401. A capacitor carries charge magnitude \(Q\) on each plate and has potential difference \(V\). Which expression gives the energy stored in it?
ⓐ. \(U=QV\)
ⓑ. \(U=\frac{Q}{2V}\)
ⓒ. \(U=\frac{1}{2}QV\)
ⓓ. \(U=\frac{V^2}{2Q}\)
Correct Answer: \(U=\frac{1}{2}QV\)
Explanation: \(\textbf{Stored energy idea:}\) During charging, the potential difference across the capacitor rises gradually from \(0\) to \(V\).
\(\textbf{Average potential:}\) The average potential difference during the charging process is \(\frac{V}{2}\).
\(\textbf{Work done:}\) The work required to transfer total charge \(Q\) is \(Q\) multiplied by average potential.
\(\textbf{Energy stored:}\) \(U=Q\left(\frac{V}{2}\right)\).
\(\textbf{Simplification:}\) \(U=\frac{1}{2}QV\).
\(\textbf{Equivalent forms:}\) Using \(Q=CV\), this can also be written as \(U=\frac{1}{2}CV^2\) or \(U=\frac{Q^2}{2C}\).
\(\textbf{Final result:}\) The stored energy is \(U=\frac{1}{2}QV\).
402. A capacitor has charge \(40\,\mu\text{C}\) and potential difference \(200\,\text{V}\). What is the energy stored in it?
ⓐ. \(4.0\times10^{-3}\,\text{J}\)
ⓑ. \(8.0\times10^{-3}\,\text{J}\)
ⓒ. \(2.0\times10^{-3}\,\text{J}\)
ⓓ. \(1.6\times10^{-2}\,\text{J}\)
Correct Answer: \(4.0\times10^{-3}\,\text{J}\)
Explanation: \(\textbf{Given:}\) \(Q=40\,\mu\text{C}=40\times10^{-6}\,\text{C}\) and \(V=200\,\text{V}\).
\(\textbf{Required:}\) Stored energy \(U\).
\(\textbf{Formula:}\) \(U=\frac{1}{2}QV\).
\(\textbf{Substitution:}\) \(U=\frac{1}{2}(40\times10^{-6})(200)\,\text{J}\).
\(\textbf{Product:}\) \(40\times200=8000\).
\(\textbf{Power of ten:}\) \((40\times10^{-6})(200)=8000\times10^{-6}=8.0\times10^{-3}\,\text{J}\).
\(\textbf{Half factor:}\) \(U=\frac{1}{2}(8.0\times10^{-3})\,\text{J}\).
\(\textbf{Final result:}\) \(U=4.0\times10^{-3}\,\text{J}\).
403. A charged parallel plate capacitor is isolated. Its plate separation is doubled while the plate area remains unchanged. What happens to its stored energy?
ⓐ. It becomes half
ⓑ. It remains unchanged
ⓒ. It becomes four times
ⓓ. It becomes double
Correct Answer: It becomes double
Explanation: \(\textbf{Isolation condition:}\) The capacitor is disconnected, so charge \(Q\) remains constant.
\(\textbf{Capacitance relation:}\) For a parallel plate capacitor, \(C=\frac{\varepsilon_0A}{d}\).
\(\textbf{Change in separation:}\) If \(d\) is doubled, the new capacitance becomes \(C'=\frac{C}{2}\).
\(\textbf{Energy formula for constant charge:}\) \(U=\frac{Q^2}{2C}\).
\(\textbf{Initial energy:}\) \(U=\frac{Q^2}{2C}\).
\(\textbf{Final energy:}\) \(U'=\frac{Q^2}{2(C/2)}\).
\(\textbf{Simplification:}\) \(U'=\frac{Q^2}{C}=2\left(\frac{Q^2}{2C}\right)=2U\).
\(\textbf{Final result:}\) The stored energy becomes double.
404. A parallel plate capacitor remains connected to a battery. If the plate separation is doubled while plate area remains unchanged, what happens to its stored energy?
ⓐ. It becomes double
ⓑ. It becomes half
ⓒ. It becomes four times
ⓓ. It remains unchanged
Correct Answer: It becomes half
Explanation: \(\textbf{Battery condition:}\) The capacitor remains connected to a battery, so \(V\) stays constant.
\(\textbf{Capacitance relation:}\) \(C=\frac{\varepsilon_0A}{d}\).
\(\textbf{Change in separation:}\) Doubling \(d\) makes the capacitance \(C'=\frac{C}{2}\).
\(\textbf{Energy formula for constant voltage:}\) \(U=\frac{1}{2}CV^2\).
\(\textbf{Initial energy:}\) \(U=\frac{1}{2}CV^2\).
\(\textbf{Final energy:}\) \(U'=\frac{1}{2}\left(\frac{C}{2}\right)V^2\).
\(\textbf{Simplification:}\) \(U'=\frac{1}{2}U\).
\(\textbf{Final result:}\) The stored energy becomes half.
405. What is the energy density of an electric field in vacuum?
ⓐ. \(u=\varepsilon_0E\)
ⓑ. \(u=\frac{E^2}{2\varepsilon_0}\)
ⓒ. \(u=\frac{1}{2}\varepsilon_0E^2\)
ⓓ. \(u=\frac{1}{2}\varepsilon_0^2E\)
Correct Answer: \(u=\frac{1}{2}\varepsilon_0E^2\)
Explanation: \(\textbf{Energy density meaning:}\) Energy density is energy stored per unit volume in the electric field.
\(\textbf{Vacuum expression:}\) For an electric field in vacuum, \(u=\frac{1}{2}\varepsilon_0E^2\).
\(\textbf{Dependence on field:}\) The energy density is proportional to the square of the field magnitude.
\(\textbf{Role of \(\varepsilon_0\):}\) The permittivity \(\varepsilon_0\) connects the field strength with stored electrostatic energy in vacuum.
\(\textbf{Unit check:}\) \(\varepsilon_0E^2\) has unit \(\text{J m}^{-3}\).
\(\textbf{Physical meaning:}\) A stronger field stores more energy per unit volume, and doubling \(E\) makes the energy density four times.
\(\textbf{Final result:}\) The energy density is \(u=\frac{1}{2}\varepsilon_0E^2\).
406. The electric field between capacitor plates is \(2.0\times10^5\,\text{V m}^{-1}\). What is the energy density in air, using \(\varepsilon_0=8.85\times10^{-12}\,\text{F m}^{-1}\)?
ⓐ. \(0.0885\,\text{J m}^{-3}\)
ⓑ. \(0.177\,\text{J m}^{-3}\)
ⓒ. \(0.354\,\text{J m}^{-3}\)
ⓓ. \(1.77\,\text{J m}^{-3}\)
Correct Answer: \(0.177\,\text{J m}^{-3}\)
Explanation: \(\textbf{Given:}\) \(E=2.0\times10^5\,\text{V m}^{-1}\) and \(\varepsilon_0=8.85\times10^{-12}\,\text{F m}^{-1}\).
\(\textbf{Required:}\) Energy density \(u\).
\(\textbf{Formula:}\) \(u=\frac{1}{2}\varepsilon_0E^2\).
\(\textbf{Square of field:}\) \(E^2=(2.0\times10^5)^2=4.0\times10^{10}\).
\(\textbf{Substitution:}\) \(u=\frac{1}{2}(8.85\times10^{-12})(4.0\times10^{10})\,\text{J m}^{-3}\).
\(\textbf{Power simplification:}\) \(10^{-12}\times10^{10}=10^{-2}\).
\(\textbf{Number simplification:}\) \(\frac{1}{2}\times8.85\times4.0=17.7\).
\(\textbf{Final calculation:}\) \(u=17.7\times10^{-2}=0.177\,\text{J m}^{-3}\).
\(\textbf{Final result:}\) The energy density is \(0.177\,\text{J m}^{-3}\).
407. A capacitor remains connected to a battery of voltage \(V\). Its capacitance increases from \(C\) to \(3C\). What is the increase in energy stored in the capacitor?
ⓐ. \(\frac{1}{2}CV^2\)
ⓑ. \(CV^2\)
ⓒ. \(\frac{3}{2}CV^2\)
ⓓ. \(2CV^2\)
Correct Answer: \(CV^2\)
Explanation: \(\textbf{Battery condition:}\) The potential difference remains \(V\).
\(\textbf{Initial energy:}\) \(U_i=\frac{1}{2}CV^2\).
\(\textbf{Final capacitance:}\) \(C_f=3C\).
\(\textbf{Final energy:}\) \(U_f=\frac{1}{2}(3C)V^2=\frac{3}{2}CV^2\).
\(\textbf{Increase in energy:}\) \(\Delta U=U_f-U_i\).
\(\textbf{Substitution:}\) \(\Delta U=\frac{3}{2}CV^2-\frac{1}{2}CV^2\).
\(\textbf{Simplification:}\) \(\Delta U=CV^2\).
\(\textbf{Final result:}\) The increase in stored energy is \(CV^2\).
408. A charged capacitor is isolated. Its capacitance increases from \(C\) to \(4C\). If its initial stored energy is \(U\), what is the final stored energy?
ⓐ. \(\frac{U}{16}\)
ⓑ. \(\frac{U}{2}\)
ⓒ. \(4U\)
ⓓ. \(\frac{U}{4}\)
Correct Answer: \(\frac{U}{4}\)
Explanation: \(\textbf{Isolation condition:}\) The capacitor is disconnected, so charge \(Q\) remains constant.
\(\textbf{Energy formula for constant charge:}\) \(U=\frac{Q^2}{2C}\).
\(\textbf{Initial energy:}\) \(U_i=\frac{Q^2}{2C}=U\).
\(\textbf{Final capacitance:}\) \(C_f=4C\).
\(\textbf{Final energy:}\) \(U_f=\frac{Q^2}{2(4C)}\).
\(\textbf{Simplification:}\) \(U_f=\frac{1}{4}\frac{Q^2}{2C}\).
\(\textbf{Using initial energy:}\) \(U_f=\frac{U}{4}\).
\(\textbf{Final result:}\) The final stored energy is \(\frac{U}{4}\).
409. A \(5.0\,\mu\text{F}\) capacitor is charged to \(100\,\text{V}\) and then disconnected from the battery. A dielectric is inserted fully so that its capacitance becomes \(20\,\mu\text{F}\). What is the final potential difference?
ⓐ. \(400\,\text{V}\)
ⓑ. \(100\,\text{V}\)
ⓒ. \(25\,\text{V}\)
ⓓ. \(20\,\text{V}\)
Correct Answer: \(25\,\text{V}\)
Explanation: \(\textbf{Initial capacitance:}\) \(C_i=5.0\,\mu\text{F}\).
\(\textbf{Initial voltage:}\) \(V_i=100\,\text{V}\).
\(\textbf{Final capacitance:}\) \(C_f=20\,\mu\text{F}\).
\(\textbf{Isolation condition:}\) The capacitor is disconnected, so charge remains constant.
\(\textbf{Initial charge:}\) \(Q=C_iV_i=(5.0\,\mu\text{F})(100\,\text{V})=500\,\mu\text{C}\).
\(\textbf{Final voltage:}\) \(V_f=\frac{Q}{C_f}\).
\(\textbf{Substitution:}\) \(V_f=\frac{500\,\mu\text{C}}{20\,\mu\text{F}}\).
\(\textbf{Calculation:}\) \(V_f=25\,\text{V}\).
\(\textbf{Final result:}\) The final potential difference is \(25\,\text{V}\).
410. Which quantity is represented by the area under a \(V\) versus \(Q\) graph for charging a capacitor from \(0\) to \(Q\)?
ⓐ. Stored energy
ⓑ. Capacitance
ⓒ. Electric field
ⓓ. Surface charge density
Correct Answer: Stored energy
Explanation: \(\textbf{Graph meaning:}\) A \(V\) versus \(Q\) graph shows how potential difference changes as charge is supplied.
\(\textbf{Small work:}\) The small work needed to add charge \(dQ\) at potential \(V\) is \(dW=V\,dQ\).
\(\textbf{Graphical interpretation:}\) The total work is the area under the \(V\)-versus-\(Q\) graph.
\(\textbf{Stored energy:}\) This work becomes electrostatic energy stored in the capacitor.
\(\textbf{For a linear capacitor:}\) The graph is a straight line from \(0\) to \(V\), so the area is \(\frac{1}{2}QV\).
\(\textbf{Equivalent forms:}\) This equals \(\frac{1}{2}CV^2\) and \(\frac{Q^2}{2C}\).
\(\textbf{Final result:}\) The area under the graph represents stored energy.
411. A capacitor stores energy \(U\) at voltage \(V\). If the voltage is doubled while capacitance remains constant, what is the new stored energy?
ⓐ. \(2U\)
ⓑ. \(U\)
ⓒ. \(\frac{U}{2}\)
ⓓ. \(4U\)
Correct Answer: \(4U\)
Explanation: \(\textbf{Energy relation:}\) \(U=\frac{1}{2}CV^2\).
\(\textbf{Constant capacitance:}\) The capacitance \(C\) remains unchanged.
\(\textbf{Voltage change:}\) The new voltage is \(V'=2V\).
\(\textbf{New energy:}\) \(U'=\frac{1}{2}C(2V)^2\).
\(\textbf{Square factor:}\) \((2V)^2=4V^2\).
\(\textbf{Simplification:}\) \(U'=4\left(\frac{1}{2}CV^2\right)\).
\(\textbf{Using initial energy:}\) \(U'=4U\).
\(\textbf{Final result:}\) The new stored energy is \(4U\).
412. A capacitor stores \(0.50\,\text{J}\) of energy with charge \(100\,\mu\text{C}\). What is its capacitance?
ⓐ. \(5.0\,\text{nF}\)
ⓑ. \(10\,\text{nF}\)
ⓒ. \(20\,\text{nF}\)
ⓓ. \(40\,\text{nF}\)
Correct Answer: \(10\,\text{nF}\)
Explanation: \(\textbf{Given:}\) \(U=0.50\,\text{J}\) and \(Q=100\,\mu\text{C}=1.0\times10^{-4}\,\text{C}\).
\(\textbf{Required:}\) Capacitance \(C\).
\(\textbf{Formula:}\) \(U=\frac{Q^2}{2C}\).
\(\textbf{Rearrangement:}\) \(C=\frac{Q^2}{2U}\).
\(\textbf{Substitution:}\) \(C=\frac{(1.0\times10^{-4})^2}{2(0.50)}\,\text{F}\).
\(\textbf{Square:}\) \((1.0\times10^{-4})^2=1.0\times10^{-8}\).
\(\textbf{Denominator:}\) \(2(0.50)=1.0\).
\(\textbf{Calculation:}\) \(C=1.0\times10^{-8}\,\text{F}=10\,\text{nF}\).
\(\textbf{Final result:}\) The capacitance is \(10\,\text{nF}\).
413. In a parallel plate capacitor, the energy stored can be written as \(U=\frac{1}{2}\varepsilon_0E^2Ad\). What does \(Ad\) represent?
ⓐ. Surface charge density
ⓑ. Electric flux through one plate
ⓒ. Volume between the capacitor plates
ⓓ. Potential difference between the plates
Correct Answer: Volume between the capacitor plates
Explanation: \(\textbf{Parallel plate geometry:}\) The plates have area \(A\) and separation \(d\).
\(\textbf{Field region:}\) For large plates, the electric field is mainly confined to the region between the plates.
\(\textbf{Volume between plates:}\) The volume of this region is area multiplied by separation.
\(\textbf{Expression:}\) \(V_{\text{volume}}=Ad\).
\(\textbf{Energy density relation:}\) The energy density is \(u=\frac{1}{2}\varepsilon_0E^2\).
\(\textbf{Total energy:}\) \(U=u\times\text{volume}=\frac{1}{2}\varepsilon_0E^2Ad\).
\(\textbf{Final result:}\) The factor \(Ad\) represents the volume of the field region between the plates.
414. A capacitor of capacitance \(8.0\,\mu\text{F}\) stores \(1.6\times10^{-3}\,\text{J}\) of energy. What is the potential difference across it?
ⓐ. \(10\,\text{V}\)
ⓑ. \(15\,\text{V}\)
ⓒ. \(25\,\text{V}\)
ⓓ. \(20\,\text{V}\)
Correct Answer: \(20\,\text{V}\)
Explanation: \(\textbf{Given:}\) \(C=8.0\,\mu\text{F}=8.0\times10^{-6}\,\text{F}\) and \(U=1.6\times10^{-3}\,\text{J}\).
\(\textbf{Required:}\) Potential difference \(V\).
\(\textbf{Energy formula:}\) \(U=\frac{1}{2}CV^2\).
\(\textbf{Rearrangement:}\) \(V^2=\frac{2U}{C}\).
\(\textbf{Substitution:}\) \(V^2=\frac{2(1.6\times10^{-3})}{8.0\times10^{-6}}\).
\(\textbf{Numerator:}\) \(2(1.6\times10^{-3})=3.2\times10^{-3}\).
\(\textbf{Division:}\) \(V^2=\frac{3.2\times10^{-3}}{8.0\times10^{-6}}=0.40\times10^3=400\).
\(\textbf{Square root:}\) \(V=\sqrt{400}=20\,\text{V}\).
\(\textbf{Final result:}\) The potential difference is \(20\,\text{V}\).
415. A charged isolated capacitor has initial energy \(U_0\). A dielectric is inserted fully, and the final energy becomes \(\frac{U_0}{5}\). What is the relative permittivity of the dielectric?
ⓐ. \(5\)
ⓑ. \(\frac{1}{5}\)
ⓒ. \(\sqrt{5}\)
ⓓ. \(25\)
Correct Answer: \(5\)
Explanation: \(\textbf{Isolation condition:}\) The capacitor is isolated, so charge \(Q\) remains constant.
\(\textbf{Energy at constant charge:}\) \(U=\frac{Q^2}{2C}\).
\(\textbf{Effect of full dielectric:}\) If the relative permittivity is \(K\), the capacitance becomes \(KC_0\).
\(\textbf{Energy comparison:}\) \(U_f=\frac{Q^2}{2KC_0}=\frac{U_0}{K}\).
\(\textbf{Given:}\) \(U_f=\frac{U_0}{5}\).
\(\textbf{Equating:}\) \(\frac{U_0}{K}=\frac{U_0}{5}\).
\(\textbf{Simplification:}\) \(K=5\).
\(\textbf{Final result:}\) The relative permittivity is \(5\).
416. A capacitor is connected to a battery and a dielectric is inserted fully between the plates. Which combination is correct?
ⓐ. \(V\) constant, \(Q\) decreases, \(U\) decreases
ⓑ. \(V\) constant, \(Q\) increases, \(U\) increases
ⓒ. \(Q\) constant, \(V\) decreases, \(U\) decreases
ⓓ. \(Q\) constant, \(V\) increases, \(U\) increases
Correct Answer: \(V\) constant, \(Q\) increases, \(U\) increases
Explanation: When a capacitor remains connected to a battery, the potential difference \(V\) is fixed by the battery. Inserting a dielectric fully increases the capacitance from \(C_0\) to \(KC_0\). Since \(Q=CV\), the charge on the plates increases because \(C\) increases while \(V\) remains constant. The stored energy under constant voltage is \(U=\frac{1}{2}CV^2\). With \(V\) fixed and \(C\) increasing, \(U\) also increases. The extra charge and extra stored energy are supplied through the battery. Therefore, the correct combination is \(V\) constant, \(Q\) increases, and \(U\) increases.
417. A \(2.0\,\mu\text{F}\) capacitor is charged to \(90\,\text{V}\), disconnected from the battery, and then a dielectric of relative permittivity \(K=3\) is fully inserted. What is the final stored energy?
ⓐ. \(2.7\times10^{-3}\,\text{J}\)
ⓑ. \(8.1\times10^{-3}\,\text{J}\)
ⓒ. \(1.62\times10^{-2}\,\text{J}\)
ⓓ. \(2.43\times10^{-2}\,\text{J}\)
Correct Answer: \(2.7\times10^{-3}\,\text{J}\)
Explanation: \(\textbf{Given:}\) \(C_0=2.0\,\mu\text{F}=2.0\times10^{-6}\,\text{F}\), \(V_0=90\,\text{V}\), and \(K=3\).
\(\textbf{Condition:}\) The capacitor is disconnected before dielectric insertion, so charge remains constant.
\(\textbf{Initial energy:}\) \(U_0=\frac{1}{2}C_0V_0^2\).
\(\textbf{Substitution:}\) \(U_0=\frac{1}{2}(2.0\times10^{-6})(90)^2\,\text{J}\).
\(\textbf{Calculation:}\) \(90^2=8100\), so \(U_0=(1.0\times10^{-6})(8100)=8.1\times10^{-3}\,\text{J}\).
\(\textbf{Dielectric effect at constant charge:}\) \(U_f=\frac{U_0}{K}\).
\(\textbf{Final calculation:}\) \(U_f=\frac{8.1\times10^{-3}}{3}=2.7\times10^{-3}\,\text{J}\).
\(\textbf{Final result:}\) The final stored energy is \(2.7\times10^{-3}\,\text{J}\).
418. A \(4.0\,\mu\text{F}\) capacitor remains connected to a \(50\,\text{V}\) battery. A dielectric of relative permittivity \(K=5\) is fully inserted. What is the final energy stored in the capacitor?
ⓐ. \(2.0\times10^{-3}\,\text{J}\)
ⓑ. \(1.0\times10^{-2}\,\text{J}\)
ⓒ. \(2.5\times10^{-2}\,\text{J}\)
ⓓ. \(5.0\times10^{-2}\,\text{J}\)
Correct Answer: \(2.5\times10^{-2}\,\text{J}\)
Explanation: \(\textbf{Given:}\) \(C_0=4.0\,\mu\text{F}=4.0\times10^{-6}\,\text{F}\), \(V=50\,\text{V}\), and \(K=5\).
\(\textbf{Battery condition:}\) The battery keeps \(V\) constant.
\(\textbf{Final capacitance:}\) \(C_f=KC_0=5(4.0\times10^{-6})=2.0\times10^{-5}\,\text{F}\).
\(\textbf{Energy formula:}\) \(U_f=\frac{1}{2}C_fV^2\).
\(\textbf{Substitution:}\) \(U_f=\frac{1}{2}(2.0\times10^{-5})(50)^2\,\text{J}\).
\(\textbf{Square:}\) \(50^2=2500\).
\(\textbf{Calculation:}\) \(U_f=(1.0\times10^{-5})(2500)=2.5\times10^{-2}\,\text{J}\).
\(\textbf{Final result:}\) The final stored energy is \(2.5\times10^{-2}\,\text{J}\).
419. A capacitor is charged and then isolated. When a dielectric slab is fully inserted, the stored energy decreases. What is the main reason for this decrease?
ⓐ. The charge on the plates becomes zero
ⓑ. The plate separation automatically becomes zero
ⓒ. The battery removes energy while voltage is fixed
ⓓ. Capacitance increases while charge is fixed
Correct Answer: Capacitance increases while charge is fixed
Explanation: For an isolated charged capacitor, there is no conducting path for charge to leave or enter the plates. Hence the free charge \(Q\) remains constant. When a dielectric is fully inserted, the capacitance increases from \(C\) to \(KC\). At constant charge, the stored energy is \(U=\frac{Q^2}{2C}\). Increasing \(C\) therefore decreases \(U\). The battery is not involved because the capacitor is disconnected. The charge does not become zero, and the plate separation need not change. The decrease in stored energy is associated with work done during dielectric insertion and the reduction of the electric field inside the capacitor.
420. A parallel plate capacitor is isolated after charging. A dielectric of relative permittivity \(K\) is inserted fully. Which combination is correct?
ⓐ. \(Q\) decreases, \(V\) decreases, \(U\) decreases
ⓑ. \(Q\) constant, \(V\) decreases, \(U\) decreases
ⓒ. \(V\) constant, \(Q\) increases, \(U\) increases
ⓓ. \(Q\) constant, \(V\) increases, \(U\) increases
Correct Answer: \(Q\) constant, \(V\) decreases, \(U\) decreases
Explanation: When the capacitor is isolated, the charge magnitude \(Q\) on each plate cannot change if leakage is neglected. Full insertion of a dielectric increases capacitance from \(C_0\) to \(KC_0\). Since \(V=\frac{Q}{C}\), the potential difference decreases to \(\frac{V_0}{K}\). The electric field between the plates also decreases by the same factor under ideal conditions. For constant charge, the stored energy is \(U=\frac{Q^2}{2C}\). Since capacitance increases, stored energy decreases. Therefore, \(Q\) remains constant, while \(V\) and \(U\) decrease.
