign of their charge would have been ignored. \(\textbf{Final answer:}\) \[ \boxed{I_{\text{net}}=5\,\mu\text{A}\text{ towards the right}} \]

Explanation: \(\textbf{Graph interpretation:}\) Charge transported during a time interval is obtained from the signed area under the current–time graph. For each constant-current interval, \[ \Delta Q=I\Delta t \] The sign of the current determines the direction in which charge is transported. \(\textbf{First interval — }0\textbf{ to }2\,\textbf{s:}\) \[ Q_1=(+4)(2)=+8\,\text{C} \] \(\textbf{Second interval — }2\textbf{ to }5\,\textbf{s:}\) \[ Q_2=(-1)(3)=-3\,\text{C} \] \(\textbf{Third interval — }5\textbf{ to }6\,\textbf{s:}\) \[ Q_3=(+2)(1)=+2\,\text{C} \] \(\textbf{Net signed charge:}\) \[ Q_{\text{net}}=Q_1+Q_2+Q_3 \] \[ Q_{\text{net}}=8-3+2=7\,\text{C} \] For the total charge transported without directional cancellation, the magnitudes of the three contributions must be added. \[ Q_{\text{total}}=|Q_1|+|Q_2|+|Q_3| \] \[ Q_{\text{total}}=8+3+2=13\,\text{C} \] The negative area reduces the net signed transfer but still contributes to the total amount of charge transported. The net result is positive, indicating a greater overall charge transfer towards the right. \(\textbf{Final answer:}\) \[ \boxed{Q_{\text{net}}=7\,\text{C},\quad Q_{\text{total}}=13\,\text{C}} \]

Explanation: \(\textbf{Given:}\) \[ Q(t)=6t-t^2 \] The function represents accumulated signed charge, not instantaneous current. \(\textbf{Current from charge variation:}\) For a time-dependent charge transfer, \[ I(t)=\frac{dQ}{dt} \] Differentiating, \[ I(t)=\frac{d}{dt}(6t-t^2) \] \[ I(t)=6-2t \] A reversal occurs when current changes sign. The current becomes zero at the transition: \[ 6-2t=0 \] \[ t=3\,\text{s} \] For \(t\lt 3\,\text{s}\), current is positive and charge moves towards the right. For \(t\gt 3\,\text{s}\), current is negative and net transport is towards the left. \(\textbf{Charge transported before reversal:}\) \[ Q(3)=6(3)-(3)^2 \] \[ Q(3)=18-9=9\,\text{C} \] Also, \[ Q(0)=0 \] Hence the rightward charge transport before reversal is \[ \Delta Q=Q(3)-Q(0)=9\,\text{C} \] At \(t=6\,\text{s}\), \(Q(6)=0\), because the subsequent reverse current cancels the earlier signed transfer. This cancellation does not mean that no charge crossed the section during the interval. \(\textbf{Final answer:}\) \[ \boxed{t=3\,\text{s},\quad Q_{\text{rightward}}=9\,\text{C}} \]

Explanation: An electron in an applied electric field experiences an electric force. Because an electron has negative charge, this force is opposite to the electric field. Between successive collisions, the electron acquires additional directed velocity due to the electric force. However, electrons do not accelerate indefinitely through the metal. They repeatedly undergo scattering events, which interrupt their directed acceleration. The average time characterising this scattering process is called the relaxation time, \(\tau\). In the elementary free-electron model, the magnitude of average drift velocity is \[ v_d=\frac{eE\tau}{m}. \] For a fixed electric field and fixed relaxation time, this relation gives a fixed average drift speed. Individual electrons can still possess rapidly changing instantaneous velocities. The constant drift velocity refers to an average over the motion of many electrons, not the uninterrupted motion of a single electron. Thus, the Assertion is correct. The Reason is also correct because collisions explain why continuous electrical acceleration does not produce an indefinitely increasing average drift speed. Therefore, the Reason correctly explains the Assertion.

Explanation: \(\textbf{Given:}\) \[ E=0.50\,\text{V m}^{-1} \] \[ \tau=1.0\times10^{-14}\,\text{s} \] \[ \frac{e}{m}=1.76\times10^{11}\,\text{C kg}^{-1} \] \(\textbf{Drift-velocity relation:}\) In the free-electron model, the magnitude of drift velocity is \[ v_d=\frac{eE\tau}{m} \] Equivalently, \[ v_d=\left(\frac{e}{m}\right)E\tau \] \(\textbf{Substitution:}\) \[ v_d=(1.76\times10^{11})(0.50)(1.0\times10^{-14}) \] \[ v_d=0.88\times10^{-3}\,\text{m s}^{-1} \] Converting to millimetres per second, \[ v_d=0.88\,\text{mm s}^{-1} \] \(\textbf{Direction of drift:}\) The applied electric field points towards the right. The electric force on an electron acts opposite to the electric field. Therefore, the electron drift velocity points towards the left. Conventional current would be towards the right. The small magnitude is consistent with the fact that drift is only a weak directed component of electron motion. \(\textbf{Final answer:}\) \[ \boxed{v_d=0.88\,\text{mm s}^{-1}\text{ towards the left}} \]

Explanation: \(\textbf{Initial relations:}\) The magnitude of acceleration of an electron between collisions is \[ a=\frac{eE}{m} \] The magnitude of average drift velocity is \[ v_d=\frac{eE\tau}{m} \] Although both quantities depend on the electric field, only the drift velocity also depends on relaxation time. \(\textbf{Changed conditions:}\) \[ E'=4E \] \[ \tau'=\frac{\tau}{2} \] \(\textbf{Acceleration comparison:}\) \[ a'=\frac{eE'}{m} \] \[ a'=\frac{e(4E)}{m}=4a \] The change in relaxation time does not alter the instantaneous electric force or acceleration between collisions. \(\textbf{Drift-velocity comparison:}\) \[ v_d'=\frac{eE'\tau'}{m} \] \[ v_d'=\frac{e(4E)(\tau/2)}{m} \] \[ v_d'=2\frac{eE\tau}{m}=2v_d \] The fourfold increase in the field enhances the drift response, but the shorter relaxation time partly offsets this increase. Thus, acceleration becomes four times its original value, whereas drift speed becomes twice its original value. \(\textbf{Final answer:}\) \[ \boxed{a'=4a,\quad v_d'=2v_d} \]

Explanation: \(\textbf{Initial observations:}\) \[ v_d=0.20\,\text{mm s}^{-1} \] \[ E=100\,\text{V m}^{-1} \] Electron mobility magnitude is defined by \[ \mu=\frac{v_d}{E} \] \(\textbf{Unit conversion:}\) \[ 0.20\,\text{mm s}^{-1} =2.0\times10^{-4}\,\text{m s}^{-1} \] \(\textbf{Mobility calculation:}\) \[ \mu=\frac{2.0\times10^{-4}}{100} \] \[ \mu=2.0\times10^{-6}\,\text{m}^2\text{V}^{-1}\text{s}^{-1} \] The SI unit follows because drift velocity has units of \(\text{m s}^{-1}\), whereas electric field has units of \(\text{V m}^{-1}\). With unchanged temperature and mobility, the drift-speed magnitude is proportional to the electric-field magnitude. \(\textbf{New field:}\) \[ E'=3(100)=300\,\text{V m}^{-1} \] \(\textbf{New drift speed:}\) \[ v_d'=\mu E' \] \[ v_d'=(2.0\times10^{-6})(300) \] \[ v_d'=6.0\times10^{-4}\,\text{m s}^{-1} \] \[ v_d'=0.60\,\text{mm s}^{-1} \] Reversing the field reverses the electron drift direction, but does not change the positive magnitude of mobility. \(\textbf{Final answer:}\) \[ \boxed{\mu=2.0\times10^{-6}\,\text{m}^2\text{V}^{-1}\text{s}^{-1},\quad v_d'=0.60\,\text{mm s}^{-1}} \]

Explanation: \(\textbf{Interpretation of graph slopes:}\) For a conductor with constant electron mobility, \[ v_d=\mu E \] Thus, the slope of the drift-speed versus electric-field graph equals the mobility magnitude. The graph has a zero intercept, so each sample has a constant drift-speed-to-field ratio. \(\textbf{Sample P:}\) At \(E_P=100\,\text{V m}^{-1}\), \[ v_P=0.40\,\text{mm s}^{-1} \] Hence, \[ \mu_P=\frac{0.40}{100} =0.0040\,\frac{\text{mm s}^{-1}}{\text{V m}^{-1}} \] At the new field of \(150\,\text{V m}^{-1}\), \[ v_P'=\mu_P E_P' \] \[ v_P'=(0.0040)(150) \] \[ v_P'=0.60\,\text{mm s}^{-1} \] \(\textbf{Sample Q:}\) The slope of line Q is \[ \mu_Q=\frac{0.25}{100} =0.0025\,\frac{\text{mm s}^{-1}}{\text{V m}^{-1}} \] To obtain the same drift speed as sample P, \[ v_Q'=v_P'=0.60\,\text{mm s}^{-1} \] Therefore, \[ E_Q'=\frac{v_Q'}{\mu_Q} \] \[ E_Q'=\frac{0.60}{0.0025} \] \[ E_Q'=240\,\text{V m}^{-1} \] Sample Q has the smaller mobility, so it requires a larger electric field to produce the same drift speed. The result is consistent with the smaller slope of line Q. \(\textbf{Final answer:}\) \[ \boxed{E_Q'=240\,\text{V m}^{-1}} \]

Explanation: \(\textbf{Governing relation:}\) For electrons of equal charge magnitude and mass, \[ v_d=\frac{eE\tau}{m} \] Therefore, drift speed is proportional to the product of electric field and relaxation time. \[ v_d\propto E\tau \] Let the initial field and relaxation time in sample Y be \(E\) and \(\tau\). \(\textbf{Initial conditions:}\) For sample X, \[ E_X=3E,\qquad \tau_X=2\tau \] For sample Y, \[ E_Y=E,\qquad \tau_Y=\tau \] The initial drift-speed ratio is \[ \frac{v_X}{v_Y} =\frac{E_X\tau_X}{E_Y\tau_Y} \] \[ \frac{v_X}{v_Y} =\frac{(3E)(2\tau)}{E\tau}=6 \] \(\textbf{Changed relaxation times:}\) In sample X, \[ \tau_X'=\frac{2\tau}{2}=\tau \] In sample Y, \[ \tau_Y'=2\tau \] \(\textbf{Changed electric fields:}\) \[ E_X'=E,\qquad E_Y'=2E \] \(\textbf{New drift-speed ratio:}\) \[ \frac{v_X'}{v_Y'} =\frac{E_X'\tau_X'}{E_Y'\tau_Y'} \] \[ \frac{v_X'}{v_Y'} =\frac{E\tau}{(2E)(2\tau)} \] \[ \frac{v_X'}{v_Y'}=\frac14 \] Initially, the drift in X is faster because it has both a larger field and a longer relaxation time. After the changes, the field–relaxation-time product in Y becomes four times that in X. The reversal of which sample has the greater drift speed is therefore physically consistent. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{v_X}{v_Y}\right)_{\!i}=6,\quad \left(\frac{v_X}{v_Y}\right)_{\!f}=\frac14} \]

Explanation: \(\textbf{Given:}\) \[ I=1.28\,\text{A},\quad n=8.0\times10^{28}\,\text{m}^{-3} \] \[ A=1.0\,\text{mm}^2=1.0\times10^{-6}\,\text{m}^2 \] \[ e=1.6\times10^{-19}\,\text{C},\quad t=5.0\,\text{s} \] \(\textbf{Stage 1 — Electron drift speed:}\) For steady electron conduction in a uniform wire, \[ I=neAv_d \] Here \(v_d\) represents the magnitude of electron drift velocity. Rearranging, \[ v_d=\frac{I}{neA} \] Substituting, \[ v_d=\frac{1.28}{(8.0\times10^{28})(1.6\times10^{-19})(1.0\times10^{-6})} \] The denominator simplifies to \[ neA=1.28\times10^4 \] Therefore, \[ v_d=\frac{1.28}{1.28\times10^4} \] \[ v_d=1.0\times10^{-4}\,\text{m s}^{-1} \] \(\textbf{Stage 2 — Average drift displacement:}\) For steady drift over time \(t\), \[ s=v_dt \] \[ s=(1.0\times10^{-4})(5.0) \] \[ s=5.0\times10^{-4}\,\text{m} \] \[ s=0.50\,\text{mm} \] \(\textbf{Stage 3 — Number of electrons crossing:}\) The magnitude of charge transported in \(5.0\,\text{s}\) is \[ Q=It \] \[ Q=(1.28)(5.0)=6.4\,\text{C} \] Each electron carries charge of magnitude \(e\), so \[ N=\frac{Q}{e} \] \[ N=\frac{6.4}{1.6\times10^{-19}} \] \[ N=4.0\times10^{19} \] A large number of electrons can cross a fixed cross-section even though the average drift displacement of each electron is very small. The distinction arises because the metal contains an enormous number of mobile charge carriers. \(\textbf{Final answer:}\) \[ \boxed{s=0.50\,\text{mm},\quad N=4.0\times10^{19}} \]

Explanation: \(\textbf{Current continuity:}\) The two portions are connected in series, with no branching or accumulation of charge in steady state. Therefore, the same current \(I\) flows through P and Q. The electron-current magnitude relation is \[ I=neAv_d \] For the same metal at a fixed temperature, \(n\) and mobility \(\mu\) are the same in both portions. Since \[ v_d=\mu E, \] we can combine these relations: \[ I=neA\mu E \] \(\textbf{Electric field in the two portions:}\) Rearranging, \[ E=\frac{I}{ne\mu A} \] Thus, for the same current and material, \[ E\propto\frac1A \] For the wider portion P, \[ E_P=\frac{I}{ne\mu(3A)} \] For the narrower portion Q, \[ E_Q=\frac{I}{ne\mu A} \] Hence, \[ \frac{E_Q}{E_P}=3 \] \(\textbf{Potential differences:}\) For each uniform portion, the magnitude of potential difference is \[ V=EL \] Since both lengths are \(L\), \[ V_P=E_PL \] \[ V_Q=E_QL=3E_PL=3V_P \] Therefore, the total potential difference is \[ V_{\text{total}}=V_P+V_Q \] \[ V_{\text{total}}=V_P+3V_P=4V_P \] The required fraction is \[ \frac{V_Q}{V_{\text{total}}} =\frac{3V_P}{4V_P} =\frac34 \] The narrower portion requires a stronger electric field to sustain the same current through its smaller cross-sectional area. Consequently, it accounts for a larger fraction of the total potential difference. \(\textbf{Final answer:}\) \[ \boxed{\frac{V_Q}{V_{\text{total}}}=\frac34} \]

Explanation: \(\textbf{Initial conditions:}\) Let the original wire length and area be \(L\) and \(A\). The original electric-field magnitude is \[ E=\frac VL \] Electron drift speed is related to the field by \[ v_d=\mu E \] The current magnitude is \[ I=neAv_d \] \(\textbf{Stage 1 — Geometrical change:}\) The wire is stretched without changing its volume. Therefore, \[ AL=A'L' \] Since \[ L'=2L, \] we obtain \[ AL=A'(2L) \] \[ A'=\frac A2 \] \(\textbf{Stage 2 — Electric-field change:}\) The applied potential difference is unchanged, but the new length is \(2L\). Hence, \[ E'=\frac{V}{L'} \] \[ E'=\frac{V}{2L}=\frac E2 \] \(\textbf{Stage 3 — Drift-speed change:}\) The electron mobility remains unchanged. Therefore, \[ v_d'=\mu E' \] \[ v_d'=\mu\left(\frac E2\right)=\frac{v_d}{2} \] \(\textbf{Stage 4 — Current change:}\) Using the microscopic current relation, \[ I'=neA'v_d' \] Substituting the new area and drift speed, \[ I'=ne\left(\frac A2\right)\left(\frac{v_d}{2}\right) \] \[ I'=\frac14neAv_d \] \[ I'=\frac I4 \] The current falls because both the cross-sectional area and drift speed become smaller. Reducing only the drift speed would incorrectly predict a current of \(I/2\), because it would ignore the geometrical change. \(\textbf{Final answer:}\) \[ \boxed{I'=\frac I4,\quad v_d'=\frac{v_d}{2}} \]

Explanation: \(\textbf{Microscopic current-density relation:}\) For electron conduction, the magnitude of current density is \[ j=nev_d \] Using \[ v_d=\mu E, \] we obtain \[ j=ne\mu E \] Current density depends on carrier number density, electron mobility and electric-field magnitude. The total current additionally depends on the cross-sectional area. \(\textbf{Stage 1 — Current-density comparison:}\) For sample X, \[ j_X=n_0e(2\mu_0)E_0 \] \[ j_X=2n_0e\mu_0E_0 \] For sample Y, \[ j_Y=(4n_0)e(\mu_0)E_0 \] \[ j_Y=4n_0e\mu_0E_0 \] Taking the ratio, \[ \frac{j_Y}{j_X} =\frac{4n_0e\mu_0E_0}{2n_0e\mu_0E_0} \] \[ \frac{j_Y}{j_X}=2 \] \(\textbf{Stage 2 — Total-current comparison:}\) For uniform current distribution, \[ I=jA \] Therefore, \[ \frac{I_Y}{I_X} =\frac{j_YA_Y}{j_XA_X} \] Substituting the area ratio, \[ \frac{I_Y}{I_X} =\left(\frac{j_Y}{j_X}\right) \left(\frac{A_Y}{A_X}\right) \] \[ \frac{I_Y}{I_X} =2\left(\frac12\right) \] \[ \frac{I_Y}{I_X}=1 \] Although sample Y has twice the current density, its cross-sectional area is half that of sample X. These two effects compensate exactly, producing equal total currents. Confusing current density with total current would incorrectly give the same ratio for both quantities. \(\textbf{Final answer:}\) \[ \boxed{\frac{j_Y}{j_X}=2,\quad\frac{I_Y}{I_X}=1} \]

Explanation: \(\textbf{Given conditions:}\) \[ L_Q=3L_P \] \[ A_Q=\frac{A_P}{2} \] The applied voltage, electron mobility and measured current are the same for both wires. \(\textbf{Microscopic current relation:}\) The magnitude of current in a uniform conductor is \[ I=neAv_d \] With electron mobility \(\mu\), \[ v_d=\mu E \] For a uniform wire carrying steady current, \[ E=\frac VL \] Combining these equations, \[ I=neA\mu\frac VL \] Thus, \[ I=\frac{ne\mu AV}{L} \] \(\textbf{Equal-current condition:}\) Since the currents are equal, \[ I_P=I_Q \] Therefore, \[ \frac{n_Pe\mu A_PV}{L_P} = \frac{n_Qe\mu A_QV}{L_Q} \] Cancelling the common factors \(e,\mu,V\), \[ \frac{n_PA_P}{L_P} = \frac{n_QA_Q}{L_Q} \] Rearranging to determine the unknown density ratio, \[ \frac{n_Q}{n_P} = \frac{A_P}{A_Q}\frac{L_Q}{L_P} \] \(\textbf{Substitution:}\) \[ \frac{n_Q}{n_P} = \left(\frac{A_P}{A_P/2}\right) \left(\frac{3L_P}{L_P}\right) \] \[ \frac{n_Q}{n_P}=2\times3 \] \[ \frac{n_Q}{n_P}=6 \] Wire Q is both longer and thinner, so the same applied voltage produces a smaller electric field and its smaller area further reduces the charge transport for a given carrier density. To sustain the same current with equal mobility, Q must therefore have a greater free-electron number density. \(\textbf{Final answer:}\) \[ \boxed{\frac{n_Q}{n_P}=6} \]

Explanation: \(\textbf{Steady-current condition:}\) The two sections are joined in series, so the same current flows through both. For electron conduction, \[ I=neAv_d \] The free-electron number densities and cross-sectional areas are different in the two sections. Therefore, current continuity must be applied using the product \(nAv_d\), not drift speed alone. \(\textbf{Stage 1 — Drift-speed comparison:}\) For section P, \[ I_P=neAv_{d,P} \] For section Q, \[ I_Q=(2n)e\left(\frac A2\right)v_{d,Q} \] Simplifying, \[ I_Q=neAv_{d,Q} \] Since \(I_P=I_Q\), \[ neAv_{d,P}=neAv_{d,Q} \] Therefore, \[ v_{d,P}=v_{d,Q} \] Hence, \[ \frac{v_{d,Q}}{v_{d,P}}=1 \] The doubling of carrier density exactly compensates for the halving of cross-sectional area. \(\textbf{Stage 2 — Electric-field comparison:}\) The mobility relation is \[ v_d=\mu E \] Therefore, \[ E=\frac{v_d}{\mu} \] For section P, \[ E_P=\frac{v_{d,P}}{2\mu} \] For section Q, \[ E_Q=\frac{v_{d,Q}}{\mu} \] Taking the ratio, \[ \frac{E_Q}{E_P} = \frac{v_{d,Q}/\mu}{v_{d,P}/(2\mu)} \] \[ \frac{E_Q}{E_P} = 2\frac{v_{d,Q}}{v_{d,P}} \] Using the previously obtained drift-speed ratio, \[ \frac{E_Q}{E_P}=2(1)=2 \] Thus, the average drift speeds are equal even though the electric fields are different. Section Q requires twice the electric field because its electron mobility is half that in section P. This demonstrates that equal steady current does not necessarily imply equal drift velocity or equal electric field in different conductor sections. \(\textbf{Final answer:}\) \[ \boxed{\frac{v_{d,Q}}{v_{d,P}}=1,\quad\frac{E_Q}{E_P}=2} \]

Explanation: \(\textbf{Charge conservation:}\) The rate of change of net charge stored in a region is the difference between the total entering and leaving conventional currents. \[ \frac{dQ}{dt}=I_{\text{in}}-I_{\text{out}} \] A positive difference produces accumulation of positive net charge, while a negative difference reduces the stored charge. \(\textbf{First interval — }0\textbf{ to }2\,\textbf{s:}\) \[ I_{\text{net},1}=5-2=+3\,\text{mA} \] The net charge accumulated during this interval is \[ \Delta Q_1=I_{\text{net},1}\Delta t_1 \] \[ \Delta Q_1=(3)(2)=+6\,\text{mC} \] Therefore, at \(t=2\,\text{s}\), the region has an excess positive charge of \(6\,\text{mC}\). \(\textbf{Second interval — Charge removal:}\) \[ I_{\text{net},2}=1-4=-3\,\text{mA} \] The region now loses net positive charge at a rate of \(3\,\text{mC s}^{-1}\). The time required to remove the accumulated \(6\,\text{mC}\) is \[ \Delta t=\frac{6\,\text{mC}}{3\,\text{mA}} \] \[ \Delta t=2\,\text{s} \] Hence, the region becomes neutral again at \[ t=2+2=4\,\text{s} \] \(\textbf{Charge remaining at }5\,\textbf{s:}\) The second interval continues for a total of \(3\,\text{s}\). \[ \Delta Q_2=(-3)(3)=-9\,\text{mC} \] The final excess charge is \[ Q_{\text{final}}=0+6-9 \] \[ Q_{\text{final}}=-3\,\text{mC} \] The negative sign indicates an excess of negative charge relative to the initially neutral state. This result follows from charge conservation; the current through different leads need not be equal when the charge stored in the region is changing. \(\textbf{Final answer:}\) \[ \boxed{t_{\text{neutral}}=4\,\text{s},\quad Q(5\,\text{s})=-3\,\text{mC}} \]

Explanation: \(\textbf{Current and transported charge:}\) The instantaneous current is given by \[ I(t)=a+bt \] The charge transported over a time interval is the area under the current–time graph. Therefore, \[ \Delta Q=\int_{t_1}^{t_2}(a+bt)\,dt \] \[ \Delta Q=a(t_2-t_1)+\frac b2(t_2^2-t_1^2) \] \(\textbf{First measurement:}\) For the interval \(0\) to \(2\,\text{s}\), \[ 8=a(2-0)+\frac b2(2^2-0^2) \] \[ 8=2a+2b \] \[ a+b=4 \] Here, \(a\) is measured in amperes and \(b\) in amperes per second, with time substituted in seconds. \(\textbf{Second measurement:}\) For the interval \(2\) to \(4\,\text{s}\), \[ 16=a(4-2)+\frac b2(4^2-2^2) \] \[ 16=2a+6b \] \[ a+3b=8 \] \(\textbf{Determine the current variation:}\) Subtracting the first equation from the second, \[ (a+3b)-(a+b)=8-4 \] \[ 2b=4 \] \[ b=2\,\text{A s}^{-1} \] Substituting into the first relation, \[ a=4-2=2\,\text{A} \] Thus, \[ I(t)=2+2t \] when \(I\) and \(t\) are expressed numerically in amperes and seconds. \(\textbf{Instantaneous current at }3\,\textbf{s:}\) \[ I(3)=2+2(3) \] \[ I(3)=8\,\text{A} \] \(\textbf{Fraction of total charge:}\) The total charge transported during \(4\,\text{s}\) is \[ Q_{\text{total}}=8+16=24\,\text{C} \] Therefore, \[ f=\frac{Q_{\text{first }2\,\text{s}}}{Q_{\text{total}}} \] \[ f=\frac8{24}=\frac13 \] The second interval carries twice as much charge as the first because the current increases with time. Using either interval's average current as the instantaneous current at \(t=3\,\text{s}\) without considering the linear variation would be incorrect. \(\textbf{Final answer:}\) \[ \boxed{I(3\,\text{s})=8\,\text{A},\quad f=\frac13} \]

Explanation: \(\textbf{Current through an inclined surface:}\) Electric current through a surface depends on the component of current density normal to that surface. For uniform current density, \[ I=\vec j\cdot\vec A \] Therefore, \[ I=jA\cos\theta \] Here, \(\theta\) is the angle between current density and the chosen area normal. It is not the angle between the current direction and the plane itself. \(\textbf{Given quantities:}\) \[ j=2.0\times10^6\,\text{A m}^{-2} \] \[ A=4.0\,\text{mm}^2=4.0\times10^{-6}\,\text{m}^2 \] \[ \theta=60^\circ \] \(\textbf{Calculate the current:}\) \[ I=(2.0\times10^6)(4.0\times10^{-6})\cos60^\circ \] \[ I=(2.0)(4.0)\left(\frac12\right) \] \[ I=4.0\,\text{A} \] \(\textbf{Electron crossing rate:}\) The magnitude of charge transported per second is \(I\). If \(N\) electrons cross per second, then \[ I=Ne \] Hence, \[ N=\frac Ie \] \[ N=\frac{4.0}{1.6\times10^{-19}} \] \[ N=2.5\times10^{19}\,\text{s}^{-1} \] The electrons themselves drift opposite to conventional current, but the number crossing the surface is positive. The oblique section has a larger actual area than the perpendicular cross-section, yet the total current is unchanged. The factor \(\cos60^\circ\) accounts for this geometrical difference. Omitting the projection factor would incorrectly predict twice the actual current. \(\textbf{Final answer:}\) \[ \boxed{I=4.0\,\text{A},\quad N=2.5\times10^{19}\,\text{s}^{-1}} \]

Explanation: \(\textbf{Microscopic conduction relation:}\) For electrons in the elementary relaxation-time model, \[ v_d=\frac{eE\tau}{m} \] The current magnitude is \[ I=neAv_d \] Combining the two relations, \[ I=\frac{ne^2A\tau E}{m} \] For a uniform wire, \[ E=\frac VL \] Therefore, \[ I=\frac{ne^2A\tau V}{mL} \] Since \(n,e,A,m,L\) remain unchanged, \[ I\propto\tau V \] \(\textbf{Experiment P — Fixed potential difference:}\) The new relaxation time is \[ \tau'=\frac{\tau}{3} \] The applied voltage remains \(V\). Thus, \[ \frac{I_P}{I} =\frac{\tau'V}{\tau V} \] \[ \frac{I_P}{I}=\frac13 \] \[ I_P=\frac I3 \] The reduced relaxation time produces a smaller drift velocity at the same electric field. \(\textbf{Experiment Q — Fixed current:}\) The current must remain equal to \(I\). Therefore, \[ \tau'V_Q=\tau V \] Substituting \(\tau'=\tau/3\), \[ \frac{\tau}{3}V_Q=\tau V \] Cancelling \(\tau\), \[ V_Q=3V \] In experiment Q, the stronger applied electric field compensates for the shorter relaxation time. Consequently, the electron drift speed and current remain at their original values. The distinction is that voltage is fixed in experiment P, whereas current is fixed in experiment Q. \(\textbf{Final answer:}\) \[ \boxed{I_P=\frac I3,\quad V_Q=3V} \]