401. A sensitive instrument is kept inside a closed hollow conductor placed in an external electrostatic field. If there is no charge inside the hollow region, what is the electric field inside the hollow region after equilibrium?
ⓐ. \(0\)
ⓑ. Equal to the external field
ⓒ. Twice the external field
ⓓ. Opposite to the external field
Correct Answer: \(0\)
Explanation: A closed conductor can shield its hollow interior from an external electrostatic field. When the external field is applied, free charges in the conductor redistribute over its surface. This redistribution creates an induced field that cancels the applied field inside the conducting material. If the cavity is closed and contains no charge, the field inside the hollow region is also zero in electrostatic equilibrium. This is the principle behind electrostatic shielding. The conductor does not remove the external field everywhere outside it. It only ensures that the field inside the conducting material and the empty closed cavity is zero.
402. A charge \(+q\) is placed inside the cavity of a neutral conductor. Which statement about electrostatic shielding is correct?
ⓐ. The cavity field is always zero even with \(+q\) inside.
ⓑ. The material has zero field; the cavity may not.
ⓒ. The conductor cannot develop induced charges on its surfaces.
ⓓ. The external field must be zero at all outside points.
Correct Answer: The material has zero field; the cavity may not.
Explanation: Electrostatic shielding guarantees zero electric field inside the conducting material in electrostatic equilibrium. If a charge \(+q\) is placed inside the cavity, it creates an electric field inside the hollow region. The conductor responds by inducing charge \(-q\) on the inner surface so that the field inside the conducting material remains zero. The cavity itself is not field-free because it contains a charge. If the conductor is neutral, charge \(+q\) appears on the outer surface. Therefore, shielding of the conducting material does not mean the cavity field is zero when a charge is present inside the cavity.
403. A closed metallic box is placed in an external electric field. Which property makes it useful for electrostatic shielding?
ⓐ. Its mass becomes very large in the field.
ⓑ. Its total charge must become positive.
ⓒ. Mobile charges cancel the internal field.
ⓓ. Its surface area becomes zero.
Correct Answer: Mobile charges cancel the internal field.
Explanation: Metals contain mobile charge carriers. When a metallic box is placed in an external electric field, these charges experience electric force and move slightly. Their redistribution produces an induced electric field inside the conductor. The motion continues until the net electric field inside the conducting material becomes zero. Once this happens, electrostatic equilibrium is reached. This cancellation of internal field is the reason a closed metallic box can shield its interior. The shielding effect does not depend on making the conductor positively charged or changing its surface area.
404. A conductor is in electrostatic equilibrium. Which condition must be true at its surface?
ⓐ. The electric field has only a tangential component.
ⓑ. The electric field just outside is normal to the surface.
ⓒ. The electric field inside is greater than the outside field.
ⓓ. The surface charge density is always zero.
Correct Answer: The electric field just outside is normal to the surface.
Explanation: In electrostatic equilibrium, free charges in a conductor are not moving. If the electric field at the surface had a tangential component, it would exert a tangential force on surface charges. Those charges would then move along the surface, contradicting equilibrium. Therefore, the tangential component of the electric field at the conductor surface must be zero. The field just outside the conductor can only be normal to the surface. The field inside the conducting material is zero. Surface charge density need not be zero; it can produce the normal external field.
405. A hollow conductor shields its empty cavity from an external electrostatic field. Which limitation of this statement is important?
ⓐ. The shielding fails if a charge is placed inside the cavity.
ⓑ. The shielding works only if the conductor is made of glass.
ⓒ. The shielding requires the external field to be zero everywhere.
ⓓ. The shielding means charge cannot appear on the conductor surface.
Correct Answer: The shielding fails if a charge is placed inside the cavity.
Explanation: A closed conductor can shield an empty cavity from external electrostatic fields. This means an external field does not produce a field inside the empty cavity after equilibrium. However, if a charge is placed inside the cavity, that charge itself produces an electric field in the cavity. The conductor still keeps the field inside the conducting material zero by inducing charges on its inner and outer surfaces. The cavity is then no longer field-free. Therefore, shielding from external fields should not be confused with removing the field due to a charge placed inside the cavity.
406. Which dimensional formula represents electric charge?
ⓐ. \([AT]\)
ⓑ. \([MLT^{-2}]\)
ⓒ. \([ML^2T^{-2}]\)
ⓓ. \([A^{-1}T]\)
Correct Answer: \([AT]\)
Explanation: \( \textbf{Basic relation:} \) Electric current is the rate of flow of charge, so \(I=\frac{q}{t}\).
\( \textbf{Solving for charge:} \) \(q=It\).
\( \textbf{Dimension of current:} \) Current has dimension \([A]\).
\( \textbf{Dimension of time:} \) Time has dimension \([T]\).
\( \textbf{Combining dimensions:} \) \([q]=[I][t]=[A][T]\).
\( \textbf{Final result:} \) The dimensional formula of electric charge is \([AT]\).
407. Which dimensional formula is correct for electric field \(E\)?
ⓐ. \([MLT^{-2}A^{-1}T^{-1}]\)
ⓑ. \([MLT^{-3}A^{-1}]\)
ⓒ. \([ML^2T^{-3}A^{-1}]\)
ⓓ. \([M^{-1}L^{-1}T^3A]\)
Correct Answer: \([MLT^{-3}A^{-1}]\)
Explanation: \( \textbf{Definition of field:} \) Electric field is force per unit charge, so \(E=\frac{F}{q}\).
\( \textbf{Dimension of force:} \) \([F]=[MLT^{-2}]\).
\( \textbf{Dimension of charge:} \) \([q]=[AT]\).
\( \textbf{Substitution into dimensions:} \) \([E]=\frac{[MLT^{-2}]}{[AT]}\).
\( \textbf{Simplification:} \) \([E]=[MLT^{-3}A^{-1}]\).
\( \textbf{Unit check:} \) This agrees with the unit \(\text{N C}^{-1}\).
\( \textbf{Final result:} \) The dimensional formula of \(E\) is \([MLT^{-3}A^{-1}]\).
408. Which unit is correctly paired with Coulomb's constant \(k\)?
ⓐ. \(\text{C}^2\text{N}^{-1}\text{m}^{-2}\)
ⓑ. \(\text{N C}^{-1}\)
ⓒ. \(\text{N m}^2\text{C}^{-2}\)
ⓓ. \(\text{N m}^2\text{C}^{-1}\)
Correct Answer: \(\text{N m}^2\text{C}^{-2}\)
Explanation: \( \textbf{Coulomb's law:} \) \(F=k\frac{|q_1q_2|}{r^2}\).
\( \textbf{Rearrange for \(k\):} \) \(k=\frac{Fr^2}{|q_1q_2|}\).
\( \textbf{Unit of force:} \) \(F\) has unit \(\text{N}\).
\( \textbf{Unit of \(r^2\):} \) \(r^2\) has unit \(\text{m}^2\).
\( \textbf{Unit of charge product:} \) \(q_1q_2\) has unit \(\text{C}^2\).
\( \textbf{Combining:} \) \([k]=\frac{\text{N m}^2}{\text{C}^2}=\text{N m}^2\text{C}^{-2}\).
\( \textbf{Final result:} \) The unit of \(k\) is \(\text{N m}^2\text{C}^{-2}\).
409. Which unit is correctly matched with vacuum permittivity \(\varepsilon_0\)?
ⓐ. \(\frac{\text{N m}^2}{\text{C}^2}\)
ⓑ. \(\frac{\text{C}^2}{\text{N m}^2}\)
ⓒ. \(\frac{\text{N}}{\text{C}}\)
ⓓ. \(\frac{\text{C}}{\text{m}}\)
Correct Answer: \(\frac{\text{C}^2}{\text{N m}^2}\)
Explanation: \( \textbf{Known relation:} \) \(k=\frac{1}{4\pi\varepsilon_0}\).
\( \textbf{Unit of \(k\):} \) \(k\) has unit \(\text{N m}^2\text{C}^{-2}\).
\( \textbf{Reciprocal relation:} \) Since \(\varepsilon_0\) is in the denominator of \(k\), its unit is the reciprocal of \(\text{N m}^2\text{C}^{-2}\).
\( \textbf{Unit of \(\varepsilon_0\):} \) \([\varepsilon_0]=\text{C}^2\text{N}^{-1}\text{m}^{-2}\).
\( \textbf{Common confusion:} \) \(\text{N m}^2\text{C}^{-2}\) is the unit of \(k\), not \(\varepsilon_0\).
\( \textbf{Final result:} \) The correct unit is \(\text{C}^2\text{N}^{-1}\text{m}^{-2}\).
410. Which dimensional check confirms that \(\Phi_E=EA\) has the unit of electric flux?
ⓐ. \((\text{N C}^{-1})(\text{m}^2)=\text{N m}^2\text{C}^{-1}\)
ⓑ. \((\text{N C}^{-1})(\text{m})=\text{N m C}^{-1}\)
ⓒ. \((\text{C m}^{-2})(\text{m}^2)=\text{C}\)
ⓓ. \((\text{N m}^2\text{C}^{-2})(\text{C})=\text{N m}^2\text{C}^{-1}\)
Correct Answer: \((\text{N C}^{-1})(\text{m}^2)=\text{N m}^2\text{C}^{-1}\)
Explanation: \( \textbf{Flux relation:} \) For a uniform field normal to a plane surface, \(\Phi_E=EA\).
\( \textbf{Unit of field:} \) Electric field has unit \(\text{N C}^{-1}\).
\( \textbf{Unit of area:} \) Area has unit \(\text{m}^2\).
\( \textbf{Unit multiplication:} \) \([\Phi_E]=(\text{N C}^{-1})(\text{m}^2)\).
\( \textbf{Simplified unit:} \) \([\Phi_E]=\text{N m}^2\text{C}^{-1}\).
\( \textbf{Final result:} \) Electric flux has unit \(\text{N m}^2\text{C}^{-1}\).
411. Which pair of physical quantity and SI unit is correctly matched?
ⓐ. Dipole moment: \(\text{N C}^{-1}\)
ⓑ. Electric field: \(\text{C m}\)
ⓒ. Electric flux: \(\text{N m}^2\text{C}^{-1}\)
ⓓ. Surface charge density: \(\text{N m}^2\text{C}^{-1}\)
Correct Answer: Electric flux: \(\text{N m}^2\text{C}^{-1}\)
Explanation: Electric flux is given by \(\Phi_E=\vec{E}\cdot\vec{A}\), so its unit is \((\text{N C}^{-1})(\text{m}^2)=\text{N m}^2\text{C}^{-1}\). Electric field has unit \(\text{N C}^{-1}\), not \(\text{C m}\). Electric dipole moment has unit \(\text{C m}\), because \(p=qd\). Linear charge density has unit \(\text{C m}^{-1}\), while surface charge density has unit \(\text{C m}^{-2}\). Correct unit matching prevents formula confusion. Therefore, electric flux paired with \(\text{N m}^2\text{C}^{-1}\) is correct.
412. Which Gaussian surface is most suitable for a charge distribution having spherical symmetry?
ⓐ. A plane sheet
ⓑ. A concentric spherical surface
ⓒ. A cylinder with arbitrary axis
ⓓ. A cube tilted at \(45^\circ\)
Correct Answer: A concentric spherical surface
Explanation: A Gaussian surface is chosen to exploit symmetry. For a spherically symmetric charge distribution, the electric field is radial. At a fixed distance from the centre, the field has the same magnitude at all points. A concentric spherical Gaussian surface has area vectors that are radial everywhere. Therefore, \(\vec{E}\) is parallel to \(d\vec{A}\), and the flux becomes \(E(4\pi r^2)\). This makes Gauss's law simple to apply. A plane, arbitrary cylinder, or tilted cube would not make the field constant and normal everywhere.
413. Which condition is most important for directly taking \(E\) outside the integral in \(\oint \vec{E}\cdot d\vec{A}\)?
ⓐ. The Gaussian surface must be very small.
ⓑ. The field must be uniform on the chosen part.
ⓒ. The enclosed charge must always be zero.
ⓓ. The surface must be made of conducting material.
Correct Answer: The field must be uniform on the chosen part.
Explanation: In Gauss-law calculations, \(E\) can be taken outside the flux integral only when symmetry makes its magnitude constant over the relevant surface part. The direction must also have a simple relation with \(d\vec{A}\), such as being parallel or perpendicular. This is why spherical, cylindrical, and pillbox surfaces are chosen for suitable charge symmetries. If \(E\) varies from point to point, the integral cannot be simplified as \(E\) times total area. The Gaussian surface is imaginary and need not be a conductor. The enclosed charge need not be zero. Therefore, constant field magnitude on the chosen surface part is essential.
414. Which symmetry and Gaussian surface pairing is incorrect?
ⓐ. Point charge: spherical Gaussian surface
ⓑ. Infinite line charge: cylindrical Gaussian surface
ⓒ. Infinite plane sheet: pillbox Gaussian surface
ⓓ. Infinite line charge: spherical Gaussian surface
Correct Answer: Infinite line charge: spherical Gaussian surface
Explanation: A point charge has spherical symmetry, so a spherical Gaussian surface is convenient. An infinite line charge has cylindrical symmetry, so a coaxial cylindrical Gaussian surface is convenient. An infinite plane sheet has planar symmetry, so a pillbox Gaussian surface is convenient. A spherical surface around an infinite line charge does not match the symmetry of the field. The field due to an infinite line charge depends on perpendicular distance from the line, not distance from a single point. Therefore, a spherical Gaussian surface does not allow direct simplification for an infinite line charge. The incorrect pairing is the infinite line charge with a spherical surface.
415. A student chooses a Gaussian surface that encloses the correct charge but does not match the symmetry of the electric field. What is the best conclusion?
ⓐ. Gauss's law becomes false for that surface.
ⓑ. Flux still follows Gauss's law, but finding \(E\) may be hard.
ⓒ. The enclosed charge automatically becomes zero.
ⓓ. The electric field must be constant on every closed surface.
Correct Answer: Flux still follows Gauss's law, but finding \(E\) may be hard.
Explanation: Gauss's law is valid for every closed surface. It states that \(\oint \vec{E}\cdot d\vec{A}=\frac{q_{\text{enc}}}{\varepsilon_0}\). Therefore, even a poorly chosen closed surface gives the correct total flux if the enclosed charge is known. However, finding the electric field from Gauss's law requires symmetry. Without symmetry, \(E\) may vary over the surface and may not have a simple angle with \(d\vec{A}\). Then the flux integral cannot be reduced to a simple expression like \(EA\). Thus, the law remains true, but the surface may not be useful for calculating \(E\).
416. Why can Gauss's law easily give \(E\) for a point charge but not usually for an arbitrary group of charges?
ⓐ. Symmetry makes \(E\) constant on a suitable spherical surface.
ⓑ. Gauss's law is valid only for a single point charge.
ⓒ. Arbitrary groups of charges produce zero electric flux always.
ⓓ. A Gaussian surface cannot enclose more than one charge.
Correct Answer: Symmetry makes \(E\) constant on a suitable spherical surface.
Explanation: Gauss's law is valid for any closed surface and any charge distribution. However, using it to find \(E\) directly requires enough symmetry. For a point charge, a spherical Gaussian surface centred on the charge makes the field magnitude constant everywhere on the sphere. The field is also parallel to the outward area vector at every point. Therefore, the flux integral simplifies to \(E(4\pi r^2)\). For an arbitrary group of charges, the field may vary in magnitude and direction over the surface. Then Gauss's law still gives total flux, but it may not easily give the field at each point.
417. Which statement is true about electric flux through a closed surface enclosing a dipole?
ⓐ. It is \(\frac{2q}{\varepsilon_0}\).
ⓑ. It is zero because the net enclosed charge is zero.
ⓒ. It is \(\frac{q}{2\varepsilon_0}\).
ⓓ. It is non-zero because the dipole moment is non-zero.
Correct Answer: It is zero because the net enclosed charge is zero.
Explanation: \( \textbf{Dipole charges:} \) An electric dipole consists of \(+q\) and \(-q\).
\( \textbf{Net enclosed charge:} \) If the closed surface encloses the whole dipole, \(q_{\text{enc}}=+q+(-q)=0\).
\( \textbf{Gauss's law:} \) \(\Phi_E=\frac{q_{\text{enc}}}{\varepsilon_0}\).
\( \textbf{Substitution:} \) \(\Phi_E=\frac{0}{\varepsilon_0}=0\).
\( \textbf{Important distinction:} \) Zero net flux does not mean the electric field is zero everywhere on the surface.
\( \textbf{Dipole moment note:} \) A dipole may have non-zero \(\vec{p}\), but Gauss's law depends on net enclosed charge for total flux.
\( \textbf{Final result:} \) The net electric flux is zero.
418. A Gaussian surface encloses only the positive charge \(+q\) of an electric dipole, while the negative charge \(-q\) lies outside. What is the net flux through the surface?
ⓐ. \(0\)
ⓑ. \(\frac{-q}{\varepsilon_0}\)
ⓒ. \(\frac{+q}{\varepsilon_0}\)
ⓓ. \(\frac{2q}{\varepsilon_0}\)
Correct Answer: \(\frac{+q}{\varepsilon_0}\)
Explanation: \( \textbf{Charges considered:} \) The Gaussian surface encloses \(+q\) only.
\( \textbf{Outside charge:} \) The charge \(-q\) is outside and does not contribute to net flux.
\( \textbf{Enclosed charge:} \) \(q_{\text{enc}}=+q\).
\( \textbf{Gauss's law:} \) \(\Phi_E=\frac{q_{\text{enc}}}{\varepsilon_0}\).
\( \textbf{Substitution:} \) \(\Phi_E=\frac{+q}{\varepsilon_0}\).
\( \textbf{Clarification:} \) The outside charge may alter the field at points on the surface, but its total flux contribution through the closed surface is zero.
\( \textbf{Final result:} \) The net flux is \(\frac{+q}{\varepsilon_0}\).
419. A cube of side \(a\) encloses a point charge \(q\) at its centre. What is the total electric flux through one face of the cube?
ⓐ. \(\frac{q}{\varepsilon_0}\)
ⓑ. \(\frac{q}{2\varepsilon_0}\)
ⓒ. \(\frac{q}{4\varepsilon_0}\)
ⓓ. \(\frac{q}{6\varepsilon_0}\)
Correct Answer: \(\frac{q}{6\varepsilon_0}\)
Explanation: \( \textbf{Total enclosed charge:} \) The cube encloses charge \(q\).
\( \textbf{Total flux by Gauss's law:} \) \(\Phi_{\text{total}}=\frac{q}{\varepsilon_0}\).
\( \textbf{Symmetry of position:} \) Since the charge is at the centre of the cube, all six faces are equivalent.
\( \textbf{Equal sharing of flux:} \) The total flux is divided equally among the six faces.
\( \textbf{Flux through one face:} \) \(\Phi_{\text{face}}=\frac{1}{6}\Phi_{\text{total}}\).
\( \textbf{Substitution:} \) \(\Phi_{\text{face}}=\frac{1}{6}\cdot\frac{q}{\varepsilon_0}\).
\( \textbf{Final result:} \) The flux through one face is \(\frac{q}{6\varepsilon_0}\).
420. A point charge \(q\) is placed at the centre of a cube. If the cube is replaced by a larger cube with the same centre, what happens to the total electric flux through the cube?
ⓐ. It remains \(\frac{q}{\varepsilon_0}\).
ⓑ. It becomes \(\frac{q}{2\varepsilon_0}\).
ⓒ. It becomes \(\frac{q}{4\varepsilon_0}\).
ⓓ. It becomes \(\frac{2q}{\varepsilon_0}\).
Correct Answer: It remains \(\frac{q}{\varepsilon_0}\).
Explanation: \( \textbf{Gauss's law:} \) \(\Phi_E=\frac{q_{\text{enc}}}{\varepsilon_0}\).
\( \textbf{Original cube:} \) It encloses the charge \(q\), so \(\Phi_E=\frac{q}{\varepsilon_0}\).
\( \textbf{Larger cube:} \) It still encloses the same charge \(q\).
\( \textbf{Surface-size effect:} \) Increasing the cube size changes the field values and areas locally, but not the net enclosed charge.
\( \textbf{Flux dependence:} \) Total closed-surface flux depends only on \(q_{\text{enc}}\).
\( \textbf{Final result:} \) The total flux remains \(\frac{q}{\varepsilon_0}\).
