Electrostatics MCQs with Answers – Part 8 (NEET Physics)

Explanation: \(\textbf{Relate voltage to plate separation:}\) For an isolated vacuum parallel-plate capacitor, \[ C=\frac{\varepsilon_0A}{d} \] Because the free charge remains fixed, \[ V=\frac QC \] Therefore, \[ V=\frac{Qd}{\varepsilon_0A} \] \(\textbf{Interpret the voltage-graph slope:}\) The coefficient of \(d\) is the measured slope: \[ s_V=\frac{Q}{\varepsilon_0A} \] This is also the magnitude of the uniform electric field between the plates. \(\textbf{Relate stored energy to separation:}\) At fixed free charge, \[ U=\frac{Q^2}{2C} \] Substituting the vacuum capacitance, \[ U=\frac{Q^2d}{2\varepsilon_0A} \] \(\textbf{Interpret the energy-graph slope:}\) Therefore, \[ s_U=\frac{Q^2}{2\varepsilon_0A} \] \(\textbf{Eliminate the unknown area:}\) From the voltage slope, \[ \frac{Q}{\varepsilon_0A}=s_V \] Thus, \[ s_U=\frac Q2s_V \] Solving, \[ Q=\frac{2s_U}{s_V} \] \(\textbf{Determine the plate area:}\) Using \[ s_V=\frac{Q}{\varepsilon_0A} \] we obtain \[ A=\frac{Q}{\varepsilon_0s_V} \] Substituting the inferred charge, \[ A=\frac{2s_U}{\varepsilon_0s_V^2} \] \(\textbf{Relate energy gradient to attractive force:}\) For an isolated ideal capacitor at fixed charge and plate area, \[ U(d)=\frac{Q^2d}{2\varepsilon_0A} \] The increase in energy per unit increase in plate separation equals the magnitude of the opposing attractive force: \[ F=\frac{dU}{dd} \] Therefore, \[ F=s_U \] \(\textbf{Check dimensions:}\) The slope \(s_V\) has units \(\mathrm{V\,m^{-1}}\). The slope \(s_U\) has units \(\mathrm{J\,m^{-1}}=\mathrm N\). Consequently, \(2s_U/s_V\) has units of charge, and the inferred area has units of \(\mathrm{m^2}\). \(\textbf{Physical interpretation:}\) The two graphs contain complementary information about the same fixed-charge capacitor. The voltage slope measures field magnitude, while the energy slope measures plate-attraction magnitude. Together, they determine both the free charge and plate area without directly measuring either quantity. \(\textbf{Final answer:}\) \[ \boxed{Q=\frac{2s_U}{s_V},\qquad A=\frac{2s_U}{\varepsilon_0s_V^2},\qquad F=s_U} \]

Explanation: \(\textbf{Determine the initial series state:}\) Initially, the two capacitances are \(C\) and \(2C\). Their equivalent capacitance is \[ C_{\mathrm{eq},i}=\frac{C(2C)}{C+2C} \] \[ C_{\mathrm{eq},i}=\frac{2C}{3} \] The battery therefore supplies initial series charge \[ Q_i=\frac{2CV_0}{3} \] \(\textbf{Determine the effect of the conducting slab:}\) The electrostatic field vanishes inside the inserted conductor. The effective vacuum thickness of the modified capacitor becomes \[ d_{\mathrm{eff}}=d-\frac d2 \] \[ d_{\mathrm{eff}}=\frac d2 \] Its new capacitance is \[ C_{\mathrm{modified}}=\frac{\varepsilon_0A}{d/2} \] Since the original capacitance was \(\varepsilon_0A/d=C\), \[ C_{\mathrm{modified}}=2C \] \(\textbf{Determine the final series equivalent:}\) Both capacitors now have capacitance \(2C\): \[ C_{\mathrm{eq},f}=\frac{(2C)(2C)}{2C+2C} \] \[ C_{\mathrm{eq},f}=C \] \(\textbf{Calculate additional battery charge:}\) The final series charge is \[ Q_f=CV_0 \] Therefore, \[ \Delta Q=Q_f-Q_i \] \[ \Delta Q=CV_0-\frac{2CV_0}{3} \] \[ \Delta Q=\frac{CV_0}{3} \] \(\textbf{Determine the modified capacitor voltage:}\) Both capacitors have the same final capacitance and equal charge. Therefore, each receives half the battery voltage: \[ V_{\mathrm{modified}}=\frac{V_0}{2} \] \(\textbf{Calculate the increase in stored energy:}\) Initially, \[ U_i=\frac12C_{\mathrm{eq},i}V_0^2 \] \[ U_i=\frac{CV_0^2}{3} \] Finally, \[ U_f=\frac12CV_0^2 \] Hence, \[ \Delta U=U_f-U_i \] \[ \Delta U=\frac{CV_0^2}{2}-\frac{CV_0^2}{3} \] \[ \Delta U=\frac{CV_0^2}{6} \] \(\textbf{Account for battery work:}\) The battery maintains \(V_0\) while supplying additional charge: \[ W_{\mathrm{battery}}=V_0\Delta Q \] \[ W_{\mathrm{battery}}=\frac{CV_0^2}{3} \] \(\textbf{Determine external mechanical work:}\) For the stated quasistatic lossless insertion, \[ \Delta U=W_{\mathrm{battery}}+W_{\mathrm{ext}} \] Therefore, \[ W_{\mathrm{ext}}=\Delta U-W_{\mathrm{battery}} \] \[ W_{\mathrm{ext}}=\frac{CV_0^2}{6}-\frac{CV_0^2}{3} \] \[ W_{\mathrm{ext}}=-\frac{CV_0^2}{6} \] \(\textbf{Physical interpretation:}\) The conducting slab increases the first capacitor's capacitance by reducing its effective field-filled gap. The battery supplies additional free charge as the series equivalent capacitance increases. The negative external work means the electrostatic system tends to draw the slab inward during insertion. \(\textbf{Final answer:}\) \[ \boxed{\Delta Q=\frac{CV_0}{3},\qquad V_{\mathrm{modified}}=\frac{V_0}{2},\qquad W_{\mathrm{ext}}=-\frac{CV_0^2}{6}} \]

Explanation: The Assertion is correct because isolation conserves the total free charge on each conducting plate, not necessarily the charge on every small surface region. Charges within a conducting plate remain mobile and can redistribute as its electrostatic environment changes. The dielectric-covered region has greater capacitance per unit plate area than the vacuum-covered region. Both regions lie between the same two equipotential conducting plates. Therefore, they experience the same potential difference and, under the stated equal-gap idealisation, the same electric-field magnitude. The normal electric displacement is related to the electric field by the permittivity of the material. For the vacuum region, its magnitude is \(\varepsilon_0E\). For the dielectric region, its magnitude is \(K\varepsilon_0E\). Consequently, the free surface charge density facing the dielectric region is greater than that facing the vacuum region. As dielectric area is introduced, the free plate charge redistributes to satisfy this condition while conserving the total. The charge density on the dielectric-facing region can therefore increase relative to the original uniform vacuum distribution. The remaining vacuum-facing region can have a lower free surface charge density. The dielectric also develops bound surface charge through polarisation, but this is not free charge conducted across the insulating material. The Reason correctly identifies the common-voltage condition and differing electric displacements that produce the redistribution. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Calculate source work during Stage P:}\) The initial capacitance is \(C\). During quasistatic charging, \[ V=\frac QC \] The electrical work equals the area under the voltage-versus-charge graph: \[ W_P=\int_0^{Q_0}V\,dQ \] \[ W_P=\frac{Q_0^2}{2C} \] Using \(V_0=Q_0/C\), \[ W_P=\frac{Q_0V_0}{2} \] This is also the stored energy immediately before dielectric insertion. \(\textbf{Track energy during dielectric insertion:}\) During Stage S, charge is fixed at \(Q_0\), while capacitance doubles: \[ C_S=2C \] The energy after insertion is \[ U_S=\frac{Q_0^2}{2(2C)} \] \[ U_S=\frac{Q_0^2}{4C} \] \[ U_S=\frac{Q_0V_0}{4} \] \(\textbf{Determine mechanical work during Stage S:}\) No electrical charge is transported by the source during this fixed-charge stage. Therefore, under the stated lossless conditions, \[ W_{\mathrm{mech}}=U_S-W_P \] \[ W_{\mathrm{mech}}=\frac{Q_0V_0}{4}-\frac{Q_0V_0}{2} \] \[ W_{\mathrm{mech}}=-\frac{Q_0V_0}{4} \] \(\textbf{Calculate source work during Stage R:}\) The capacitance now remains \(2C\). Consequently, \[ V=\frac{Q}{2C} \] The source increases charge from \(Q_0\) to \(2Q_0\). Its electrical work is \[ W_R=\int_{Q_0}^{2Q_0}\frac{Q}{2C}\,dQ \] \[ W_R=\frac{(2Q_0)^2-Q_0^2}{4C} \] \[ W_R=\frac{3Q_0^2}{4C} \] \[ W_R=\frac{3Q_0V_0}{4} \] \(\textbf{Combine the charging-source work:}\) \[ W_{\mathrm{source}}=W_P+W_R \] \[ W_{\mathrm{source}}=\frac{Q_0V_0}{2}+\frac{3Q_0V_0}{4} \] \[ W_{\mathrm{source}}=\frac{5Q_0V_0}{4} \] \(\textbf{Calculate final stored energy:}\) The final charge is \(2Q_0\), and the final capacitance is \(2C\): \[ U_f=\frac{(2Q_0)^2}{2(2C)} \] \[ U_f=\frac{Q_0^2}{C} \] \[ U_f=Q_0V_0 \] \(\textbf{Check the full energy balance:}\) \[ W_{\mathrm{source}}+W_{\mathrm{mech}}=\frac{5Q_0V_0}{4}-\frac{Q_0V_0}{4} \] \[ W_{\mathrm{source}}+W_{\mathrm{mech}}=Q_0V_0=U_f \] \(\textbf{Physical interpretation:}\) The voltage-versus-charge graph has different slopes before and after dielectric insertion. The vertical change at fixed charge represents a capacitance modification, not electrical charge transfer. Electrical charging work and mechanical insertion work must be included separately when accounting for the final stored energy. \(\textbf{Final answer:}\) \[ \boxed{W_{\mathrm{source}}=\frac{5Q_0V_0}{4},\qquad W_{\mathrm{mech}}=-\frac{Q_0V_0}{4},\qquad U_f=Q_0V_0} \]

Explanation: \(\textbf{Determine the initial series state:}\) Two equal capacitors in series have equivalent capacitance \[ C_i=\frac{C^2}{2C} \] \[ C_i=\frac C2 \] The battery initially supplies \[ Q_i=C_iV_0 \] \[ Q_i=\frac{CV_0}{2} \] The initial energy is \[ U_i=\frac12C_iV_0^2 \] \[ U_i=\frac{CV_0^2}{4} \] \(\textbf{Identify the final capacitor topology:}\) The newly connected \(3C\) capacitor lies in parallel with the original first capacitor \(C\). Their equivalent capacitance is \[ C_{AX}=C+3C \] \[ C_{AX}=4C \] This group is in series with the unchanged capacitor \(C\) between X and B. \(\textbf{Determine the final equivalent capacitance:}\) \[ C_f=\frac{(4C)C}{4C+C} \] \[ C_f=\frac{4C}{5} \] \(\textbf{Determine additional battery charge:}\) The battery voltage remains \(V_0\). Thus, \[ Q_f=C_fV_0 \] \[ Q_f=\frac{4CV_0}{5} \] The additional battery charge is \[ \Delta Q=Q_f-Q_i \] \[ \Delta Q=\frac{4CV_0}{5}-\frac{CV_0}{2} \] \[ \Delta Q=\frac{3CV_0}{10} \] \(\textbf{Determine the voltage across A–X:}\) The equivalent capacitor \(4C\) carries series charge \(Q_f\). Therefore, \[ V_{AX}=\frac{Q_f}{4C} \] \[ V_{AX}=\frac{4CV_0/5}{4C} \] \[ V_{AX}=\frac{V_0}{5} \] \(\textbf{Calculate the final stored energy:}\) \[ U_f=\frac12C_fV_0^2 \] \[ U_f=\frac25CV_0^2 \] The increase in stored energy is \[ \Delta U=\frac25CV_0^2-\frac14CV_0^2 \] \[ \Delta U=\frac{3CV_0^2}{20} \] \(\textbf{Determine battery work and dissipation:}\) The battery supplies additional charge \(\Delta Q\) at fixed voltage: \[ W_{\mathrm{battery}}=V_0\Delta Q \] \[ W_{\mathrm{battery}}=\frac{3CV_0^2}{10} \] The dissipated energy is \[ H=W_{\mathrm{battery}}-\Delta U \] \[ H=\frac{3CV_0^2}{10}-\frac{3CV_0^2}{20} \] \[ H=\frac{3CV_0^2}{20} \] \(\textbf{Physical interpretation:}\) Adding a parallel branch increases the capacitance of the A–X group. That modified group remains in series with the second capacitor, so the original series voltage division changes. The new connection draws battery charge and redistributes stored electrostatic energy. The calculated final values do not match any listed option.

Explanation: \(\textbf{Establish the original capacitor state:}\) The initial free charge is \[ Q_i=C_0V_0 \] The original stored energy is \[ U_i=\frac12C_0V_0^2 \] \(\textbf{Determine the capacitance after insertion:}\) Complete filling by dielectric constant \(K=3\) changes the capacitance to \[ C_P=3C_0 \] During Stage P, the battery maintains voltage \(V_0\). Therefore, \[ Q_P=3C_0V_0 \] \(\textbf{Calculate additional battery charge:}\) \[ \Delta Q_P=Q_P-Q_i \] \[ \Delta Q_P=3C_0V_0-C_0V_0 \] \[ \Delta Q_P=2C_0V_0 \] \(\textbf{Calculate energy delivered by the battery:}\) \[ W_{\mathrm{battery}}=V_0\Delta Q_P \] \[ W_{\mathrm{battery}}=2C_0V_0^2 \] The battery is disconnected before Stage R, so it performs no further work. \(\textbf{Determine the stored energy after insertion:}\) \[ U_P=\frac12C_PV_0^2 \] \[ U_P=\frac32C_0V_0^2 \] Therefore, \[ U_P-U_i=C_0V_0^2 \] \(\textbf{Calculate mechanical work during insertion:}\) For a lossless quasistatic operation, \[ U_P-U_i=W_{\mathrm{battery}}+W_P \] Thus, \[ W_P=C_0V_0^2-2C_0V_0^2 \] \[ W_P=-C_0V_0^2 \] The field tends to draw the dielectric inward. \(\textbf{Track isolation before withdrawal:}\) After Stage Q, the free charge remains fixed at \[ Q_R=3C_0V_0 \] Removing the dielectric restores capacitance \(C_0\). \(\textbf{Calculate final stored energy:}\) \[ U_f=\frac{Q_R^2}{2C_0} \] \[ U_f=\frac{(3C_0V_0)^2}{2C_0} \] \[ U_f=\frac92C_0V_0^2 \] Therefore, \[ \frac{U_f}{U_i}=\frac{9C_0V_0^2/2}{C_0V_0^2/2} \] \[ \frac{U_f}{U_i}=9 \] \(\textbf{Calculate mechanical work during withdrawal:}\) No battery is connected during Stage R. Therefore, \[ W_R=U_f-U_P \] \[ W_R=\frac92C_0V_0^2-\frac32C_0V_0^2 \] \[ W_R=3C_0V_0^2 \] \(\textbf{Check the complete energy balance:}\) The net external mechanical work is \[ W_P+W_R=2C_0V_0^2 \] Together with battery work, \[ W_{\mathrm{battery}}+W_P+W_R=4C_0V_0^2 \] This equals the overall increase in stored energy: \[ U_f-U_i=4C_0V_0^2 \] \(\textbf{Physical interpretation:}\) Dielectric insertion while battery-connected increases free plate charge. Disconnecting the battery then preserves this larger charge during withdrawal. Consequently, the final voltage and stored energy can greatly exceed their original values, even though the capacitor returns to its original geometry and dielectric configuration. \(\textbf{Final answer:}\) \[ \boxed{W_P=-C_0V_0^2,\qquad W_R=3C_0V_0^2,\qquad \frac{U_f}{U_i}=9,\qquad W_{\mathrm{battery}}=2C_0V_0^2} \]

Explanation: \(\textbf{Determine the capacitance change:}\) The vacuum capacitance is \[ C=\frac{\varepsilon_0A}{d} \] Doubling separation at fixed area gives \[ C_f=\frac{\varepsilon_0A}{2d} \] \[ \frac{C_f}{C_i}=\frac12 \] \(\textbf{Apply the battery-voltage condition:}\) The battery maintains the same potential difference \(V_0\). Initially, \[ E_i=\frac{V_0}{d} \] Finally, \[ E_f=\frac{V_0}{2d} \] Therefore, \[ \frac{E_f}{E_i}=\frac12 \] \(\textbf{Compare electric torque magnitudes:}\) For a dipole at angle \(\theta\) to a uniform field, \[ \tau=pE|\sin\theta| \] The dipole magnitude and angle remain fixed. Thus, \[ \frac{\tau_f}{\tau_i}=\frac{E_f}{E_i} \] \[ \frac{\tau_f}{\tau_i}=\frac12 \] \(\textbf{Compare stored capacitor energies:}\) At fixed battery voltage, \[ U=\frac12CV_0^2 \] Therefore, \[ \frac{U_f}{U_i}=\frac{C_f}{C_i} \] \[ \frac{U_f}{U_i}=\frac12 \] \(\textbf{Determine the midpoint potential drop initially:}\) The field is uniform between ideal large parallel plates. From the positive plate to the midpoint of the initial gap, \[ \Delta V_{\mathrm{half},i}=E_i\frac d2 \] \[ \Delta V_{\mathrm{half},i}=\frac{V_0}{2} \] \(\textbf{Determine the midpoint potential drop finally:}\) The final gap is \(2d\). The halfway point is now distance \(d\) from the positive plate: \[ \Delta V_{\mathrm{half},f}=E_fd \] \[ \Delta V_{\mathrm{half},f}=\frac{V_0}{2d}d \] \[ \Delta V_{\mathrm{half},f}=\frac{V_0}{2} \] Therefore, \[ \frac{\Delta V_{\mathrm{half},f}}{\Delta V_{\mathrm{half},i}}=1 \] \(\textbf{Physical interpretation:}\) Doubling plate separation at fixed battery voltage weakens the field and the torque on a fixed dipole. It also reduces capacitance and stored energy. However, a point halfway across either uniform-field gap remains halfway through the same total potential difference. \(\textbf{Final answer:}\) \[ \boxed{\frac{C_f}{C_i}=\frac12,\qquad \frac{\tau_f}{\tau_i}=\frac12,\qquad \frac{U_f}{U_i}=\frac12,\qquad \frac{\Delta V_{\mathrm{half},f}}{\Delta V_{\mathrm{half},i}}=1} \]

Explanation: Statement I is correct because the capacitor in P is isolated. Its free plate charge remains fixed while separation increases. Increasing separation reduces capacitance, and at fixed charge the stored energy is inversely proportional to capacitance. Therefore, doubling separation doubles stored energy. Statement II is correct because the battery in Q maintains the original voltage while plate area doubles. Capacitance is proportional to plate area, so it doubles. Both stored charge and energy are proportional to capacitance at fixed voltage, so they also double. The ideal gap field remains \(V/d\), which is unchanged because both voltage and separation remain unchanged. Statement III is incorrect because halving plate area at fixed total charge doubles free surface charge density. The gap-field magnitude consequently doubles. The attractive electrostatic pressure, proportional to the square of field magnitude, becomes four times its previous value. However, the acting area is halved, so the total attractive force becomes twice its original magnitude, not half. Statement IV is correct because halving plate separation at fixed battery voltage doubles capacitance. The battery supplies enough additional charge to double the free plate-charge magnitude. Stored energy also doubles because capacitance doubles at constant voltage. The field magnitude doubles because the same voltage is maintained over half the original separation. These results illustrate why charge, field, capacitance, energy and force must not be assumed to change in the same proportion under different electrical constraints. Statements I, II and IV are correct.

Explanation: \(\textbf{Apply the isolated-capacitor condition:}\) The capacitor remains disconnected from external circuits. Its free plate-charge magnitude therefore remains \[ Q_f=Q_0 \] The dielectric can become polarised without transporting free charge across the gap. \(\textbf{Infer the dielectric constant from bound charge:}\) For complete filling by a homogeneous linear dielectric, \[ |Q_b|=\frac{K-1}{K}Q_0 \] The measured magnitude is \[ |Q_b|=\frac{3Q_0}{4} \] Therefore, \[ \frac{K-1}{K}=\frac34 \] \[ 1-\frac1K=\frac34 \] \[ \frac1K=\frac14 \] Hence, \[ K=4 \] \(\textbf{Determine the modified capacitance:}\) The original vacuum capacitance is \[ C_0=\frac{\varepsilon_0A}{d} \] After insertion and separation increase, \[ C_f=\frac{K\varepsilon_0A}{2d} \] Substituting \(K=4\), \[ C_f=\frac{4\varepsilon_0A}{2d} \] \[ C_f=2C_0 \] \(\textbf{Compare the terminal voltages:}\) At fixed charge, \[ V=\frac QC \] Therefore, \[ \frac{V_f}{V_0}=\frac{C_0}{C_f} \] \[ \frac{V_f}{V_0}=\frac12 \] \(\textbf{Compare the electric fields:}\) Initially, \[ E_0=\frac{V_0}{d} \] Finally, \[ E_f=\frac{V_f}{2d} \] Using \(V_f=V_0/2\), \[ E_f=\frac{V_0/2}{2d} \] \[ E_f=\frac{V_0}{4d} \] Thus, \[ \frac{E_f}{E_0}=\frac14 \] \(\textbf{Compare stored electrostatic energies:}\) At fixed charge, \[ U=\frac{Q_0^2}{2C} \] Therefore, \[ \frac{U_f}{U_0}=\frac{C_0}{C_f} \] \[ \frac{U_f}{U_0}=\frac12 \] \(\textbf{Physical interpretation:}\) The measured polarisation charge reveals the dielectric constant independently of the altered plate separation. Although separation doubles, the dielectric increases capacitance sufficiently for the final capacitance to exceed its original value. At fixed free charge, this results in lower voltage, weaker field and reduced stored energy. \(\textbf{Final answer:}\) \[ \boxed{K=4,\qquad \frac{V_f}{V_0}=\frac12,\qquad \frac{E_f}{E_0}=\frac14,\qquad \frac{U_f}{U_0}=\frac12} \]

Explanation: \(\textbf{Interpret the measured charge–voltage slopes:}\) For an ideal linear capacitor, \[ Q=CV \] Therefore, the slope of its \(Q\)-versus-\(V\) graph equals capacitance: \[ C_P=2\,\mu\mathrm{F} \] \[ C_S=4\,\mu\mathrm{F} \] \(\textbf{Determine the original series capacitance:}\) \[ C_{\mathrm{eq},i}=\frac{C_PC_S}{C_P+C_S} \] \[ C_{\mathrm{eq},i}=\frac{(2)(4)}{2+4}\,\mu\mathrm{F} \] \[ C_{\mathrm{eq},i}=\frac43\,\mu\mathrm{F} \] \(\textbf{Calculate the initial series charge:}\) The battery voltage is \(18\,\mathrm{V}\): \[ Q_i=C_{\mathrm{eq},i}V \] \[ Q_i=\frac43(18)\,\mu\mathrm{C} \] \[ Q_i=24\,\mu\mathrm{C} \] \(\textbf{Determine P's modified capacitance:}\) Complete dielectric filling multiplies its capacitance by \(K=3\): \[ C_{P,f}=3(2\,\mu\mathrm{F}) \] \[ C_{P,f}=6\,\mu\mathrm{F} \] Capacitor S remains at \(4\,\mu\mathrm{F}\). \(\textbf{Recalculate the series equivalent:}\) \[ C_{\mathrm{eq},f}=\frac{(6)(4)}{6+4}\,\mu\mathrm{F} \] \[ C_{\mathrm{eq},f}=2.4\,\mu\mathrm{F} \] \(\textbf{Determine the final series charge:}\) The battery remains connected at \(18\,\mathrm{V}\): \[ Q_f=C_{\mathrm{eq},f}V \] \[ Q_f=2.4(18)\,\mu\mathrm{C} \] \[ Q_f=43.2\,\mu\mathrm{C} \] \(\textbf{Calculate the additional battery charge:}\) \[ \Delta Q=Q_f-Q_i \] \[ \Delta Q=43.2-24 \] \[ \Delta Q=19.2\,\mu\mathrm{C} \] \(\textbf{Determine the final voltage across P:}\) \[ V_{P,f}=\frac{Q_f}{C_{P,f}} \] \[ V_{P,f}=\frac{43.2\,\mu\mathrm{C}}{6\,\mu\mathrm{F}} \] \[ V_{P,f}=7.2\,\mathrm{V} \] As a check, the unchanged capacitor receives the remaining \(10.8\,\mathrm{V}\), giving the required total of \(18\,\mathrm{V}\). \(\textbf{Physical interpretation:}\) The original graph slopes identify the physical capacitances without requiring a separate geometric measurement. Dielectric insertion modifies P's capacitance, not the battery voltage. The series charge therefore changes even though the intermediate conducting junction remains neutral. \(\textbf{Final answer:}\) \[ \boxed{Q_i=24\,\mu\mathrm{C},\qquad \Delta Q=19.2\,\mu\mathrm{C},\qquad V_{P,f}=7.2\,\mathrm{V}} \]

Explanation: \(\textbf{Establish the original stored state:}\) The initial free plate-charge magnitude is \[ Q_0=CV_0 \] The original stored energy is \[ U_0=\frac12CV_0^2 \] \(\textbf{Apply isolation during plate separation:}\) The battery is disconnected during P. Consequently, \[ Q_P=Q_0 \] Doubling the plate separation halves the capacitance: \[ C_P=\frac C2 \] \(\textbf{Determine the capacitor voltage after P:}\) \[ V_P=\frac{Q_0}{C/2} \] \[ V_P=2V_0 \] \(\textbf{Calculate the energy after mechanical separation:}\) \[ U_P=\frac{Q_0^2}{2C_P} \] \[ U_P=\frac{(CV_0)^2}{2(C/2)} \] \[ U_P=CV_0^2 \] \(\textbf{Determine the external work during P:}\) No battery is connected and the operation is quasistatic: \[ W_P=U_P-U_0 \] \[ W_P=CV_0^2-\frac12CV_0^2 \] \[ W_P=\frac12CV_0^2 \] \(\textbf{Identify charge flow during reconnection:}\) The capacitor begins Stage S at voltage \(2V_0\), exceeding the battery voltage \(V_0\). It therefore returns charge to the battery. Its final capacitance remains \(C/2\), so \[ Q_f=\frac C2V_0 \] \(\textbf{Calculate charge returned to the battery:}\) \[ Q_{\mathrm{returned}}=Q_P-Q_f \] \[ Q_{\mathrm{returned}}=CV_0-\frac12CV_0 \] \[ Q_{\mathrm{returned}}=\frac12CV_0 \] \(\textbf{Determine battery energy received:}\) \[ E_{\mathrm{battery}}=V_0Q_{\mathrm{returned}} \] \[ E_{\mathrm{battery}}=\frac12CV_0^2 \] \(\textbf{Calculate the final capacitor energy:}\) \[ U_f=\frac12C_fV_0^2 \] \[ U_f=\frac12\frac C2V_0^2 \] \[ U_f=\frac14CV_0^2 \] \(\textbf{Determine resistor heating:}\) During S, the capacitor releases \[ U_P-U_f=CV_0^2-\frac14CV_0^2 \] \[ U_P-U_f=\frac34CV_0^2 \] Part reaches the battery, and the remainder becomes heat: \[ H=(U_P-U_f)-E_{\mathrm{battery}} \] \[ H=\frac34CV_0^2-\frac12CV_0^2 \] \[ H=\frac14CV_0^2 \] \(\textbf{Physical interpretation:}\) Mechanical separation increases the energy of the isolated capacitor. Reconnection to a battery at lower voltage reverses the direction of charge transfer. The released capacitor energy is shared between battery energy gain and resistor heating. \(\textbf{Final answer:}\) \[ \boxed{W_P=\frac12CV_0^2,\qquad E_{\mathrm{battery}}=\frac12CV_0^2,\qquad H=\frac14CV_0^2} \]

Explanation: \(\textbf{Apply the measured parallel constraint:}\) The parallel measurement fixes the sum of the two unknown capacitances at \(8.0\,\mu\mathrm{F}\): \[ C_1+C_2=8.0\,\mu\mathrm{F} \] For two ideal capacitors, the series equivalent is their product divided by their sum: \[ C_{\mathrm{series}}=\frac{C_1C_2}{C_1+C_2} \] \(\textbf{Determine the largest possible series capacitance:}\) For a fixed positive sum, the product of two positive numbers is greatest when those numbers are equal. This gives the largest possible series equivalent for the measured parallel capacitance. In the present case, the equal values would each be \(4.0\,\mu\mathrm{F}\). Their maximum product is \[ C_1C_2\leq(4.0\,\mu\mathrm{F})^2=16.0\,(\mu\mathrm{F})^2 \] The corresponding greatest possible series equivalent is \[ C_{\mathrm{series,max}}=\frac{16.0\,(\mu\mathrm{F})^2}{8.0\,\mu\mathrm{F}} \] \[ C_{\mathrm{series,max}}=2.0\,\mu\mathrm{F} \] If the two individual capacitances are unequal while their sum remains \(8.0\,\mu\mathrm{F}\), their product is smaller. The series equivalent must consequently be less than \(2.0\,\mu\mathrm{F}\). \(\textbf{Test the reported series reading:}\) The claimed reading \(2.4\,\mu\mathrm{F}\) lies above this maximum and cannot be obtained from any two positive ideal capacitances satisfying the parallel measurement: \[ 2.4\,\mu\mathrm{F}\gt 2.0\,\mu\mathrm{F} \] The example values \(2.0\,\mu\mathrm{F}\) and \(6.0\,\mu\mathrm{F}\) would produce a series equivalent of only \(1.5\,\mu\mathrm{F}\): \[ C_{\mathrm{series}}=\frac{(2.0)(6.0)}{2.0+6.0}\,\mu\mathrm{F} \] \[ C_{\mathrm{series}}=1.5\,\mu\mathrm{F} \] \(\textbf{Physical conclusion:}\) The correct response is therefore to identify an incompatible measurement rather than force an algebraic solution with nonphysical capacitance values. The reasoning uses both series and parallel relations and the restriction that ordinary ideal capacitances are positive. The given pair of meter readings cannot simultaneously be correct.

Explanation: \(\textbf{Determine the initial charges:}\) For an isolated conducting sphere, \[ Q=CV \] The smaller sphere initially carries \[ Q_{1,i}=C(3V_0) \] \[ Q_{1,i}=3CV_0 \] The larger sphere initially carries \[ Q_{2,i}=2C(-V_0) \] \[ Q_{2,i}=-2CV_0 \] \(\textbf{Apply conservation of charge:}\) The combined initial charge is \[ Q_{\mathrm{total}}=3CV_0-2CV_0 \] \[ Q_{\mathrm{total}}=CV_0 \] No charge leaves the two-sphere system. \(\textbf{Determine the final common potential:}\) After connection, both spheres have the same potential \(V_f\). Therefore, \[ (C+2C)V_f=CV_0 \] \[ V_f=\frac{V_0}{3} \] \(\textbf{Calculate the final sphere charges:}\) For the smaller sphere, \[ Q_{1,f}=CV_f \] \[ Q_{1,f}=\frac{CV_0}{3} \] For the larger sphere, \[ Q_{2,f}=2CV_f \] \[ Q_{2,f}=\frac{2CV_0}{3} \] \(\textbf{Determine transferred charge:}\) The smaller sphere loses \[ Q_{\mathrm{transferred}}=Q_{1,i}-Q_{1,f} \] \[ Q_{\mathrm{transferred}}=3CV_0-\frac{CV_0}{3} \] \[ Q_{\mathrm{transferred}}=\frac{8CV_0}{3} \] The larger sphere gains the same signed charge. \(\textbf{Calculate the initial self-energy:}\) \[ U_i=\frac12C(3V_0)^2+\frac12(2C)(-V_0)^2 \] \[ U_i=\frac92CV_0^2+CV_0^2 \] \[ U_i=\frac{11}{2}CV_0^2 \] \(\textbf{Calculate the final self-energy:}\) \[ U_f=\frac12(C+2C)V_f^2 \] \[ U_f=\frac12(3C)\left(\frac{V_0}{3}\right)^2 \] \[ U_f=\frac16CV_0^2 \] \(\textbf{Calculate dissipation:}\) \[ H=U_i-U_f \] \[ H=\left(\frac{11}{2}-\frac16\right)CV_0^2 \] \[ H=\frac{16}{3}CV_0^2 \] \(\textbf{Physical interpretation:}\) Connected conductors attain equal potential, not equal charge, unless their capacitances are equal. The larger sphere carries twice the smaller sphere's final charge. The very different initial potentials create substantial redistribution and energy dissipation even though total charge is conserved. \(\textbf{Final answer:}\) \[ \boxed{V_f=\frac{V_0}{3},\qquad Q_{\mathrm{transferred}}=\frac{8CV_0}{3},\qquad H=\frac{16}{3}CV_0^2} \]

Explanation: \(\textbf{Model the slab and vacuum regions:}\) The dielectric slab occupies thickness \[ t=\frac d3 \] The vacuum region occupies thickness \[ d-t=\frac{2d}{3} \] These regions occur consecutively along the electric-field direction. \(\textbf{Write the partial-slab capacitance:}\) \[ C_f=\frac{\varepsilon_0A}{d-t+t/K} \] Substituting the specified thickness, \[ C_f=\frac{\varepsilon_0A}{2d/3+d/(3K)} \] Using \(C_0=\varepsilon_0A/d\), \[ \frac{C_f}{C_0}=\frac1{2/3+1/(3K)} \] \(\textbf{Infer the dielectric constant:}\) The measured capacitance ratio is \(4/3\): \[ \frac43=\frac1{2/3+1/(3K)} \] Therefore, \[ \frac34=\frac23+\frac1{3K} \] \[ \frac1{3K}=\frac1{12} \] Hence, \[ K=4 \] \(\textbf{Determine the fields in both materials:}\) The total free plate charge remains \(Q_0\). The normal free-charge displacement is consequently the same through the slab and vacuum region. Therefore, \[ E_v=\frac{Q_0}{\varepsilon_0A} \] and \[ E_d=\frac{Q_0}{K\varepsilon_0A} \] Thus, \[ \frac{E_d}{E_v}=\frac14 \] \(\textbf{Compare electric-field energy densities:}\) The energy density in a linear dielectric is \[ u=\frac12\varepsilon E^2 \] Hence, \[ u_v=\frac12\varepsilon_0E_v^2 \] and \[ u_d=\frac12K\varepsilon_0E_d^2 \] Their ratio is \[ \frac{u_d}{u_v}=K\left(\frac{E_d}{E_v}\right)^2 \] \[ \frac{u_d}{u_v}=4\left(\frac14\right)^2 \] \[ \frac{u_d}{u_v}=\frac14 \] \(\textbf{Calculate the energy change:}\) The capacitor is isolated: \[ U=\frac{Q_0^2}{2C} \] Therefore, \[ \frac{U_f}{U_0}=\frac{C_0}{C_f} \] \[ \frac{U_f}{U_0}=\frac34 \] \(\textbf{Determine mechanical work:}\) The insertion is quasistatic and lossless, with no battery work: \[ W_{\mathrm{ext}}=U_f-U_0 \] \[ W_{\mathrm{ext}}=-\frac{U_0}{4} \] \(\textbf{Physical interpretation:}\) The capacitance measurement determines the dielectric constant through the effective thickness of the gap. The dielectric's field energy density is lower than that of the vacuum region even though its permittivity is higher. The negative external work indicates that the electrostatic system tends to draw the dielectric inward. \(\textbf{Final answer:}\) \[ \boxed{K=4,\qquad \frac{u_d}{u_v}=\frac14,\qquad W_{\mathrm{ext}}=-\frac{U_0}{4}} \]

Explanation: \(\textbf{Determine the initial electric field:}\) For an ideal isolated vacuum capacitor, \[ E_1=\frac{Q}{\varepsilon_0A} \] The field points from the positive plate towards the negative plate. \(\textbf{Determine the modified field:}\) In the second experiment, the plate area becomes \(A/2\). The free plate charge remains \(Q\): \[ E_2=\frac{Q}{\varepsilon_0(A/2)} \] \[ E_2=2E_1 \] \(\textbf{Compare the particle accelerations:}\) The electric force on the small positive particle is \[ F_e=qE \] Its acceleration is \[ a=\frac{qE}{m} \] Therefore, \[ \frac{a_2}{a_1}=\frac{E_2}{E_1} \] \[ \frac{a_2}{a_1}=2 \] \(\textbf{Determine the final speeds:}\) The particle travels distance \(d\) from rest under constant acceleration: \[ v^2=2ad \] Therefore, \[ \frac{v_2^2}{v_1^2}=\frac{a_2}{a_1} \] \[ \frac{v_2}{v_1}=\sqrt2 \] \(\textbf{Compare transit times:}\) From \[ d=\frac12at^2 \] we obtain \[ t=\sqrt{\frac{2d}{a}} \] Therefore, \[ \frac{t_2}{t_1}=\sqrt{\frac{a_1}{a_2}} \] \[ \frac{t_2}{t_1}=\frac1{\sqrt2} \] \(\textbf{Compare plate-attraction forces:}\) The attractive force between isolated ideal capacitor plates is \[ F=\frac{Q^2}{2\varepsilon_0A} \] At fixed charge, the force is inversely proportional to plate area: \[ \frac{F_2}{F_1}=\frac{A}{A/2} \] \[ \frac{F_2}{F_1}=2 \] \(\textbf{Physical interpretation:}\) Halving area at fixed charge increases free surface charge density and the electric field. A test particle consequently accelerates more strongly and crosses the unchanged gap faster. The attractive force between the plates also increases, but it must be calculated using the field of the opposite plate rather than the total gap field acting on a test charge. \(\textbf{Final answer:}\) \[ \boxed{\frac{v_2}{v_1}=\sqrt2,\qquad \frac{t_2}{t_1}=\frac1{\sqrt2},\qquad \frac{F_2}{F_1}=2} \]

Explanation: Statement I is correct because capacitance is a property of the conductor geometry and the intervening material for an ideal linear capacitor. Changing battery voltage does not change that fixed capacitance. Doubling voltage consequently doubles free plate charge through \(Q=CV\). Statement II is correct because increasing plate separation from \(d\) to \(3d/2\) reduces vacuum capacitance by the factor \(2/3\). The capacitor has been disconnected before this mechanical change, so its free plate charge remains \(Q_0\). Statement III is incorrect because the electric field of an ideal isolated vacuum parallel-plate capacitor depends on its free surface charge density. With unchanged free charge and plate area, that field remains \(Q_0/(\varepsilon_0A)\). Increasing separation changes the potential difference rather than the ideal field magnitude. Statement IV is correct for two different reasons. In S, capacitance is fixed and voltage doubles, so energy, proportional to the square of voltage, becomes four times its original value. In R, charge is fixed while capacitance falls to two-thirds, so energy increases by the inverse factor \(3/2\). These comparisons show why a capacitor's voltage, field, stored charge and energy cannot be predicted correctly without identifying whether a battery remains connected. Statements I, II and IV are correct.

Explanation: \(\textbf{Apply the series charge condition:}\) In Experiment P, both capacitors carry the same charge magnitude \(Q\). Their individual voltages are \[ V_1=\frac QC \] \[ V_2=\frac{Q}{2C} \] Thus, \[ V_1=2V_2 \] \(\textbf{Determine the first rating reached in series:}\) The first capacitor can withstand at most \(V_b\). Therefore, its maximum allowed charge is \[ Q_{1,\max}=CV_b \] The second capacitor can withstand \(2V_b\), giving maximum allowed charge \[ Q_{2,\max}=(2C)(2V_b) \] \[ Q_{2,\max}=4CV_b \] The first capacitor is the limiting device: \[ Q_{P,\max}=CV_b \] \(\textbf{Calculate maximum series energy:}\) At this charge, \[ U_P=\frac{Q_{P,\max}^2}{2C}+\frac{Q_{P,\max}^2}{2(2C)} \] \[ U_P=\frac{CV_b^2}{2}+\frac{CV_b^2}{4} \] \[ U_P=\frac{3CV_b^2}{4} \] \(\textbf{Determine the parallel voltage limit:}\) In Experiment S, both capacitors experience the same battery voltage. The smaller individual voltage rating is \(V_b\). Therefore, \[ V_{S,\max}=V_b \] \(\textbf{Calculate maximum parallel energy:}\) The parallel equivalent capacitance is \[ C_S=C+2C=3C \] Therefore, \[ U_S=\frac12C_SV_b^2 \] \[ U_S=\frac32CV_b^2 \] \(\textbf{Compare maximum energies:}\) \[ \frac{U_S}{U_P}=\frac{3CV_b^2/2}{3CV_b^2/4} \] \[ \frac{U_S}{U_P}=2 \] \(\textbf{Physical interpretation:}\) The safe operating voltage of a combination depends on the individual capacitor ratings and actual voltage distribution. A higher-rated capacitor cannot compensate for a lower-rated capacitor if the latter reaches breakdown first. Under the specified ratings, the parallel arrangement stores twice as much energy at its own safe maximum as the series arrangement at its safe maximum. \(\textbf{Final answer:}\) \[ \boxed{U_P=\frac{3CV_b^2}{4},\qquad U_S=\frac{3CV_b^2}{2},\qquad \frac{U_S}{U_P}=2} \]

Explanation: \(\textbf{Identify the initial capacitances:}\) A is initially vacuum-filled: \[ C_{A,i}=C \] B contains the complete dielectric: \[ C_{B,i}=3C \] Both carry the same fixed free charge magnitude \(Q\). \(\textbf{Calculate their initial voltages:}\) \[ V_{A,i}=\frac QC \] \[ V_{B,i}=\frac{Q}{3C} \] \(\textbf{Determine the final capacitances:}\) After transfer, A contains the dielectric and B is vacuum-filled: \[ C_{A,f}=3C \] \[ C_{B,f}=C \] \(\textbf{Determine their final voltages:}\) Neither capacitor exchanges free charge with another object. Thus, \[ V_{A,f}=\frac{Q}{3C} \] \[ V_{B,f}=\frac QC \] Hence, \[ \frac{V_{A,f}}{V_{A,i}}=\frac13 \] and \[ \frac{V_{B,f}}{V_{B,i}}=3 \] \(\textbf{Calculate A's energy change:}\) At fixed charge, \[ U=\frac{Q^2}{2C} \] Therefore, \[ \Delta U_A=\frac{Q^2}{2(3C)}-\frac{Q^2}{2C} \] \[ \Delta U_A=-\frac{Q^2}{3C} \] \(\textbf{Calculate B's energy change:}\) \[ \Delta U_B=\frac{Q^2}{2C}-\frac{Q^2}{2(3C)} \] \[ \Delta U_B=+\frac{Q^2}{3C} \] \(\textbf{Determine net mechanical work:}\) No battery supplies or receives energy during the operation. For a lossless quasistatic process, \[ W_{\mathrm{ext,total}}=\Delta U_A+\Delta U_B \] \[ W_{\mathrm{ext,total}}=-\frac{Q^2}{3C}+\frac{Q^2}{3C} \] \[ W_{\mathrm{ext,total}}=0 \] \(\textbf{Physical interpretation:}\) Inserting dielectric into A reduces its energy and allows negative external mechanical work. Withdrawing dielectric from B requires positive external work of equal magnitude. The two operations can therefore exchange mechanical energy without requiring net external work, even though both capacitor voltages change substantially. \(\textbf{Final answer:}\) \[ \boxed{W_{\mathrm{ext,total}}=0,\qquad \frac{V_{A,f}}{V_{A,i}}=\frac13,\qquad \frac{V_{B,f}}{V_{B,i}}=3} \]

Explanation: \(\textbf{Compare the capacitances:}\) For vacuum parallel plates, \[ C=\frac{\varepsilon_0A}{d} \] For P, \[ C_P=\frac{\varepsilon_0A}{d} \] For S, \[ C_S=\frac{\varepsilon_0(3A)}{3d} \] \[ C_S=C_P \] Thus, both capacitors store the same free charge at the same applied voltage. \(\textbf{Compare electric fields:}\) The batteries maintain voltage \(V_0\). For P, \[ E_P=\frac{V_0}{d} \] For S, \[ E_S=\frac{V_0}{3d} \] Therefore, \[ \frac{E_S}{E_P}=\frac13 \] \(\textbf{Determine final particle speeds:}\) The electrostatic work across the complete gap is \[ W=qV_0 \] The particle starts from rest: \[ \frac12mv^2=qV_0 \] Therefore, in either capacitor, \[ v=\sqrt{\frac{2qV_0}{m}} \] Hence, \[ \frac{v_S}{v_P}=1 \] \(\textbf{Determine the transit-time dependence:}\) The acceleration magnitude is \[ a=\frac{qE}{m} \] A particle starting from rest and crossing gap \(D\) requires time \[ t=\sqrt{\frac{2D}{a}} \] Substituting \(E=V_0/D\), \[ t=\sqrt{\frac{2mD^2}{qV_0}} \] Thus, at fixed voltage, transit time is proportional to gap separation: \[ \frac{t_S}{t_P}=\frac{3d}{d} \] \[ \frac{t_S}{t_P}=3 \] \(\textbf{Compare plate-attraction forces:}\) For a battery-connected vacuum capacitor, \[ F=\frac12\varepsilon_0A E^2 \] Therefore, \[ \frac{F_S}{F_P}=\frac{3A}{A}\left(\frac{E_S}{E_P}\right)^2 \] \[ \frac{F_S}{F_P}=3\left(\frac13\right)^2 \] \[ \frac{F_S}{F_P}=\frac13 \] \(\textbf{Physical interpretation:}\) The two capacitors have equal capacitance despite different plate areas and separations. Particles crossing their complete gaps gain the same kinetic energy because the maintained voltages are equal. However, their accelerations and travel times differ, and the attractive plate forces need not be equal. \(\textbf{Final answer:}\) \[ \boxed{\frac{v_S}{v_P}=1,\qquad \frac{t_S}{t_P}=3,\qquad \frac{F_S}{F_P}=\frac13} \]

Explanation: \(\textbf{Determine the reference vacuum field:}\) At fixed free plate charge, \[ E_0=\frac{Q}{\varepsilon_0A} \] The original voltage is \[ V_0=E_0d \] \(\textbf{Determine the fields after dielectric insertion:}\) The free charge remains fixed in both experiments. The remaining vacuum region has field magnitude \[ E_v=E_0 \] The dielectric region has field magnitude \[ E_d=\frac{E_0}{K} \] With \(K=2\), \[ E_d=\frac{E_0}{2} \] \(\textbf{Determine the final terminal voltage:}\) In either experiment, half the gap is vacuum and half is dielectric. The total potential difference is \[ V_f=E_v\frac d2+E_d\frac d2 \] \[ V_f=E_0\frac d2+\frac{E_0}{2}\frac d2 \] \[ V_f=\frac34E_0d \] Therefore, \[ V_f=\frac{3V_0}{4} \] \(\textbf{Calculate work to the midpoint in P:}\) In P, the dielectric occupies the first half of the gap. The potential drop from the positive plate to the midpoint is \[ \Delta V_P=E_d\frac d2 \] \[ \Delta V_P=\frac{E_0d}{4} \] \[ \Delta V_P=\frac{V_0}{4} \] The field work on \(+q\) is \[ W_P=\frac{qV_0}{4} \] \(\textbf{Calculate work to the midpoint in S:}\) In S, the first half of the gap is vacuum: \[ \Delta V_S=E_v\frac d2 \] \[ \Delta V_S=\frac{V_0}{2} \] Therefore, \[ W_S=\frac{qV_0}{2} \] \(\textbf{Compare midpoint works:}\) \[ \frac{W_S}{W_P}=\frac{qV_0/2}{qV_0/4} \] \[ \frac{W_S}{W_P}=2 \] \(\textbf{Compare the stored energies:}\) The same dielectric thickness and permittivity occur across the same total gap in both arrangements. Their equivalent capacitances are equal. With the same isolated free charge, \[ U_S=U_P \] Thus, \[ U_S-U_P=0 \] \(\textbf{Physical interpretation:}\) Moving a dielectric slab from one side of the gap to the other does not change the total effective gap thickness. It can nevertheless change the potential at a specified interior position because that point may lie in different materials. Equal terminal capacitance does not require identical potential distributions at corresponding interior points. \(\textbf{Final answer:}\) \[ \boxed{\frac{W_S}{W_P}=2,\qquad V_f=\frac{3V_0}{4},\qquad U_S-U_P=0} \]