Electrostatic Potential and Capacitance MCQs with Answers – Part 4 (Class 12 Physics)

Explanation: \( \textbf{Graph idea:} \) Stored energy equals the area under the \(V\)-versus-\(Q\) graph. \( \textbf{Graph shape:} \) The line from the origin to \((Q,V)\) forms a triangle. \( \textbf{Area formula:} \) \[ U=\frac{1}{2}QV \] \( \textbf{Given values:} \) \[ Q=80\,\mu\text{C}=80\times10^{-6}\,\text{C} \] \[ V=25\,\text{V} \] \( \textbf{Substitution:} \) \[ U=\frac{1}{2}(80\times10^{-6})(25) \] \( \textbf{Multiplication:} \) \[ (80\times10^{-6})(25)=2000\times10^{-6}=2.0\times10^{-3}\,\text{J} \] \( \textbf{Apply factor} \) \(\frac{1}{2}\): \[ U=1.0\times10^{-3}\,\text{J} \] \( \textbf{Final answer:} \) The stored energy is \(1.0\times10^{-3}\,\text{J}\). The graph area is triangular because the voltage rises gradually from zero.

Explanation: \( \textbf{Parallel equivalent capacitance:} \) \[ C_{\text{eq}}=2\,\mu\text{F}+4\,\mu\text{F}=6\,\mu\text{F} \] \( \textbf{Battery voltage:} \) \[ V=10\,\text{V} \] \( \textbf{Energy formula for known} \) \(C_{\text{eq}}\) \( \textbf{and} \) \(V\): \[ U=\frac{1}{2}C_{\text{eq}}V^2 \] \( \textbf{Substitution:} \) \[ U=\frac{1}{2}(6\times10^{-6})(10)^2 \] \( \textbf{Calculation:} \) \[ U=\frac{1}{2}(6\times10^{-6})(100) \] \[ U=3.0\times10^{-4}\,\text{J} \] \( \textbf{Final answer:} \) The total stored energy is \(3.0\times10^{-4}\,\text{J}\). The same result is obtained by adding the energies of the two parallel capacitors separately.

Explanation: The stored energy of a capacitor can first be written as \(U=\frac{1}{2}QV\). Since charge, capacitance, and potential difference are related by \(Q=CV\), the same expression can be rewritten in different forms. Replacing \(Q\) by \(CV\) gives \(U=\frac{1}{2}CV^2\). Replacing \(V\) by \(\frac{Q}{C}\) gives \(U=\frac{Q^2}{2C}\). The three forms describe the same stored electrostatic energy, not three separate energies. The best form to use depends on which quantities are given or kept fixed.

Explanation: \( \textbf{Given data:} \) \(Q=40\,\mu\text{C}=40\times10^{-6}\,\text{C}\) and \(V=30\,\text{V}\). \( \textbf{Required quantity:} \) Stored energy \(U\). \( \textbf{Use the direct charge-voltage form:} \) \[ U=\frac{1}{2}QV \] \( \textbf{Substitution:} \) \[ U=\frac{1}{2}(40\times10^{-6})(30) \] \( \textbf{First multiply} \) \(Q\) \( \textbf{and} \) \(V\): \[ (40\times10^{-6})(30)=1200\times10^{-6}=1.2\times10^{-3}\,\text{J} \] \( \textbf{Apply the factor} \) \(\frac{1}{2}\): \[ U=6.0\times10^{-4}\,\text{J} \] \( \textbf{Final answer:} \) The stored energy is \(6.0\times10^{-4}\,\text{J}\). The value \(1.2\times10^{-3}\,\text{J}\) misses the factor \(\frac{1}{2}\), which appears because the capacitor voltage rises gradually during charging.

Explanation: For fixed capacitance, stored energy is \(U=\frac{1}{2}CV^2\). Since \(C\) is constant, \(U\) is proportional to \(V^2\). A quantity proportional to the square of the horizontal-axis variable gives a parabolic graph. The graph passes through the origin because an uncharged capacitor at \(V=0\) stores no energy. A straight line would describe \(Q\) versus \(V\), not \(U\) versus \(V\). The curvature shows that doubling voltage makes the stored energy four times larger.

Explanation: With fixed capacitance, the stored energy is \(U=\frac{1}{2}CV^2\). If \(V\) is replaced by \(2V\), the new energy is \[ U'=\frac{1}{2}C(2V)^2 \] This gives \[ U'=\frac{1}{2}C(4V^2)=4U \] The energy depends on the square of the potential difference. It does not simply double when the voltage doubles. This quadratic dependence is why high voltage can greatly increase the energy stored in a capacitor.

Explanation: If \(V\) is fixed, the useful formula is \(U=\frac{1}{2}CV^2\). Doubling \(C\) then doubles \(U\), so row P is correct. If \(Q\) is fixed, the useful formula is \(U=\frac{Q^2}{2C}\). Under fixed \(Q\), doubling \(C\) would make the energy half, not double. The fixed quantity must be identified before predicting energy change. Battery-connected and disconnected capacitors can therefore show opposite energy trends for the same change in capacitance.

Explanation: \( \textbf{Condition:} \) The capacitor is disconnected, so \(Q\) remains constant. \( \textbf{Dielectric effect:} \) Full insertion makes the capacitance \(C'=KC_0\). \( \textbf{Energy formula for fixed charge:} \) \[ U=\frac{Q^2}{2C} \] \( \textbf{New energy:} \) \[ U'=\frac{Q^2}{2KC_0} \] \( \textbf{Compare with original energy:} \) \[ U_0=\frac{Q^2}{2C_0} \] \[ U'=\frac{U_0}{K} \] \( \textbf{For} \) \(K=4\): \[ U'=\frac{U_0}{4} \] \( \textbf{Final answer:} \) The new energy is \(\frac{U_0}{4}\). The energy decreases because the same charge is stored at a smaller potential difference after dielectric insertion.

Explanation: \( \textbf{Battery condition:} \) The potential difference \(V\) remains constant. \( \textbf{Dielectric effect:} \) \[ C'=KC_0 \] \( \textbf{Energy formula for fixed voltage:} \) \[ U=\frac{1}{2}CV^2 \] \( \textbf{Final energy:} \) \[ U'=\frac{1}{2}(KC_0)V^2 \] \( \textbf{Compare with initial energy:} \) \[ U_0=\frac{1}{2}C_0V^2 \] \[ U'=KU_0 \] \( \textbf{For} \) \(K=3\): \[ U'=3U_0 \] \( \textbf{Final answer:} \) The final stored energy is \(3U_0\). The battery supplies extra charge and energy while keeping the voltage fixed.

Explanation: The Assertion is true for a disconnected charged capacitor. When the dielectric is inserted fully, the capacitance increases from \(C_0\) to \(KC_0\). Since the capacitor is isolated, the free charge \(Q\) cannot change. The stored energy is \(U=\frac{Q^2}{2C}\), so increasing \(C\) with fixed \(Q\) lowers \(U\). The Reason states exactly the two conditions needed for the energy decrease. The decrease in energy is associated with mechanical work done during dielectric insertion.

Explanation: Energy stored in a capacitor can be viewed as energy stored in the electric field between its plates. For vacuum, the electric energy density is \(u=\frac{1}{2}\varepsilon_0E^2\). This gives energy per unit volume, so its SI unit is \(\text{J m}^{-3}\). The dependence is on \(E^2\), so reversing the direction of the electric field does not make the energy density negative. The expression with \(E^2\) divided by \(\varepsilon_0\) is not the vacuum field energy density. The factor \(\varepsilon_0\) reflects the permittivity of free space.

Explanation: \( \textbf{Start with capacitor energy:} \) \[ U=\frac{1}{2}CV^2 \] \( \textbf{Use parallel-plate capacitance:} \) \[ C=\frac{\varepsilon_0A}{d} \] \( \textbf{Use field-potential relation:} \) \[ V=Ed \] \( \textbf{Substitute both relations:} \) \[ U=\frac{1}{2}\left(\frac{\varepsilon_0A}{d}\right)(Ed)^2 \] \( \textbf{Simplify:} \) \[ U=\frac{1}{2}\varepsilon_0AE^2d \] \( \textbf{Volume between plates:} \) \[ \text{Volume}=Ad \] \( \textbf{Energy density:} \) \[ u=\frac{U}{Ad}=\frac{1}{2}\varepsilon_0E^2 \] \( \textbf{Final answer:} \) The field energy density is \(\frac{1}{2}\varepsilon_0E^2\). The area and separation disappear because energy density is energy per unit volume.

Explanation: The energy density in vacuum is \(u=\frac{1}{2}\varepsilon_0E^2\). Since \(\varepsilon_0\) is constant, \(u\propto E^2\). If \(E\) becomes \(2E\), the new energy density is \[ u'=\frac{1}{2}\varepsilon_0(2E)^2 \] This gives \[ u'=4\left(\frac{1}{2}\varepsilon_0E^2\right)=4u \] The energy density is not sensitive to the direction sign of \(E\), only to the square of its magnitude. A stronger field stores disproportionately more energy per unit volume.

Explanation: The expression \(\frac{1}{2}QV\) gives the total electrostatic energy stored in a charged capacitor. The expression \(\frac{1}{2}\varepsilon_0E^2\) is energy density, not capacitance. The expression \(\frac{\varepsilon_0A}{d}\) is the capacitance of an ideal parallel-plate capacitor, not energy density. The expression \(CV\) gives charge \(Q\), not stored energy. The units also separate these meanings: energy is in \(\text{J}\), capacitance is in \(\text{F}\), and energy density is in \(\text{J m}^{-3}\).

Explanation: \( \textbf{Given data:} \) \(E=3.0\times10^4\,\text{N C}^{-1}\) and \(\varepsilon_0=8.85\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}\). \( \textbf{Required quantity:} \) Energy density \(u\). \( \textbf{Formula:} \) \[ u=\frac{1}{2}\varepsilon_0E^2 \] \( \textbf{Square the field:} \) \[ E^2=(3.0\times10^4)^2=9.0\times10^8 \] \( \textbf{Substitution:} \) \[ u=\frac{1}{2}(8.85\times10^{-12})(9.0\times10^8) \] \( \textbf{Multiply powers and numbers:} \) \[ (8.85)(9.0)\times10^{-4}=79.65\times10^{-4}=7.965\times10^{-3} \] \( \textbf{Apply} \) \(\frac{1}{2}\): \[ u=3.9825\times10^{-3}\,\text{J m}^{-3} \] \( \textbf{Final answer:} \) The energy density is approximately \(3.98\times10^{-3}\,\text{J m}^{-3}\). The square of \(E\) must be taken before multiplying by \(\varepsilon_0\).

Explanation: \( \textbf{Initial condition:} \) One capacitor of capacitance \(C\) has voltage \(100\,\text{V}\), and the identical second capacitor is uncharged. \( \textbf{Initial total charge:} \) \[ Q_i=C(100\,\text{V}) \] \( \textbf{After connection:} \) The capacitors are in parallel and have common final voltage \(V_f\). \( \textbf{Equivalent capacitance after connection:} \) \[ C_{\text{eq}}=C+C=2C \] \( \textbf{Charge conservation:} \) \[ Q_i=Q_f \] \[ C(100\,\text{V})=(2C)V_f \] \( \textbf{Solve for} \) \(V_f\): \[ V_f=50\,\text{V} \] \( \textbf{Final answer:} \) The final common potential difference is \(50\,\text{V}\). Charge is shared between two identical capacitors, so the final voltage is half the initial value.

Explanation: When a charged capacitor is connected to an uncharged capacitor, charge flows through the connecting wire until both reach the same potential. The initial electrostatic energy is not usually equal to the final stored capacitor energy. The difference is dissipated during the transient process, mainly as heat in resistance and some electromagnetic radiation. Energy conservation is not violated because the missing stored energy has gone into other forms. The final capacitance arrangement only describes the final storage state, not the whole redistribution process. This is why capacitor connection problems must distinguish charge conservation from stored-energy conservation.

Explanation: \( \textbf{Initial charge on the first capacitor:} \) \[ Q_i=CV \] \[ Q_i=(6\,\mu\text{F})(20\,\text{V})=120\,\mu\text{C} \] \( \textbf{After connection:} \) The two identical capacitors are in parallel, so their equivalent capacitance is \[ C_{\text{eq}}=6\,\mu\text{F}+6\,\mu\text{F}=12\,\mu\text{F} \] \( \textbf{Charge conservation:} \) The total charge available to the connected capacitor system remains \(120\,\mu\text{C}\). \( \textbf{Final common voltage:} \) \[ V_f=\frac{Q_i}{C_{\text{eq}}} \] \[ V_f=\frac{120\,\mu\text{C}}{12\,\mu\text{F}}=10\,\text{V} \] \( \textbf{Final stored energy:} \) \[ U_f=\frac{1}{2}C_{\text{eq}}V_f^2 \] \[ U_f=\frac{1}{2}(12\times10^{-6})(10)^2 \] \[ U_f=6\times10^{-6}\times100 \] \[ U_f=6.0\times10^{-4}\,\text{J} \] \( \textbf{Final answer:} \) The final energy stored in the two capacitors is \(0.6\times10^{-3}\,\text{J}\). Only half the initial stored energy remains in the capacitors; the rest is dissipated during charge sharing.

Explanation: A Van de Graaff generator is an electrostatic device designed to build up very high potentials. It transfers charge continuously to a large conducting dome. Because the dome has capacitance, the accumulated charge raises its potential according to \(V=\frac{Q}{C}\). The conducting dome spreads charge over its outer surface in electrostatic equilibrium. The device is not a rectifier, bridge, or magnetic-field apparatus. Its operation belongs to electrostatic potential, capacitance, and charge accumulation.

Explanation: A large conducting dome has capacitance and can be raised to high potential by adding charge. Its smooth shape avoids very small radii of curvature, where surface charge density and electric field would become very large. Strong electric fields near sharp points can ionize air and cause leakage of charge. In electrostatic equilibrium, charge resides on the outer surface of the conductor, not throughout the metal volume. The dome is an equipotential conductor. A smooth large surface helps maintain high voltage more effectively than a sharp conductor.