501. Which part of a Van de Graaff generator sprays charge onto the moving belt near the lower pulley?
ⓐ. The upper collecting comb
ⓑ. The lower spray comb
ⓒ. The hollow dome surface
ⓓ. The insulating support column
Correct Answer: The lower spray comb
Explanation: In a Van de Graaff generator, the lower comb is connected to a high-voltage source. Its sharp points produce a strong electric field near the belt. This strong field ionises nearby air and helps spray charge onto the moving insulating belt. The belt then carries the charge upward. The upper comb collects the charge from the belt and transfers it to the conducting dome. The insulating support mainly provides mechanical support and prevents leakage. The dome stores the transferred charge on its outer surface. Therefore, the lower spray comb is responsible for charging the belt.
502. Why are sharp combs used in a Van de Graaff generator?
ⓐ. They reduce the electric field near their tips
ⓑ. They make the belt conducting throughout
ⓒ. They create strong fields at their tips
ⓓ. They keep the dome potential exactly zero
Correct Answer: They create strong fields at their tips
Explanation: Sharp points on a conductor have a small radius of curvature. Charge density becomes large near such sharp points. Since the electric field just outside a conductor is related to surface charge density by \(E=\frac{\sigma}{\varepsilon_0}\), the field near sharp points becomes very strong. This strong local field helps ionise the surrounding air. The ionised air allows charge to be sprayed onto the belt or collected from it. Without sharp combs, charge transfer would be much less effective. The combs do not make the belt conducting; the belt remains insulating. Hence sharp combs are used to create strong local fields for charge transfer.
503. In a Van de Graaff generator, what happens to charge brought to the inside of the hollow conducting dome?
ⓐ. It remains fixed on the inner surface
ⓑ. It is destroyed by the electric field
ⓒ. It moves to the outer surface of the dome
ⓓ. It stays uniformly inside the metal volume
Correct Answer: It moves to the outer surface of the dome
Explanation: A hollow conducting dome is a conductor in electrostatic equilibrium after charge redistribution. Excess charge supplied to a conductor resides on its outer surface. If charge is delivered to the inside of the dome, free charges in the conductor rearrange until the conducting material has zero internal electric field. The excess charge then appears on the outer surface. This is the key reason repeated charge delivery to the inside can keep increasing the dome charge. The charge is not destroyed, and it does not remain distributed through the metal volume. It also does not stay permanently on the inner wall when no internal isolated charge is present. Therefore, the transferred charge moves to the outer surface of the dome.
504. A spherical Van de Graaff dome of radius \(0.50\,\text{m}\) is at potential \(3.0\times10^6\,\text{V}\). What is the electric field just outside its surface, approximately?
ⓐ. \(1.5\times10^6\,\text{V m}^{-1}\)
ⓑ. \(3.0\times10^6\,\text{V m}^{-1}\)
ⓒ. \(4.5\times10^6\,\text{V m}^{-1}\)
ⓓ. \(6.0\times10^6\,\text{V m}^{-1}\)
Correct Answer: \(6.0\times10^6\,\text{V m}^{-1}\)
Explanation: \(\textbf{Given:}\) Radius \(R=0.50\,\text{m}\) and potential \(V=3.0\times10^6\,\text{V}\).
\(\textbf{Required:}\) Electric field just outside the spherical dome.
\(\textbf{Relation for conducting sphere:}\) \(E=\frac{V}{R}\).
\(\textbf{Why it applies:}\) For a charged conducting sphere, \(V=\frac{kQ}{R}\) and \(E=\frac{kQ}{R^2}\), so \(E=\frac{V}{R}\).
\(\textbf{Substitution:}\) \(E=\frac{3.0\times10^6}{0.50}\,\text{V m}^{-1}\).
\(\textbf{Calculation:}\) Dividing by \(0.50\) is the same as multiplying by \(2\).
\(\textbf{Result:}\) \(E=6.0\times10^6\,\text{V m}^{-1}\).
\(\textbf{Final result:}\) The electric field just outside the dome is \(6.0\times10^6\,\text{V m}^{-1}\).
505. Why is the maximum potential of a Van de Graaff generator limited in air?
ⓐ. Air breaks down at too large an electric field
ⓑ. The dome capacitance becomes exactly zero
ⓒ. The belt stops being insulating at all voltages
ⓓ. Charge cannot reside on a spherical conductor
Correct Answer: Air breaks down at too large an electric field
Explanation: As charge accumulates on the dome, the electric potential of the dome increases. The electric field near the dome surface also increases. If the field becomes greater than the dielectric strength of air, air gets ionised. Ionised air becomes conducting and allows charge to leak away from the dome. This discharge limits further rise of potential. Sparks may occur when the field becomes sufficiently large. The capacitance of the dome does not become zero, and charge can reside on the outer surface of a spherical conductor. Therefore, air breakdown is the main practical limit on maximum potential.
506. A spherical dome has radius \(1.0\,\text{m}\). If air breakdown occurs near \(3.0\times10^6\,\text{V m}^{-1}\), what is the approximate maximum potential before breakdown?
ⓐ. \(1.0\times10^6\,\text{V}\)
ⓑ. \(3.0\times10^6\,\text{V}\)
ⓒ. \(6.0\times10^6\,\text{V}\)
ⓓ. \(9.0\times10^6\,\text{V}\)
Correct Answer: \(3.0\times10^6\,\text{V}\)
Explanation: \(\textbf{Given:}\) Dome radius \(R=1.0\,\text{m}\) and breakdown field \(E_{\max}=3.0\times10^6\,\text{V m}^{-1}\).
\(\textbf{Required:}\) Approximate maximum potential \(V_{\max}\).
\(\textbf{Surface field relation:}\) For a spherical conductor, \(E=\frac{V}{R}\).
\(\textbf{Rearrangement:}\) \(V=ER\).
\(\textbf{Substitution:}\) \(V_{\max}=(3.0\times10^6)(1.0)\,\text{V}\).
\(\textbf{Calculation:}\) \(V_{\max}=3.0\times10^6\,\text{V}\).
\(\textbf{Physical meaning:}\) A larger radius permits a higher potential for the same surface field limit.
\(\textbf{Final result:}\) The approximate maximum potential is \(3.0\times10^6\,\text{V}\).
507. Why is a large spherical dome preferred in a Van de Graaff generator?
ⓐ. It decreases capacitance and increases leakage
ⓑ. It makes charge remain inside the hollow space
ⓒ. It reduces surface field for a given potential
ⓓ. It prevents any charge from reaching the outer surface
Correct Answer: It reduces surface field for a given potential
Explanation: For a spherical conducting dome, the surface field is approximately \(E=\frac{V}{R}\). For the same potential \(V\), increasing the radius \(R\) reduces the electric field at the surface. A lower surface field reduces the chance of air breakdown and charge leakage. A larger dome also has larger capacitance because \(C=4\pi\varepsilon_0R\). This allows more charge to be stored for a given potential. The dome still carries charge on its outer surface. The large spherical shape also avoids sharp regions where charge density and field would become excessive. Hence a large spherical dome helps achieve high potential safely by reducing surface field.
508. If a Van de Graaff dome has radius \(R\), which proportionality best describes its capacitance in air?
ⓐ. \(C\propto R\)
ⓑ. \(C\propto R^2\)
ⓒ. \(C\propto \frac{1}{R}\)
ⓓ. \(C\propto \frac{1}{R^2}\)
Correct Answer: \(C\propto R\)
Explanation: \(\textbf{Dome model:}\) The dome may be approximated as an isolated conducting sphere.
\(\textbf{Capacitance of sphere:}\) \(C=4\pi\varepsilon_0R\) in air or vacuum approximately.
\(\textbf{Proportionality:}\) Since \(4\pi\varepsilon_0\) is constant, \(C\propto R\).
\(\textbf{Meaning:}\) A larger dome has larger capacitance.
\(\textbf{Potential effect:}\) For a given charge \(Q\), \(V=\frac{Q}{C}\), so a larger capacitance gives lower potential for the same charge.
\(\textbf{Practical value:}\) A larger radius also reduces surface field at a given potential.
\(\textbf{Final result:}\) The capacitance is directly proportional to \(R\).
509. A Van de Graaff dome receives charge at a steady rate \(I\). If its capacitance is \(C\) and leakage is neglected, how does its potential change with time from an initially uncharged state?
ⓐ. \(V=\frac{It}{C}\)
ⓑ. \(V=ICt\)
ⓒ. \(V=\frac{C}{It}\)
ⓓ. \(V=\frac{I}{Ct^2}\)
Correct Answer: \(V=\frac{It}{C}\)
Explanation: \(\textbf{Charge rate:}\) Current is the rate of charge transfer, so \(I=\frac{dQ}{dt}\).
\(\textbf{Steady rate:}\) If the dome starts uncharged and receives charge at constant rate \(I\), then \(Q=It\).
\(\textbf{Capacitance relation:}\) \(V=\frac{Q}{C}\).
\(\textbf{Substitution:}\) \(V=\frac{It}{C}\).
\(\textbf{Meaning:}\) If leakage is neglected, the potential rises linearly with time.
\(\textbf{Practical note:}\) In real air, leakage and breakdown eventually limit the voltage.
\(\textbf{Final result:}\) The potential is \(V=\frac{It}{C}\).
510. A Van de Graaff dome has capacitance \(10\,\text{pF}\) and receives charge at \(2.0\,\mu\text{A}\). Neglecting leakage, what potential is reached after \(5.0\,\text{s}\)?
ⓐ. \(1.0\times10^5\,\text{V}\)
ⓑ. \(5.0\times10^5\,\text{V}\)
ⓒ. \(1.0\times10^6\,\text{V}\)
ⓓ. \(2.0\times10^6\,\text{V}\)
Correct Answer: \(1.0\times10^6\,\text{V}\)
Explanation: \(\textbf{Given:}\) \(C=10\,\text{pF}=10\times10^{-12}\,\text{F}\), \(I=2.0\,\mu\text{A}=2.0\times10^{-6}\,\text{A}\), and \(t=5.0\,\text{s}\).
\(\textbf{Required:}\) Potential \(V\).
\(\textbf{Charge delivered:}\) \(Q=It\).
\(\textbf{Substitution for charge:}\) \(Q=(2.0\times10^{-6})(5.0)\,\text{C}=1.0\times10^{-5}\,\text{C}\).
\(\textbf{Potential relation:}\) \(V=\frac{Q}{C}\).
\(\textbf{Substitution:}\) \(V=\frac{1.0\times10^{-5}}{10\times10^{-12}}\,\text{V}\).
\(\textbf{Simplification:}\) \(10\times10^{-12}=1.0\times10^{-11}\).
\(\textbf{Calculation:}\) \(V=1.0\times10^6\,\text{V}\).
\(\textbf{Final result:}\) The potential reached is \(1.0\times10^6\,\text{V}\).
511. Which statement correctly explains electrostatic shielding by the dome of a Van de Graaff generator?
ⓐ. The field inside conducting metal is zero
ⓑ. The field inside the metal is always greater than the outside field
ⓒ. The dome stores all excess charge in its hollow interior
ⓓ. The dome becomes an insulator when the belt moves
Correct Answer: The field inside conducting metal is zero
Explanation: A conducting dome has mobile charges that redistribute until electrostatic equilibrium is reached. In this state, the electric field inside the conducting material is zero. This property is closely related to electrostatic shielding. The hollow region can be protected from external electrostatic fields when no charge is placed inside it. Excess charge supplied to the dome appears on the outer surface. The field outside may be large, but the conducting material itself has zero electric field. The dome does not become an insulator, and it does not store excess charge throughout the hollow interior. Therefore, shielding is based on zero electric field inside a conductor in electrostatic equilibrium.
512. A Van de Graaff generator is commonly used to accelerate charged particles. Which energy relation is used when a particle of charge \(q\) moves through potential difference \(V\)?
ⓐ. \(K=qV\)
ⓑ. \(K=\frac{q}{V}\)
ⓒ. \(K=\frac{V}{q}\)
ⓓ. \(K=qV^2\)
Correct Answer: \(K=qV\)
Explanation: \(\textbf{Physical situation:}\) A charged particle is accelerated through a potential difference \(V\).
\(\textbf{Work-energy idea:}\) Electric work done on the particle becomes kinetic energy if other losses are neglected.
\(\textbf{Electric work:}\) Work done by the electric field has magnitude \(qV\) for charge magnitude \(q\).
\(\textbf{Energy gained:}\) The kinetic energy gained is \(K=qV\).
\(\textbf{Unit check:}\) \(\text{C}\cdot\text{V}=\text{J}\).
\(\textbf{Application:}\) A high-voltage generator can therefore give large kinetic energy to charged particles.
\(\textbf{Final result:}\) The energy gained is \(K=qV\).
513. A proton is accelerated through a potential difference of \(2.0\times10^6\,\text{V}\). What kinetic energy does it gain in electron-volt notation?
ⓐ. \(1.0\,\text{MeV}\)
ⓑ. \(2.0\,\text{MeV}\)
ⓒ. \(4.0\,\text{MeV}\)
ⓓ. \(8.0\,\text{MeV}\)
Correct Answer: \(2.0\,\text{MeV}\)
Explanation: \(\textbf{Given:}\) A proton has charge magnitude \(e\), and the potential difference is \(2.0\times10^6\,\text{V}\).
\(\textbf{Energy relation:}\) \(K=qV\).
\(\textbf{Electron-volt meaning:}\) A charge \(e\) moving through \(1\,\text{V}\) gains \(1\,\text{eV}\).
\(\textbf{For \(2.0\times10^6\,\text{V}\):}\) A proton gains \(2.0\times10^6\,\text{eV}\).
\(\textbf{Conversion:}\) \(10^6\,\text{eV}=1\,\text{MeV}\).
\(\textbf{Final energy:}\) \(2.0\times10^6\,\text{eV}=2.0\,\text{MeV}\).
\(\textbf{Final result:}\) The kinetic energy gained is \(2.0\,\text{MeV}\).
514. Why does charge leakage from a Van de Graaff dome become more serious near sharp projections?
ⓐ. Sharp projections raise local charge density and field
ⓑ. Sharp projections make the dome capacitance infinite
ⓒ. Sharp projections force charge into the hollow interior
ⓓ. Sharp projections make air a perfect insulator
Correct Answer: Sharp projections raise local charge density and field
Explanation: On a conductor, charge density becomes larger at regions with smaller radius of curvature. Sharp projections have very small radius of curvature. Therefore, the local surface charge density near a sharp projection becomes large. Since the electric field just outside a conductor is \(E=\frac{\sigma}{\varepsilon_0}\), the local field also becomes large. A large field can ionise air more easily, causing corona discharge or leakage. This reduces the ability of the dome to maintain high potential. A smooth spherical surface is preferred because it avoids excessive local fields. Hence leakage is more serious near sharp projections.
515. Which design feature helps a Van de Graaff generator reach a higher potential before air breakdown?
ⓐ. A small dome with sharp corners
ⓑ. A large smooth spherical dome
ⓒ. A conducting belt exposed to the dome
ⓓ. A pointed dome connected to earth
Correct Answer: A large smooth spherical dome
Explanation: A large smooth spherical dome reduces excessive electric field concentration. For a sphere, the surface field is approximately \(E=\frac{V}{R}\), so a larger radius allows a larger potential before the surface field reaches the breakdown value of air. Smoothness also matters because sharp points concentrate charge and create very large local fields. Such local fields cause corona discharge and charge leakage. A pointed dome would break down more easily. Earthing the dome would prevent it from rising to high potential. Therefore, a large smooth spherical dome helps the generator reach higher potential.
516. A Van de Graaff dome is charged to potential \(V\). If its radius is doubled while the same maximum safe surface field is allowed, what happens to the maximum safe potential?
ⓐ. It becomes half
ⓑ. It remains unchanged
ⓒ. It becomes double
ⓓ. It becomes four times
Correct Answer: It becomes double
Explanation: \(\textbf{Surface field relation:}\) For a spherical conducting dome, \(E=\frac{V}{R}\).
\(\textbf{Maximum safe condition:}\) The largest safe field \(E_{\max}\) is fixed by breakdown of air.
\(\textbf{Rearrangement:}\) \(V_{\max}=E_{\max}R\).
\(\textbf{Radius change:}\) If \(R\) is doubled, \(R'=2R\).
\(\textbf{New maximum potential:}\) \(V'_{\max}=E_{\max}(2R)\).
\(\textbf{Simplification:}\) \(V'_{\max}=2E_{\max}R=2V_{\max}\).
\(\textbf{Final result:}\) The maximum safe potential becomes double.
517. Which statement best distinguishes a dielectric from a conductor in an external electric field?
ⓐ. A dielectric has free charges that move through its entire volume
ⓑ. A dielectric polarizes by bound-charge displacement or alignment
ⓒ. A dielectric always makes the net electric field inside it exactly zero
ⓓ. A dielectric stores charge only on its outer conducting surface
Correct Answer: A dielectric polarizes by bound-charge displacement or alignment
Explanation: A dielectric is an insulating material, so it does not have free charges that move through the material as in a conductor. In an external electric field, its bound positive and negative charges are slightly displaced or its permanent dipoles partly align. This produces polarization rather than conduction. The induced bound charges create an internal field that opposes the applied field, reducing the net field in many simple capacitor situations. The field inside a dielectric is generally reduced, not necessarily zero. Zero internal electrostatic field is a conductor property in electrostatic equilibrium, not a general dielectric property.
518. Which pair correctly describes non-polar and polar dielectrics before an external electric field is applied?
ⓐ. Non-polar molecules have permanent dipoles; polar molecules have no dipoles
ⓑ. Non-polar molecules are conductors; polar molecules are insulators
ⓒ. Non-polar molecules have separated charge centres; polar molecules have coincident charge centres
ⓓ. Non-polar molecules have coincident charge centres; polar molecules have permanent dipoles
Correct Answer: Non-polar molecules have coincident charge centres; polar molecules have permanent dipoles
Explanation: In a non-polar molecule, the centres of positive and negative charge coincide when no external field is applied. Such a molecule has no permanent dipole moment, but an external field can induce a dipole moment. In a polar molecule, the centres of positive and negative charge are naturally separated. Hence polar molecules possess permanent dipole moments even without an external field. In an external field, polar molecules tend to align partially with the field, though thermal agitation prevents perfect alignment. Both polar and non-polar dielectrics are insulating materials, not conductors.
519. Polarization \(\vec{P}\) of a dielectric is defined as dipole moment per unit volume. What is the SI unit of \(\vec{P}\)?
ⓐ. \(\text{C m}^{-2}\)
ⓑ. \(\text{C m}^{-1}\)
ⓒ. \(\text{C m}^{2}\)
ⓓ. \(\text{N C}^{-1}\)
Correct Answer: \(\text{C m}^{-2}\)
Explanation: \(\textbf{Definition:}\) Polarization is dipole moment per unit volume.
\(\textbf{Dipole moment unit:}\) Electric dipole moment has unit \(\text{C m}\).
\(\textbf{Volume unit:}\) Volume has unit \(\text{m}^3\).
\(\textbf{Unit of polarization:}\) \(\frac{\text{C m}}{\text{m}^3}=\text{C m}^{-2}\).
\(\textbf{Meaning:}\) This unit is the same as surface charge density, which is consistent with polarization producing bound surface charge.
\(\textbf{Final result:}\) The SI unit of polarization is \(\text{C m}^{-2}\).
520. When a dielectric is placed in an external electric field, why is the net electric field inside it usually reduced?
ⓐ. The dielectric creates free current opposite to the field
ⓑ. The dielectric converts all field energy into heat instantly
ⓒ. Bound charges produce an induced field in the same direction as the applied field
ⓓ. Bound charges produce an induced field opposite to the applied field
Correct Answer: Bound charges produce an induced field opposite to the applied field
Explanation: In a dielectric, charges are bound to atoms or molecules and cannot move freely through the material. An external field slightly displaces the positive and negative charge centres or partially aligns molecular dipoles. This creates polarization and bound surface charges. The field produced by these bound charges is opposite to the applied field inside the dielectric. Therefore, the resultant field is smaller than the original applied field in many standard capacitor situations. The dielectric does not need free conduction current to reduce the field. The reduction is due to polarization of bound charges.
