Electrostatics MCQs with Answers – Part 6 (NEET Physics)

Explanation: \(\textbf{Apply the constant-voltage condition:}\) The battery remains connected throughout the process. Therefore, \[ V_i=V_f=V_0 \] The initial capacitance is \[ C_i=C_0 \] Doubling separation halves capacitance: \[ C_f=\frac{C_0}{2} \] \(\textbf{Calculate initial and final plate charges:}\) Initially, \[ Q_i=C_0V_0 \] Finally, \[ Q_f=C_fV_0 \] \[ Q_f=\frac{C_0V_0}{2} \] Charge therefore flows back from the capacitor towards the battery. \(\textbf{Calculate battery work:}\) At fixed ideal battery voltage, \[ W_{\mathrm{battery}}=V_0(Q_f-Q_i) \] Substituting, \[ W_{\mathrm{battery}}=V_0\left(\frac{C_0V_0}{2}-C_0V_0\right) \] \[ W_{\mathrm{battery}}=-\frac{C_0V_0^2}{2} \] The negative sign means the battery receives energy. \(\textbf{Determine the stored-energy change:}\) Initially, \[ U_i=\frac12C_0V_0^2 \] Finally, \[ U_f=\frac12\left(\frac{C_0}{2}\right)V_0^2 \] \[ U_f=\frac14C_0V_0^2 \] Therefore, \[ \frac{U_f}{U_i}=\frac12 \] The energy change is \[ \Delta U=-\frac14C_0V_0^2 \] \(\textbf{Apply complete energy accounting:}\) For a quasistatic lossless process, \[ \Delta U=W_{\mathrm{battery}}+W_{\mathrm{ext}} \] Thus, \[ W_{\mathrm{ext}}=\Delta U-W_{\mathrm{battery}} \] \[ W_{\mathrm{ext}}=-\frac14C_0V_0^2+\frac12C_0V_0^2 \] \[ W_{\mathrm{ext}}=+\frac14C_0V_0^2 \] \(\textbf{Physical interpretation:}\) The plates attract, so an external agent must perform positive work to separate them. The capacitor simultaneously returns charge and energy to the battery. The capacitor's stored energy decreases even though positive mechanical work is supplied. \(\textbf{Final answer:}\) \[ \boxed{W_{\mathrm{battery}}=-\frac12C_0V_0^2,\qquad W_{\mathrm{ext}}=+\frac14C_0V_0^2,\qquad \frac{U_f}{U_i}=\frac12} \]

Explanation: \(\textbf{Identify the parallel-area arrangement:}\) The dielectric fills half the plate area through the complete plate separation. The remaining half retains vacuum. The two regions are therefore equivalent to two capacitors connected in parallel. They share the battery voltage \(V_0\). \(\textbf{Determine the vacuum-half capacitance:}\) If the total plate area is \(A\) and separation is \(d\), \[ C_0=\frac{\varepsilon_0A}{d} \] The vacuum half contributes \[ C_{\mathrm{vac}}=\frac{\varepsilon_0(A/2)}d \] \[ C_{\mathrm{vac}}=\frac{C_0}{2} \] \(\textbf{Determine the dielectric-half capacitance:}\) With \(K=4\), \[ C_{\mathrm{dielectric}}=\frac{4\varepsilon_0(A/2)}d \] \[ C_{\mathrm{dielectric}}=2C_0 \] \(\textbf{Calculate final capacitance:}\) The two regions add in parallel: \[ C_f=C_{\mathrm{vac}}+C_{\mathrm{dielectric}} \] \[ C_f=\frac{C_0}{2}+2C_0 \] \[ C_f=\frac{5C_0}{2} \] \(\textbf{Determine the battery's charge transfer:}\) The initial charge is \[ Q_i=C_0V_0 \] At fixed voltage, the final charge is \[ Q_f=\frac{5C_0V_0}{2} \] Therefore, \[ \Delta Q=Q_f-Q_i \] \[ \Delta Q=\frac{3C_0V_0}{2} \] \(\textbf{Calculate battery work:}\) \[ W_{\mathrm{battery}}=V_0\Delta Q \] \[ W_{\mathrm{battery}}=\frac{3C_0V_0^2}{2} \] \(\textbf{Determine the stored-energy change:}\) Initially, \[ U_i=\frac12C_0V_0^2 \] Finally, \[ U_f=\frac12\left(\frac{5C_0}{2}\right)V_0^2 \] \[ U_f=\frac54C_0V_0^2 \] Thus, \[ \Delta U=\frac34C_0V_0^2 \] \(\textbf{Calculate the external mechanical work:}\) Energy accounting gives \[ \Delta U=W_{\mathrm{battery}}+W_{\mathrm{ext}} \] Therefore, \[ W_{\mathrm{ext}}=\frac34C_0V_0^2-\frac32C_0V_0^2 \] \[ W_{\mathrm{ext}}=-\frac34C_0V_0^2 \] \(\textbf{Physical interpretation:}\) The dielectric is drawn into the capacitor by electrostatic attraction. The external agent restraining its slow insertion therefore does negative work. The battery supplies more energy than the increase in stored field energy, with the difference accounting for the mechanical work delivered by the field. \(\textbf{Final answer:}\) \[ \boxed{C_f=\frac{5C_0}{2},\qquad W_{\mathrm{battery}}=\frac32C_0V_0^2,\qquad W_{\mathrm{ext}}=-\frac34C_0V_0^2} \]

Explanation: \(\textbf{Identify the dielectric and vacuum layers:}\) The dielectric fills the complete plate area but only half the separation. Electric-field lines pass successively through the vacuum and dielectric regions. The arrangement therefore behaves like two capacitive layers in series. \(\textbf{Determine the effective separation:}\) For layers normal to the electric field, the equivalent capacitance is \[ C_f=\frac{\varepsilon_0A}{d_{\mathrm{vac}}+d_{\mathrm{dielectric}}/K} \] Here, \[ d_{\mathrm{vac}}=\frac d2 \] and \[ d_{\mathrm{dielectric}}=\frac d2 \] Using \(K=2\), \[ C_f=\frac{\varepsilon_0A}{d/2+(d/2)/2} \] \[ C_f=\frac{\varepsilon_0A}{3d/4} \] Therefore, \[ C_f=\frac{4\varepsilon_0A}{3d} \] \[ C_f=\frac43C_0 \] \(\textbf{Apply charge conservation:}\) The capacitor is isolated. Therefore, its plate-charge magnitude remains \(Q\). The electrostatic energy must be evaluated using \[ U=\frac{Q^2}{2C} \] \(\textbf{Compare initial and final energies:}\) Initially, \[ U_i=\frac{Q^2}{2C_0} \] Finally, \[ U_f=\frac{Q^2}{2(4C_0/3)} \] \[ U_f=\frac{3Q^2}{8C_0} \] Therefore, \[ \frac{U_f}{U_i}=\frac{3Q^2/(8C_0)}{Q^2/(2C_0)} \] \[ \frac{U_f}{U_i}=\frac34 \] \(\textbf{Determine mechanical work:}\) There is no connected battery to exchange energy. In the quasistatic lossless insertion, \[ W_{\mathrm{ext}}=\Delta U \] Thus, \[ W_{\mathrm{ext}}=U_f-U_i \] \[ W_{\mathrm{ext}}=\frac34U_i-U_i \] \[ W_{\mathrm{ext}}=-\frac14U_i \] \(\textbf{Physical interpretation:}\) The dielectric reduces the effective separation for capacitance purposes without changing the actual plate separation. It increases capacitance and decreases stored energy at fixed charge. The negative external work indicates that the electric field tends to draw the dielectric into the gap. \(\textbf{Final answer:}\) \[ \boxed{C_f=\frac43C_0,\qquad \frac{U_f}{U_i}=\frac34,\qquad W_{\mathrm{ext}}=-\frac14U_i} \]

Explanation: \(\textbf{Calculate the initial series capacitance:}\) Initially, all three capacitors are in series: \[ \frac1{C_{\mathrm{eq},i}}=\frac1{2C}+\frac1{3C}+\frac1{6C} \] Combining, \[ \frac1{C_{\mathrm{eq},i}}=\frac{3+2+1}{6C} \] \[ \frac1{C_{\mathrm{eq},i}}=\frac1C \] Therefore, \[ C_{\mathrm{eq},i}=C \] \(\textbf{Determine the initial battery charge:}\) \[ Q_i=C_{\mathrm{eq},i}V_0 \] \[ Q_i=CV_0 \] Each individual series capacitor initially carries this charge magnitude. \(\textbf{Interpret the short-circuit operation:}\) Closing the switch connects the two terminals of the \(3C\) capacitor directly. Its final voltage becomes zero. The \(3C\) capacitor therefore has no final stored charge in the ideal static state. The remaining \(2C\) and \(6C\) capacitors form a series combination across the battery. \(\textbf{Determine the new equivalent capacitance:}\) \[ C_{\mathrm{eq},f}=\frac{(2C)(6C)}{2C+6C} \] \[ C_{\mathrm{eq},f}=\frac{12C^2}{8C} \] \[ C_{\mathrm{eq},f}=\frac{3C}{2} \] \(\textbf{Determine the final series charge:}\) The battery remains connected, maintaining \(V_0\): \[ Q_f=C_{\mathrm{eq},f}V_0 \] \[ Q_f=\frac{3CV_0}{2} \] \(\textbf{Calculate additional battery charge:}\) \[ \Delta Q=Q_f-Q_i \] \[ \Delta Q=\frac{3CV_0}{2}-CV_0 \] \[ \Delta Q=\frac{CV_0}{2} \] \(\textbf{Determine the final voltage across }2C\textbf{:}\) Both remaining series capacitors carry charge magnitude \(Q_f\). Therefore, \[ V_{2C,f}=\frac{Q_f}{2C} \] \[ V_{2C,f}=\frac{3CV_0/2}{2C} \] \[ V_{2C,f}=\frac{3V_0}{4} \] The \(6C\) capacitor carries the remaining voltage \(V_0/4\). \(\textbf{Physical interpretation:}\) Shorting one series capacitor changes the topology of the network. Because the battery remains connected, the series charge need not remain at its initial value. The battery supplies additional charge as the equivalent capacitance increases. \(\textbf{Final answer:}\) \[ \boxed{\Delta Q=\frac{CV_0}{2},\qquad V_{2C,f}=\frac{3V_0}{4}} \]

Explanation: \(\textbf{Identify the shared capacitance change:}\) Both capacitors have identical geometry and receive the same complete dielectric filling. Therefore, \[ C_f=KC_0 \] With \(K=3\), \[ C_f=3C_0 \] Their final electrical states differ because their external connections differ. \(\textbf{Calculate the energy in Experiment P:}\) The capacitor is isolated, so \[ Q_P=Q_0 \] For fixed charge, \[ U_P=\frac{Q_0^2}{2C_f} \] \[ U_P=\frac{Q_0^2}{6C_0} \] Initially, \[ U_0=\frac{Q_0^2}{2C_0} \] Therefore, \[ U_P=\frac{U_0}{3} \] \(\textbf{Calculate the energy in Experiment S:}\) The battery maintains \(V_0\). For fixed voltage, \[ U_S=\frac12C_fV_0^2 \] \[ U_S=\frac12(3C_0)V_0^2 \] Since \(U_0=C_0V_0^2/2\), \[ U_S=3U_0 \] \(\textbf{Compare the stored energies:}\) \[ \frac{U_P}{U_S}=\frac{U_0/3}{3U_0} \] \[ \frac{U_P}{U_S}=\frac19 \] \(\textbf{Determine bound charge in Experiment P:}\) For complete filling, the polarisation-bound surface charge magnitude is \[ Q_b=\frac{K-1}{K}Q_{\mathrm{free}} \] In Experiment P, \[ Q_{b,P}=\frac23Q_0 \] \(\textbf{Determine bound charge in Experiment S:}\) At fixed voltage, the final free plate charge is \[ Q_{\mathrm{free},S}=C_fV_0 \] \[ Q_{\mathrm{free},S}=3Q_0 \] Therefore, \[ Q_{b,S}=\frac23(3Q_0) \] \[ Q_{b,S}=2Q_0 \] \(\textbf{Compare the bound charges:}\) \[ \frac{Q_{b,S}}{Q_{b,P}}=\frac{2Q_0}{2Q_0/3} \] \[ \frac{Q_{b,S}}{Q_{b,P}}=3 \] \(\textbf{Physical interpretation:}\) The dielectric constant alone does not determine the final capacitor energy. The electrical boundary condition determines whether charge or voltage remains fixed. The battery-connected experiment develops greater free plate charge and correspondingly greater bound surface charge. \(\textbf{Final answer:}\) \[ \boxed{\frac{U_P}{U_S}=\frac19,\qquad \frac{Q_{b,S}}{Q_{b,P}}=3} \]

Explanation: \(\textbf{Reduce the initial series branch:}\) The two initial \(C\) capacitors are in series. Their equivalent capacitance is \[ C_{s,i}=\frac{C\cdot C}{C+C} \] \[ C_{s,i}=\frac C2 \] \(\textbf{Determine initial complete capacitance:}\) The series branch is in parallel with the \(2C\) branch. Therefore, \[ C_{\mathrm{eq},i}=2C+\frac C2 \] \[ C_{\mathrm{eq},i}=\frac{5C}{2} \] \(\textbf{Determine the conserved node charge:}\) The battery initially maintains voltage \(V_0\). The total free charge magnitude on node A is \[ Q_A=C_{\mathrm{eq},i}V_0 \] \[ Q_A=\frac{5CV_0}{2} \] The battery is then disconnected, leaving A and B electrically isolated. Their total node charges cannot change. \(\textbf{Reduce the modified series branch:}\) After dielectric insertion, the series-branch capacitances become \(3C\) and \(C\). Its equivalent capacitance is \[ C_{s,f}=\frac{(3C)(C)}{3C+C} \] \[ C_{s,f}=\frac{3C}{4} \] \(\textbf{Determine final complete capacitance:}\) The other parallel branch remains \(2C\): \[ C_{\mathrm{eq},f}=2C+\frac{3C}{4} \] \[ C_{\mathrm{eq},f}=\frac{11C}{4} \] \(\textbf{Calculate the final common voltage:}\) The isolated node charge remains \(Q_A\): \[ V_f=\frac{Q_A}{C_{\mathrm{eq},f}} \] \[ V_f=\frac{5CV_0/2}{11C/4} \] \[ V_f=\frac{10V_0}{11} \] \(\textbf{Determine the charge of the modified series branch:}\) The series branch experiences voltage \(V_f\). Its charge magnitude is \[ Q_s=C_{s,f}V_f \] \[ Q_s=\frac{3C}{4}\left(\frac{10V_0}{11}\right) \] \[ Q_s=\frac{15CV_0}{22} \] The modified \(3C\) capacitor carries the same series charge magnitude: \[ Q_{3C}=Q_s \] \(\textbf{Physical interpretation:}\) The isolated parallel network conserves total charge on each external node, not necessarily the charge on each branch. Changing one capacitance allows charge to redistribute between branches while the common voltage changes. Within the initially neutral series branch, the two individual capacitor charges remain equal in magnitude. \(\textbf{Final answer:}\) \[ \boxed{V_f=\frac{10V_0}{11},\qquad Q_{3C}=\frac{15CV_0}{22}} \]

Explanation: \(\textbf{Interpret the measured series voltages:}\) Initially uncharged capacitors connected in series acquire equal charge magnitudes: \[ Q_1=Q_2=Q_s \] Their voltages are \[ V_1=\frac{Q_s}{C_1} \] \[ V_2=\frac{Q_s}{C_2} \] \(\textbf{Use the given voltage ratio:}\) The measurement gives \[ \frac{V_1}{V_2}=3 \] Therefore, \[ \frac{Q_s/C_1}{Q_s/C_2}=3 \] \[ \frac{C_2}{C_1}=3 \] Thus, \[ C_2=3C_1 \] \(\textbf{Apply the measured equivalent capacitance:}\) For two series capacitors, \[ C_{\mathrm{eq},s}=\frac{C_1C_2}{C_1+C_2} \] Substituting \(C_2=3C_1\), \[ C_{\mathrm{eq},s}=\frac{3C_1^2}{4C_1} \] \[ C_{\mathrm{eq},s}=\frac{3C_1}{4} \] The measured equivalent is \(3\,\mu\mathrm{F}\): \[ \frac{3C_1}{4}=3\,\mu\mathrm{F} \] Therefore, \[ C_1=4\,\mu\mathrm{F} \] and \[ C_2=12\,\mu\mathrm{F} \] \(\textbf{Determine the parallel equivalent:}\) After discharge and reconnection, the capacitances add: \[ C_{\mathrm{eq},p}=C_1+C_2 \] \[ C_{\mathrm{eq},p}=16\,\mu\mathrm{F} \] \(\textbf{Compare battery charge transfers:}\) Both experiments use the same battery voltage \(V_0\). The charge supplied to an initially uncharged equivalent capacitor is \[ Q=C_{\mathrm{eq}}V_0 \] Thus, \[ \frac{Q_p}{Q_s}=\frac{C_{\mathrm{eq},p}}{C_{\mathrm{eq},s}} \] \[ \frac{Q_p}{Q_s}=\frac{16\,\mu\mathrm{F}}{3\,\mu\mathrm{F}} \] \[ \frac{Q_p}{Q_s}=\frac{16}{3} \] \(\textbf{Physical interpretation:}\) The series voltage ratio identifies the inverse capacitance ratio. The equivalent-capacitance measurement then determines both individual values. Changing the connection from series to parallel substantially increases the charge drawn from the same battery. \(\textbf{Final answer:}\) \[ \boxed{C_1=4\,\mu\mathrm{F},\qquad C_2=12\,\mu\mathrm{F},\qquad \frac{Q_p}{Q_s}=\frac{16}{3}} \]

Explanation: \(\textbf{Identify the conductor's internal-field condition:}\) A conducting slab at electrostatic equilibrium has zero electric field within its metal. Therefore, no potential drop occurs across its thickness. The capacitor voltage is established entirely across the remaining vacuum gaps. \(\textbf{Determine the effective gap thickness:}\) The original plate separation is \(d\). The inserted slab occupies thickness \[ t=\frac d2 \] The combined vacuum-gap thickness is \[ d_{\mathrm{eff}}=d-t \] \[ d_{\mathrm{eff}}=\frac d2 \] \(\textbf{Calculate the final capacitance:}\) \[ C_f=\frac{\varepsilon_0A}{d_{\mathrm{eff}}} \] \[ C_f=\frac{\varepsilon_0A}{d/2} \] \[ C_f=2C_0 \] \(\textbf{Determine the charge supplied by the battery:}\) The voltage remains \(V_0\). Initially, \[ Q_i=C_0V_0 \] Finally, \[ Q_f=2C_0V_0 \] Therefore, \[ \Delta Q=Q_f-Q_i \] \[ \Delta Q=C_0V_0 \] \(\textbf{Calculate the battery work:}\) \[ W_{\mathrm{battery}}=V_0\Delta Q \] \[ W_{\mathrm{battery}}=C_0V_0^2 \] \(\textbf{Determine the increase in stored energy:}\) Initially, \[ U_i=\frac12C_0V_0^2 \] Finally, \[ U_f=\frac12(2C_0)V_0^2 \] \[ U_f=C_0V_0^2 \] Thus, \[ \Delta U=\frac12C_0V_0^2 \] \(\textbf{Calculate the external mechanical work:}\) Energy conservation gives \[ \Delta U=W_{\mathrm{battery}}+W_{\mathrm{ext}} \] Therefore, \[ W_{\mathrm{ext}}=\frac12C_0V_0^2-C_0V_0^2 \] \[ W_{\mathrm{ext}}=-\frac12C_0V_0^2 \] \(\textbf{Physical interpretation:}\) The conducting slab removes the potential drop across the region occupied by metal. At fixed battery voltage, this increases capacitance and causes additional battery charge to flow. The electric field draws the slab inward; the external agent restraining quasistatic insertion therefore does negative work. \(\textbf{Final answer:}\) \[ \boxed{C_f=2C_0,\qquad \Delta Q=C_0V_0,\qquad W_{\mathrm{ext}}=-\frac12C_0V_0^2} \]

Explanation: \(\textbf{Determine the capacitance change:}\) For a vacuum parallel-plate capacitor, \[ C=\frac{\varepsilon_0A}{d} \] After reconstruction, \[ A_f=\frac A2,\qquad d_f=\frac d2 \] Therefore, \[ C_f=\frac{\varepsilon_0(A/2)}{d/2} \] \[ C_f=\frac{\varepsilon_0A}{d} \] \[ C_f=C_i \] \(\textbf{Determine the field change:}\) The plate charge remains \(Q\). For isolated large plates, \[ E=\frac{Q}{\varepsilon_0A} \] Therefore, \[ E_f=\frac{Q}{\varepsilon_0(A/2)} \] \[ E_f=2E_i \] \(\textbf{Determine the voltage change:}\) The voltage magnitude is \[ V=Ed \] Hence, \[ V_f=E_fd_f \] \[ V_f=(2E_i)\left(\frac d2\right) \] \[ V_f=E_id \] \[ V_f=V_i \] \(\textbf{Compare stored energies:}\) The charge is fixed and capacitance is unchanged: \[ U=\frac{Q^2}{2C} \] Therefore, \[ U_f=U_i \] \(\textbf{Determine the force change:}\) The attractive force magnitude is \[ F=\frac{Q^2}{2\varepsilon_0A} \] Initially, \[ F_i=\frac{Q^2}{2\varepsilon_0A} \] Finally, \[ F_f=\frac{Q^2}{2\varepsilon_0(A/2)} \] \[ F_f=2F_i \] \(\textbf{Physical interpretation:}\) Halving area and separation compensates exactly in the capacitance formula. However, halving the area doubles the surface charge density and therefore doubles the field. The smaller gap compensates for the stronger field in determining voltage. The stored energy remains unchanged, while the attractive force doubles. \(\textbf{Final answer:}\) \[ \boxed{\frac{C_f}{C_i}=1,\quad \frac{E_f}{E_i}=2,\quad \frac{V_f}{V_i}=1,\quad \frac{U_f}{U_i}=1,\quad \frac{F_f}{F_i}=2} \]

Explanation: \(\textbf{Relate the fields across the interface:}\) No free charge is deposited at the dielectric interface. For the ideal layered geometry, the normal electric displacement has the same magnitude in both layers: \[ \varepsilon_0K_PE_P=\varepsilon_0K_SE_S \] Using \(K_P=2\) and \(K_S=4\), \[ 2E_P=4E_S \] Therefore, \[ E_P=2E_S \] \(\textbf{Apply the battery voltage:}\) Each dielectric has thickness \(a\). The total voltage is the sum of the drops across the layers: \[ V_0=E_Pa+E_Sa \] Substituting \(E_P=2E_S\), \[ V_0=3aE_S \] Thus, \[ E_S=\frac{V_0}{3a} \] and \[ E_P=\frac{2V_0}{3a} \] \(\textbf{Determine the positive-plate free charge:}\) At the positive plate, \[ \sigma_f=\varepsilon_0K_PE_P \] Therefore, \[ \sigma_f=\varepsilon_0(2)\frac{2V_0}{3a} \] \[ \sigma_f=\frac{4\varepsilon_0V_0}{3a} \] The plate charge magnitude is \[ Q_f=\sigma_fA \] \[ Q_f=\frac{4\varepsilon_0AV_0}{3a} \] \(\textbf{Determine polarisation in P:}\) For a linear dielectric, \[ P_P=\varepsilon_0(K_P-1)E_P \] \[ P_P=\varepsilon_0(1)\frac{2V_0}{3a} \] \[ P_P=\frac{2\varepsilon_0V_0}{3a} \] \(\textbf{Determine polarisation in S:}\) \[ P_S=\varepsilon_0(K_S-1)E_S \] \[ P_S=\varepsilon_0(3)\frac{V_0}{3a} \] \[ P_S=\frac{\varepsilon_0V_0}{a} \] \(\textbf{Combine the bound interface charges:}\) Both polarisations point from the positive plate towards the negative plate. The right face of P carries positive bound charge density \(+P_P\). The left face of S carries negative bound charge density \(-P_S\). Thus, \[ \sigma_{b,\mathrm{int}}=P_P-P_S \] \[ \sigma_{b,\mathrm{int}}=\frac{2\varepsilon_0V_0}{3a}-\frac{3\varepsilon_0V_0}{3a} \] \[ \sigma_{b,\mathrm{int}}=-\frac{\varepsilon_0V_0}{3a} \] \(\textbf{Physical interpretation:}\) The layers share the same free-charge condition, not the same electric field. The lower-permittivity layer supports the larger field. Their unequal polarisations produce a nonzero net bound charge at the interface. \(\textbf{Final answer:}\) \[ \boxed{E_P=\frac{2V_0}{3a},\qquad E_S=\frac{V_0}{3a},\qquad Q_f=\frac{4\varepsilon_0AV_0}{3a},\qquad \sigma_{b,\mathrm{int}}=-\frac{\varepsilon_0V_0}{3a}} \]

Explanation: \(\textbf{Determine the first particle's kinetic energy:}\) Initially, the capacitor is vacuum-filled. A positive particle released near the positive plate is accelerated towards the negative plate. The electric field converts its decrease in electric potential energy into kinetic energy: \[ \frac12mv_0^2=qV_0 \] Therefore, \[ v_0^2=\frac{2qV_0}{m} \] \(\textbf{Apply the isolated-capacitor condition:}\) The battery remains disconnected. The capacitor's free plate-charge magnitude is consequently fixed: \[ Q_f=Q_0 \] Complete insertion of a dielectric with \(K=4\) changes the capacitance to \[ C_f=4C_0 \] \(\textbf{Determine the new potential difference:}\) \[ V_f=\frac{Q_f}{C_f} \] \[ V_f=\frac{Q_0}{4C_0} \] Since \[ V_0=\frac{Q_0}{C_0} \] we obtain \[ V_f=\frac{V_0}{4} \] \(\textbf{Calculate the particle's kinetic energy in the second experiment:}\) The particle is again released from rest. Its final kinetic energy is \[ \frac12mv_f^2=qV_f \] Substituting, \[ \frac12mv_f^2=\frac{qV_0}{4} \] \(\textbf{Compare the final speeds:}\) Divide the second energy equation by the first: \[ \frac{v_f^2}{v_0^2}=\frac{qV_0/4}{qV_0} \] \[ \frac{v_f^2}{v_0^2}=\frac14 \] Therefore, \[ \frac{v_f}{v_0}=\frac12 \] \(\textbf{Determine the bound charge:}\) For complete dielectric filling, \[ |Q_b|=\frac{K-1}{K}Q_{\mathrm{free}} \] Here, \[ K=4,\qquad Q_{\mathrm{free}}=Q_0 \] Thus, \[ |Q_b|=\frac34Q_0 \] \(\textbf{Physical interpretation:}\) The dielectric reduces the field and potential difference of an isolated charged capacitor. The positive particle consequently gains less kinetic energy in the second experiment. Its speed decreases by the square root of the energy ratio, rather than by the voltage ratio itself. The dielectric's bound charges arise from polarisation while the capacitor's free plate charge remains fixed. \(\textbf{Final answer:}\) \[ \boxed{\frac{v_f}{v_0}=\frac12,\qquad |Q_b|=\frac{3Q_0}{4}} \]

Explanation: \(\textbf{Connect field energy density with total energy:}\) For an ideal vacuum parallel-plate capacitor, the field is approximately uniform between the plates. Its electric-field energy density is \[ u=\frac12\varepsilon_0E^2 \] The field occupies volume \[ \mathcal V=Aa \] Therefore, total stored energy is \[ U=uAa \] \(\textbf{Infer the plate area:}\) The given measurements are \[ U=12U_0,\qquad u=3u_0 \] Substituting, \[ 12U_0=(3u_0)Aa \] Hence, \[ A=\frac{4U_0}{u_0a} \] \(\textbf{Determine the electric-field magnitude:}\) From the energy-density relation, \[ 3u_0=\frac12\varepsilon_0E^2 \] Therefore, \[ E^2=\frac{6u_0}{\varepsilon_0} \] \[ E=\sqrt{\frac{6u_0}{\varepsilon_0}} \] \(\textbf{Calculate the potential difference:}\) The field is uniform and the plate separation is \(a\). Thus, \[ V=Ea \] \[ V=a\sqrt{\frac{6u_0}{\varepsilon_0}} \] \(\textbf{Relate energy density to plate attraction:}\) The magnitude of the attractive force between ideal oppositely charged parallel plates is \[ F=\frac12\varepsilon_0E^2A \] Since \(u=\varepsilon_0E^2/2\), \[ F=uA \] Substituting the inferred area, \[ F=(3u_0)\frac{4U_0}{u_0a} \] \[ F=\frac{12U_0}{a} \] \(\textbf{Physical check:}\) The area expression has dimensions of area because energy divided by energy density gives volume. The force expression has dimensions of energy per distance. A nonzero stored field energy is associated with attraction between the oppositely charged plates. \(\textbf{Final answer:}\) \[ \boxed{A=\frac{4U_0}{u_0a},\qquad F=\frac{12U_0}{a},\qquad V=a\sqrt{\frac{6u_0}{\varepsilon_0}}} \]

Explanation: \(\textbf{Identify the initially stored charge:}\) Only the original \(C\) capacitor is charged. Its charge magnitude is \[ Q_0=CV_0 \] The second capacitor initially stores no charge. The total charge on the connected positive node must remain \(Q_0\) because there is no battery or leakage path. \(\textbf{Determine the voltage after charge sharing:}\) Immediately after the first parallel connection, \[ C_{\mathrm{eq},1}=C+2C \] \[ C_{\mathrm{eq},1}=3C \] Hence, \[ V_1=\frac{Q_0}{3C} \] \[ V_1=\frac{V_0}{3} \] The initially uncharged capacitor acquires charge through the connecting wires. \(\textbf{Apply the dielectric change:}\) The original capacitor's capacitance increases from \(C\) to \(2C\). The other capacitor remains \(2C\). Consequently, \[ C_{\mathrm{eq},f}=2C+2C \] \[ C_{\mathrm{eq},f}=4C \] \(\textbf{Use conservation of total node charge again:}\) Both capacitors remain connected in parallel. Their individual charges can change, but the total positive-node charge cannot: \[ Q_f=Q_0=CV_0 \] Therefore, \[ V_f=\frac{Q_f}{C_{\mathrm{eq},f}} \] \[ V_f=\frac{CV_0}{4C} \] \[ V_f=\frac{V_0}{4} \] \(\textbf{Calculate the original stored energy:}\) \[ U_i=\frac12CV_0^2 \] \(\textbf{Calculate the final stored energy:}\) \[ U_f=\frac12C_{\mathrm{eq},f}V_f^2 \] Substituting, \[ U_f=\frac12(4C)\left(\frac{V_0}{4}\right)^2 \] \[ U_f=\frac{CV_0^2}{8} \] \(\textbf{Compare the energies:}\) \[ \frac{U_f}{U_i}=\frac{CV_0^2/8}{CV_0^2/2} \] \[ \frac{U_f}{U_i}=\frac14 \] \(\textbf{Physical interpretation:}\) The first operation redistributes charge between two capacitors and dissipates some of the original field energy. The second operation increases the total capacitance at fixed total charge, reducing the remaining stored energy. Charge conservation therefore does not imply conservation of stored electrostatic energy. \(\textbf{Final answer:}\) \[ \boxed{V_f=\frac{V_0}{4},\qquad \frac{U_f}{U_i}=\frac14} \]

Explanation: \(\textbf{Voltage–charge relation:}\) The capacitor has a fixed ideal capacitance throughout both charging intervals. Its voltage is therefore proportional to its stored charge through \[ V=\frac{Q}{C} \] Doubling the charge from \(Q_0\) to \(2Q_0\) doubles the potential difference from \(V_0\) to \(2V_0\). The work needed to transfer an additional small positive charge increases as the potential difference rises. Consequently, the work for a finite charging interval is represented by the area under the voltage-versus-charge graph. \(\textbf{Work during the first interval:}\) During the first interval, the graph forms a triangle of base \(Q_0\) and height \(V_0\). The work is the area of that triangle: \[ W_1=\frac12 Q_0V_0 \] \(\textbf{Work during the second interval:}\) During the second interval, the relevant region is a trapezium extending from voltage \(V_0\) to \(2V_0\) over the same charge increment \(Q_0\). The trapezium area is \[ W_2=\frac12(V_0+2V_0)Q_0 \] \[ W_2=\frac{3Q_0V_0}{2} \] The second interval therefore requires three times as much work as the first interval: \[ \frac{W_2}{W_1}=\frac{3Q_0V_0/2}{Q_0V_0/2}=3 \] \(\textbf{Stored-energy consistency check:}\) This agrees with the stored-energy relation \[ U=\frac{Q^2}{2C} \] which makes energy proportional to the square of charge at fixed capacitance. The total energy at \(2Q_0\) is four times the energy at \(Q_0\), and the additional energy is consequently three times the initial energy. The correct comparison concerns the incremental work over the specified intervals, not merely the final energy ratio. \(\textbf{Final answer:}\) \[ \boxed{\frac{W_2}{W_1}=3,\qquad V_f=2V_0} \]

Explanation: \(\textbf{Use the measured partial-slab capacitance:}\) For a dielectric slab of thickness \(t\) inserted between parallel plates, \[ C=\frac{\varepsilon_0A}{d-t+t/K} \] The first slab has thickness \[ t=\frac d2 \] Therefore, \[ C_{\mathrm{half}}=\frac{\varepsilon_0A}{d/2+d/(2K)} \] \(\textbf{Express this in terms of vacuum capacitance:}\) Using \(C_0=\varepsilon_0A/d\), \[ C_{\mathrm{half}}=\frac{2KC_0}{K+1} \] The measurement gives \[ \frac{2KC_0}{K+1}=\frac32C_0 \] Cancelling \(C_0\), \[ \frac{2K}{K+1}=\frac32 \] Cross-multiplying, \[ 4K=3K+3 \] Therefore, \[ K=3 \] \(\textbf{Determine the full-slab capacitance:}\) When the dielectric fills the entire gap, \[ C_f=KC_0 \] Thus, \[ C_f=3C_0 \] \(\textbf{Determine the energy with the half-thickness slab:}\) The capacitor remains disconnected, so its charge stays fixed. At fixed charge, \[ U\propto\frac1C \] The vacuum energy is \(U_0\). Therefore, \[ U_{\mathrm{half}}=U_0\frac{C_0}{3C_0/2} \] \[ U_{\mathrm{half}}=\frac23U_0 \] \(\textbf{Determine the energy after complete filling:}\) \[ U_f=U_0\frac{C_0}{3C_0} \] \[ U_f=\frac13U_0 \] \(\textbf{Calculate the work during replacement:}\) The external agent works quasistatically, with no battery or dissipative loss. Therefore, \[ W_{\mathrm{ext}}=U_f-U_{\mathrm{half}} \] \[ W_{\mathrm{ext}}=\frac13U_0-\frac23U_0 \] \[ W_{\mathrm{ext}}=-\frac13U_0 \] \(\textbf{Physical interpretation:}\) The partial-slab capacitance measurement determines the dielectric constant. Replacing the partial filling by complete filling further increases capacitance. At fixed free plate charge, stored energy decreases, and the negative external work indicates field-assisted insertion. \(\textbf{Final answer:}\) \[ \boxed{K=3,\qquad C_f=3C_0,\qquad W_{\mathrm{ext}}=-\frac13U_0} \]

Explanation: \(\textbf{Inspect the ratios in the two branches:}\) The upper branch contains \(C\) followed by \(2C\). The lower branch contains \(3C\) followed by \(6C\). Both branches have the same capacitance ratio between their first and second capacitors. This suggests that junctions X and Y may attain the same potential. \(\textbf{Calculate the upper series charge:}\) Ignoring the bridge momentarily, the upper branch has equivalent capacitance \[ C_u=\frac{C(2C)}{C+2C} \] \[ C_u=\frac{2C}{3} \] Its series charge magnitude is \[ Q_u=C_uV_0 \] \[ Q_u=\frac{2CV_0}{3} \] \(\textbf{Determine the potential at X:}\) The voltage across the first capacitor \(C\) is \[ V_{AX}=\frac{Q_u}{C} \] \[ V_{AX}=\frac{2V_0}{3} \] Therefore, \[ V_X=V_A-V_{AX} \] \[ V_X=\frac{V_0}{3} \] \(\textbf{Calculate the lower series charge:}\) The lower equivalent capacitance is \[ C_l=\frac{(3C)(6C)}{3C+6C} \] \[ C_l=2C \] Thus, \[ Q_l=2CV_0 \] \(\textbf{Determine the potential at Y:}\) The voltage across the \(3C\) capacitor is \[ V_{AY}=\frac{Q_l}{3C} \] \[ V_{AY}=\frac{2V_0}{3} \] Therefore, \[ V_Y=V_0-\frac{2V_0}{3} \] \[ V_Y=\frac{V_0}{3} \] \(\textbf{Check the bridge capacitor:}\) The calculated junction potentials are equal: \[ V_X=V_Y \] Hence, \[ V_{XY}=0 \] The bridge capacitor stores no charge in equilibrium: \[ Q_{\mathrm{bridge}}=(5C)V_{XY}=0 \] Because this zero-charge bridge satisfies the junction conditions, it does not alter the two calculated branch voltages. \(\textbf{Calculate the total equivalent capacitance:}\) The upper and lower branches are effectively in parallel: \[ C_{\mathrm{eq}}=C_u+C_l \] \[ C_{\mathrm{eq}}=\frac{2C}{3}+2C \] \[ C_{\mathrm{eq}}=\frac{8C}{3} \] \(\textbf{Physical interpretation:}\) The bridge capacitor connects points already at the same potential. Its capacitance does not contribute to the terminal equivalent under the specified balanced arrangement. The result follows from checking node potentials rather than from visual proximity of components. \(\textbf{Final answer:}\) \[ \boxed{V_{XY}=0,\qquad C_{\mathrm{eq}}=\frac{8C}{3}} \]

Explanation: \(\textbf{Apply the equal-charge property of series capacitors:}\) Both capacitors have equal charge magnitudes in the initial series arrangement. For a capacitor carrying charge \(Q\), \[ U=\frac{Q^2}{2C} \] Thus, at equal charge, stored energies are inversely proportional to capacitances. \(\textbf{Calculate the initial energy ratio:}\) The initial capacitances are \(C\) and \(3C\). Therefore, \[ \frac{U_{1,i}}{U_{2,i}}=\frac{3C}{C} \] \[ \frac{U_{1,i}}{U_{2,i}}=3 \] The smaller initial capacitor stores more energy because it has the larger voltage at the same series charge. \(\textbf{Apply the dielectric change:}\) The first capacitance increases to \[ C_{1,f}=4C \] The second remains \[ C_{2,f}=3C \] The battery remains connected, but this does not invalidate equal final charge magnitudes in the series branch. \(\textbf{Calculate the final energy ratio:}\) \[ \frac{U_{1,f}}{U_{2,f}}=\frac{C_{2,f}}{C_{1,f}} \] \[ \frac{U_{1,f}}{U_{2,f}}=\frac{3C}{4C} \] \[ \frac{U_{1,f}}{U_{2,f}}=\frac34 \] \(\textbf{Determine the initial equivalent capacitance:}\) \[ C_{\mathrm{eq},i}=\frac{C(3C)}{C+3C} \] \[ C_{\mathrm{eq},i}=\frac{3C}{4} \] \(\textbf{Determine the final equivalent capacitance:}\) \[ C_{\mathrm{eq},f}=\frac{(4C)(3C)}{4C+3C} \] \[ C_{\mathrm{eq},f}=\frac{12C}{7} \] \(\textbf{Compare total energies at fixed voltage:}\) For the complete combination, \[ U_{\mathrm{total}}=\frac12C_{\mathrm{eq}}V_0^2 \] The battery holds \(V_0\) fixed in both configurations. Therefore, \[ \frac{U_{\mathrm{total},f}}{U_{\mathrm{total},i}}=\frac{C_{\mathrm{eq},f}}{C_{\mathrm{eq},i}} \] Substituting, \[ \frac{U_{\mathrm{total},f}}{U_{\mathrm{total},i}}=\frac{12C/7}{3C/4} \] \[ \frac{U_{\mathrm{total},f}}{U_{\mathrm{total},i}}=\frac{16}{7} \] \(\textbf{Physical interpretation:}\) The energy distribution between individual series capacitors follows their inverse capacitance ratio. The total stored energy follows the equivalent capacitance because the battery fixes the total voltage. These two comparisons need not change in the same proportion. \(\textbf{Final answer:}\) \[ \boxed{\frac{U_{1,i}}{U_{2,i}}=3,\qquad \frac{U_{1,f}}{U_{2,f}}=\frac34,\qquad \frac{U_{\mathrm{total},f}}{U_{\mathrm{total},i}}=\frac{16}{7}} \]

Explanation: Statement I is correct because the energy-density comparison depends on the electrical boundary condition. In P, the capacitor is isolated, so free plate charge remains fixed and the internal electric field decreases by the factor \(K\). Its energy density is \(u=\varepsilon E^2/2\), with \(\varepsilon=K\varepsilon_0\), giving a final value \(1/K\) times the original vacuum energy density. In S, the battery maintains the same voltage across the same plate separation, so the electric-field magnitude remains equal to its initial value. The dielectric energy density is then \(K\varepsilon_0E^2/2\), which is \(K\) times the original vacuum value. Statement II is therefore also correct. Statement III is correct because disconnecting the battery isolates the free plate charge in P. In S, the battery remains connected and supplies the additional charge needed to preserve the voltage after capacitance increases. Statement IV is incorrect because an ordinary dielectric generally reduces the field under fixed-charge conditions rather than eliminating it. Complete cancellation of electrostatic field within the material is a conductor's equilibrium condition, not a general consequence of electrical insulation. The comparison shows that the same dielectric insertion can change energy density in opposite directions depending on whether charge or voltage is fixed. Statements I, II and III are correct.

Explanation: \(\textbf{Identify the two area regions:}\) The plate area is divided into two equal portions. Each portion spans the same battery-connected plates, so the two regions behave as parallel capacitors. However, the right half contains two successive layers along the field direction. Its capacitance must therefore be calculated using the partial-slab relation. \(\textbf{Calculate the vacuum-half capacitance:}\) The left region has area \(A/2\) and separation \(d\): \[ C_L=\frac{\varepsilon_0(A/2)}d \] \[ C_L=\frac{C_0}{2} \] \(\textbf{Determine the effective thickness of the right region:}\) The right region contains vacuum thickness \(d/2\) and dielectric thickness \(d/2\). For dielectric constant \(K=3\), \[ d_{\mathrm{eff}}=\frac d2+\frac{d/2}{3} \] \[ d_{\mathrm{eff}}=\frac{2d}{3} \] \(\textbf{Calculate the right-region capacitance:}\) \[ C_R=\frac{\varepsilon_0(A/2)}{2d/3} \] \[ C_R=\frac{3\varepsilon_0A}{4d} \] \[ C_R=\frac{3C_0}{4} \] \(\textbf{Combine the parallel regions:}\) \[ C_f=C_L+C_R \] \[ C_f=\frac{C_0}{2}+\frac{3C_0}{4} \] \[ C_f=\frac{5C_0}{4} \] \(\textbf{Determine the battery charge transfer:}\) The initial positive-plate charge is \[ Q_i=C_0V_0 \] The final charge is \[ Q_f=C_fV_0 \] \[ Q_f=\frac{5C_0V_0}{4} \] Therefore, \[ \Delta Q=Q_f-Q_i \] \[ \Delta Q=\frac{C_0V_0}{4} \] \(\textbf{Compare the stored energies:}\) At constant battery voltage, \[ U=\frac12CV_0^2 \] Therefore, \[ \frac{U_f}{U_i}=\frac{C_f}{C_0} \] \[ \frac{U_f}{U_i}=\frac54 \] \(\textbf{Physical interpretation:}\) The arrangement contains both parallel area regions and successive dielectric layers. Using the correct geometric reduction in each part is essential. The battery supplies additional free charge as the effective capacitance increases. \(\textbf{Final answer:}\) \[ \boxed{C_f=\frac{5C_0}{4},\qquad \Delta Q=\frac{C_0V_0}{4},\qquad \frac{U_f}{U_i}=\frac54} \]

Explanation: \(\textbf{Calculate the initial capacitor charges:}\) For the first capacitor, \[ Q_{1,i}=C(4V_0) \] \[ Q_{1,i}=4CV_0 \] For the second, \[ Q_{2,i}=(3C)(2V_0) \] \[ Q_{2,i}=6CV_0 \] \(\textbf{Use the like-polarity connection:}\) Positive plates are connected to positive plates. The conserved positive-node charge is \[ Q_{\mathrm{total}}=Q_{1,i}+Q_{2,i} \] \[ Q_{\mathrm{total}}=10CV_0 \] The combined parallel capacitance is \[ C_{\mathrm{total}}=C+3C \] \[ C_{\mathrm{total}}=4C \] \(\textbf{Determine the final common voltage:}\) \[ V_f=\frac{Q_{\mathrm{total}}}{C_{\mathrm{total}}} \] \[ V_f=\frac{10CV_0}{4C} \] \[ V_f=\frac{5V_0}{2} \] \(\textbf{Track charge redistribution:}\) The final charge on the first capacitor is \[ Q_{1,f}=CV_f \] \[ Q_{1,f}=\frac{5CV_0}{2} \] It initially carried \(4CV_0\). Therefore, the magnitude of charge it loses is \[ Q_{\mathrm{transferred}}=4CV_0-\frac{5CV_0}{2} \] \[ Q_{\mathrm{transferred}}=\frac{3CV_0}{2} \] The second capacitor gains the same charge magnitude. \(\textbf{Calculate the initial total energy:}\) \[ U_i=\frac12C(4V_0)^2+\frac12(3C)(2V_0)^2 \] \[ U_i=8CV_0^2+6CV_0^2 \] \[ U_i=14CV_0^2 \] \(\textbf{Calculate the final stored energy:}\) \[ U_f=\frac12C_{\mathrm{total}}V_f^2 \] \[ U_f=\frac12(4C)\left(\frac{5V_0}{2}\right)^2 \] \[ U_f=\frac{25CV_0^2}{2} \] \(\textbf{Determine the energy decrease:}\) \[ U_{\mathrm{lost}}=U_i-U_f \] \[ U_{\mathrm{lost}}=14CV_0^2-\frac{25CV_0^2}{2} \] \[ U_{\mathrm{lost}}=\frac{3CV_0^2}{2} \] \(\textbf{Physical interpretation:}\) Charge flows from the capacitor initially at higher voltage towards the one initially at lower voltage. Their common final voltage is fixed by charge conservation, not by averaging the initial voltages arithmetically. The decrease in field energy appears in other forms during redistribution; total physical energy remains conserved. \(\textbf{Final answer:}\) \[ \boxed{V_f=\frac{5V_0}{2},\qquad Q_{\mathrm{transferred}}=\frac{3CV_0}{2},\qquad U_{\mathrm{lost}}=\frac{3CV_0^2}{2}} \]