401. A charged conductor is in electrostatic equilibrium. Just outside its surface, the electric field must be
ⓐ. normal to the surface
ⓑ. tangential to the surface
ⓒ. zero for every charged conductor
ⓓ. directed along any arbitrary line on the surface
Correct Answer: normal to the surface
Explanation: In electrostatic equilibrium, charges on a conductor are at rest. If the electric field had a tangential component at the surface, free charges would feel a force along the surface and would continue moving. Therefore, the tangential component must be zero. The remaining field just outside the conductor is normal to the surface. The field inside the conducting material is zero, but the field just outside a charged conductor need not be zero. This normal-field condition is a direct consequence of free charge mobility on conductors.
402. A small pillbox Gaussian surface is drawn across the surface of a charged conductor in electrostatic equilibrium. The field just outside is \(E\), the surface charge density is \(\sigma\), and the field inside the conductor is zero. Gauss’s law gives
ⓐ. \(E=\frac{\sigma}{\varepsilon_0}\)
ⓑ. \(E=\frac{\sigma}{2\varepsilon_0}\)
ⓒ. \(E=\frac{2\sigma}{\varepsilon_0}\)
ⓓ. \(E=\frac{\varepsilon_0}{\sigma}\)
Correct Answer: \(E=\frac{\sigma}{\varepsilon_0}\)
Explanation: For a charged conductor, the electric field inside the conducting material is zero. A thin pillbox across the surface encloses charge \(q_{\text{enc}}=\sigma A\), where \(A\) is the area of its outer flat face. The inner face contributes no flux because the field inside the conductor is zero. The curved surface can be made very small, so its flux is negligible in the local argument. The outside face contributes flux \(EA\). Gauss’s law gives \(EA=\frac{\sigma A}{\varepsilon_0}\), so \(E=\frac{\sigma}{\varepsilon_0}\).
403. Near a positively charged conductor surface, the electric field just outside is directed
ⓐ. tangential to the surface
ⓑ. inward normal to the surface
ⓒ. outward normal to the surface
ⓓ. along the surface only where \(\sigma\) is large
Correct Answer: outward normal to the surface
Explanation: A positive surface charge produces an electric field directed away from the surface. For a conductor in electrostatic equilibrium, the field cannot have a tangential component, otherwise free charges would move along the surface. Therefore, the field is normal to the surface. Since the charge is positive, the normal direction is outward. For a negatively charged conductor, the field just outside would be inward normal. The sign of \(\sigma\) decides inward or outward, while conductor equilibrium removes the tangential component.
404. A conducting body has a sharply curved tip and a broad rounded region. In electrostatic equilibrium, the surface charge density is usually larger near the sharper tip. This means the electric field just outside the sharper tip is
ⓐ. weaker because sharp tips cannot hold charge
ⓑ. stronger; \(E=\frac{\sigma}{\varepsilon_0}\)
ⓒ. zero because the tip is part of a conductor
ⓓ. parallel to the conductor surface
Correct Answer: stronger; \(E=\frac{\sigma}{\varepsilon_0}\)
Explanation: On a conductor, excess charge redistributes over the outer surface. Regions of higher curvature, such as sharp tips, usually have larger surface charge density. Just outside a conductor, the local field magnitude is related to surface charge density by \(E=\frac{\sigma}{\varepsilon_0}\). Therefore, a larger \(\sigma\) near the tip corresponds to a stronger electric field just outside that region. The field inside the conductor is zero, but the outside field near a charged surface can be large. The field remains normal to the surface, not parallel to it.
405. Study the statements about a charged conductor in electrostatic equilibrium.
I. The electric field inside the conductor is zero.
II. The electric field just outside the surface is normal to the surface.
III. A tangential electric field at the surface would make free charges move.
IV. The field just outside every charged conductor surface is always zero.
The supported statements are
ⓐ. I, II, III and IV
ⓑ. I and IV only
ⓒ. II and IV only
ⓓ. I, II and III
Correct Answer: I, II and III
Explanation: Statement I is true because free charges in a conductor redistribute until the internal electric field becomes zero. Statement II is true because the surface field cannot have a tangential component in electrostatic equilibrium. Statement III explains the reason: a tangential component would exert force on mobile charges and cause surface current. Statement IV is false because a charged conductor generally has a non-zero electric field just outside its surface. The field just outside is \(E=\frac{\sigma}{\varepsilon_0}\), so it is zero only where the local surface charge density is zero.
406. A graph of \(E\) versus \(r\) for a uniformly charged thin spherical shell of radius \(R\) is compared with a graph for an infinite line charge. The correct comparison is
ⓐ. shell: \(E\propto r^{-1}\) for \(r\lt R\); line: \(E\propto r^{-2}\)
ⓑ. shell: inside \(0\), outside \(r^{-2}\); line: \(r^{-1}\)
ⓒ. shell: \(E=0\) for \(r\gt R\); line: \(E\propto r^0\)
ⓓ. shell: \(E\propto r\) outside; line: \(E\propto r^2\)
Correct Answer: shell: inside \(0\), outside \(r^{-2}\); line: \(r^{-1}\)
Explanation: A uniformly charged thin spherical shell has zero electric field inside it. Outside the shell, it behaves like a point charge at the centre, so the field varies as \(r^{-2}\). An infinitely long uniformly charged line has cylindrical symmetry. Gauss’s law gives \(E=\frac{\lambda}{2\pi\varepsilon_0r}\), so its field varies as \(r^{-1}\). These graph shapes are different because the charge distributions have different symmetry. A single inverse-square pattern should not be assigned to every source.
407. A graph has three labelled curves for electric field magnitude \(E\) against distance \(r\). Curve P falls as \(r^{-2}\), curve Q falls as \(r^{-1}\), and curve R is horizontal. The most suitable source matching is
ⓐ. P: point charge, Q: infinite line charge, R: infinite plane sheet
ⓑ. P: infinite plane sheet, Q: point charge, R: infinite line charge
ⓒ. P: infinite line charge, Q: infinite plane sheet, R: point charge
ⓓ. P: conductor interior, Q: point charge, R: spherical shell interior
Correct Answer: P: point charge, Q: infinite line charge, R: infinite plane sheet
Explanation: A point charge gives \(E\propto\frac{1}{r^2}\), so it matches curve P. An infinite line charge gives \(E\propto\frac{1}{r}\), so it matches curve Q. An ideal infinite plane sheet gives \(E=\frac{\sigma}{2\varepsilon_0}\), independent of distance, so it matches a horizontal graph. The graph comparison is a compact way to recognize source symmetry. Spherical spreading, cylindrical spreading, and planar symmetry produce different distance dependences. The conductor interior has \(E=0\), not a general horizontal non-zero sheet field.
408. A conductor surface has local surface charge density \(3.0\times10^{-8}\,\text{C m}^{-2}\). Taking \(\varepsilon_0=8.85\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}\), the field just outside the surface is closest to
ⓐ. \(1.7\times10^3\,\text{N C}^{-1}\)
ⓑ. \(3.4\times10^3\,\text{N C}^{-1}\)
ⓒ. \(2.7\times10^{-19}\,\text{N C}^{-1}\)
ⓓ. \(3.0\times10^{-8}\,\text{N C}^{-1}\)
Correct Answer: \(3.4\times10^3\,\text{N C}^{-1}\)
Explanation: \( \textbf{Given surface charge density:} \) \(\sigma=3.0\times10^{-8}\,\text{C m}^{-2}\).
\( \textbf{Permittivity:} \) \(\varepsilon_0=8.85\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}\).
\( \textbf{Conductor surface field relation:} \)
\[
E=\frac{\sigma}{\varepsilon_0}
\]
\( \textbf{Substitution:} \)
\[
E=\frac{3.0\times10^{-8}}{8.85\times10^{-12}}\,\text{N C}^{-1}
\]
\( \textbf{Power of ten:} \)
\[
\frac{10^{-8}}{10^{-12}}=10^4
\]
\( \textbf{Numerical part:} \)
\[
\frac{3.0}{8.85}\approx0.339
\]
\( \textbf{Calculation:} \)
\[
E\approx0.339\times10^4\,\text{N C}^{-1}
\]
\( \textbf{Result:} \)
\[
E\approx3.4\times10^3\,\text{N C}^{-1}
\]
\( \textbf{Final answer:} \) The field just outside is approximately \(3.4\times10^3\,\text{N C}^{-1}\).
409. A charged conductor surface and an isolated infinite non-conducting sheet have the same surface charge density \(\sigma\). The field just outside the conductor is larger than the field on one side of the sheet because
ⓐ. the conductor field is always tangential
ⓑ. the non-conducting sheet has no electric field at all
ⓒ. inside field is zero; flux leaves only outside face
ⓓ. Gauss’s law applies only when both faces give flux
Correct Answer: inside field is zero; flux leaves only outside face
Explanation: For an isolated infinite non-conducting sheet, field exists on both sides and the pillbox flux is \(2EA\). This gives \(E=\frac{\sigma}{2\varepsilon_0}\) on either side. For a conductor surface in electrostatic equilibrium, the field inside the conductor is zero. A pillbox straddling the conductor surface has flux only through the outside face, so \(EA=\frac{\sigma A}{\varepsilon_0}\). This gives \(E=\frac{\sigma}{\varepsilon_0}\) just outside. The factor of \(2\) difference comes from whether one or two faces contribute to flux.
410. A positive charge \(q=2.0\times10^{-6}\,\text{C}\) is placed in a uniform electric field of magnitude \(3.0\times10^4\,\text{N C}^{-1}\). If its mass is \(4.0\times10^{-6}\,\text{kg}\), the acceleration magnitude is
ⓐ. \(2.4\times10^{-16}\,\text{m s}^{-2}\)
ⓑ. \(6.0\times10^{-2}\,\text{m s}^{-2}\)
ⓒ. \(1.5\times10^4\,\text{m s}^{-2}\)
ⓓ. \(7.5\times10^9\,\text{m s}^{-2}\)
Correct Answer: \(1.5\times10^4\,\text{m s}^{-2}\)
Explanation: \( \textbf{Given charge:} \) \(q=2.0\times10^{-6}\,\text{C}\).
\( \textbf{Electric field:} \) \(E=3.0\times10^4\,\text{N C}^{-1}\).
\( \textbf{Mass:} \) \(m=4.0\times10^{-6}\,\text{kg}\).
\( \textbf{Force on charge:} \)
\[
F=qE
\]
\( \textbf{Substitution for force:} \)
\[
F=(2.0\times10^{-6})(3.0\times10^4)\,\text{N}
\]
\( \textbf{Force value:} \)
\[
F=6.0\times10^{-2}\,\text{N}
\]
\( \textbf{Newton's second law:} \)
\[
F=ma
\]
\( \textbf{Acceleration:} \)
\[
a=\frac{F}{m}
\]
\( \textbf{Substitution:} \)
\[
a=\frac{6.0\times10^{-2}}{4.0\times10^{-6}}\,\text{m s}^{-2}
\]
\( \textbf{Calculation:} \)
\[
a=1.5\times10^4\,\text{m s}^{-2}
\]
\( \textbf{Final answer:} \) The acceleration magnitude is \(1.5\times10^4\,\text{m s}^{-2}\).
411. A point charge produces field \(E\) at distance \(r\) in vacuum. The source charge is doubled, the distance is tripled, and the space is filled with a dielectric of constant \(K=2\). The new field magnitude is
ⓐ. \(\frac{2E}{9}\)
ⓑ. \(\frac{E}{9}\)
ⓒ. \(\frac{E}{18}\)
ⓓ. \(\frac{9E}{2}\)
Correct Answer: \(\frac{E}{9}\)
Explanation: \( \textbf{Original point-charge field dependence:} \)
\[
E\propto\frac{|Q|}{r^2}
\]
\( \textbf{Charge change:} \) Doubling the source charge gives a factor \(2\).
\( \textbf{Distance change:} \) Tripling distance gives a denominator factor \(3^2=9\).
\( \textbf{Dielectric effect:} \) A dielectric of constant \(K=2\) divides the field by \(2\).
\( \textbf{Combined ratio:} \)
\[
\frac{E'}{E}=\frac{2}{9\times2}
\]
\( \textbf{Simplification:} \)
\[
\frac{E'}{E}=\frac{1}{9}
\]
\( \textbf{New field:} \)
\[
E'=\frac{E}{9}
\]
\( \textbf{Final answer:} \) The new field magnitude is \(\frac{E}{9}\).
412. Read the case below.
A small positive charge is placed at a point where two fields act simultaneously. One field is \(4.0\times10^3\,\text{N C}^{-1}\) toward east, and the other is \(3.0\times10^3\,\text{N C}^{-1}\) toward north. The charge magnitude is \(2.0\,\mu\text{C}\).
The magnitude of the net electric force on the charge is
ⓐ. \(7.0\times10^{-3}\,\text{N}\)
ⓑ. \(1.4\times10^{-2}\,\text{N}\)
ⓒ. \(2.0\times10^{-3}\,\text{N}\)
ⓓ. \(1.0\times10^{-2}\,\text{N}\)
Correct Answer: \(1.0\times10^{-2}\,\text{N}\)
Explanation: \( \textbf{Field components:} \) \(E_x=4.0\times10^3\,\text{N C}^{-1}\), \(E_y=3.0\times10^3\,\text{N C}^{-1}\).
\( \textbf{Net field magnitude:} \)
\[
E_{\text{net}}=\sqrt{E_x^2+E_y^2}
\]
\( \textbf{Substitution:} \)
\[
E_{\text{net}}=\sqrt{(4.0\times10^3)^2+(3.0\times10^3)^2}
\]
\( \textbf{Use the \(3\)-\(4\)-\(5\) pattern:} \)
\[
E_{\text{net}}=5.0\times10^3\,\text{N C}^{-1}
\]
\( \textbf{Charge magnitude:} \)
\[
q=2.0\,\mu\text{C}=2.0\times10^{-6}\,\text{C}
\]
\( \textbf{Force magnitude relation:} \)
\[
F=qE_{\text{net}}
\]
\( \textbf{Substitution:} \)
\[
F=(2.0\times10^{-6})(5.0\times10^3)\,\text{N}
\]
\( \textbf{Calculation:} \)
\[
F=10.0\times10^{-3}\,\text{N}=1.0\times10^{-2}\,\text{N}
\]
\( \textbf{Final answer:} \) The net force magnitude is \(1.0\times10^{-2}\,\text{N}\).
413. A tilted surface has area \(0.40\,\text{m}^2\) and is placed in a uniform field \(250\,\text{N C}^{-1}\). The field makes \(60^\circ\) with the surface plane. The electric flux through the surface is
ⓐ. \(100\,\text{N m}^2\text{C}^{-1}\)
ⓑ. \(50\,\text{N m}^2\text{C}^{-1}\)
ⓒ. \(86.6\,\text{N m}^2\text{C}^{-1}\)
ⓓ. \(0\,\text{N m}^2\text{C}^{-1}\)
Correct Answer: \(86.6\,\text{N m}^2\text{C}^{-1}\)
Explanation: \( \textbf{Area:} \) \(A=0.40\,\text{m}^2\).
\( \textbf{Electric field:} \) \(E=250\,\text{N C}^{-1}\).
\( \textbf{Given angle:} \) The field makes \(60^\circ\) with the surface plane.
\( \textbf{Angle with area vector:} \) The area vector is normal to the plane, so \(\theta=90^\circ-60^\circ=30^\circ\).
\( \textbf{Flux formula:} \)
\[
\Phi_E=EA\cos\theta
\]
\( \textbf{Substitution:} \)
\[
\Phi_E=(250)(0.40)\cos30^\circ
\]
\( \textbf{First product:} \)
\[
EA=100\,\text{N m}^2\text{C}^{-1}
\]
\( \textbf{Use \(\cos30^\circ=\frac{\sqrt{3}}{2}\approx0.866\):} \)
\[
\Phi_E=100(0.866)
\]
\( \textbf{Result:} \)
\[
\Phi_E=86.6\,\text{N m}^2\text{C}^{-1}
\]
\( \textbf{Final answer:} \) The electric flux is \(86.6\,\text{N m}^2\text{C}^{-1}\).
414. A point charge \(+q\) is kept at the centre of a spherical Gaussian surface. The field is first found from Gauss’s law and then a charge \(-q_0\) is placed on the surface. The force on \(-q_0\) is
ⓐ. tangential to the spherical surface
ⓑ. away from the centre with magnitude \(\frac{1}{4\pi\varepsilon_0}\frac{qq_0}{r^2}\)
ⓒ. zero because Gauss’s law gives only flux
ⓓ. toward the centre with magnitude \(\frac{1}{4\pi\varepsilon_0}\frac{qq_0}{r^2}\)
Correct Answer: toward the centre with magnitude \(\frac{1}{4\pi\varepsilon_0}\frac{qq_0}{r^2}\)
Explanation: The field due to the central positive charge \(+q\) at radius \(r\) is radially outward. Its magnitude is \(E=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\). The force on a charge placed in the field is \(\vec{F}=q_{\text{placed}}\vec{E}\). For the negative charge \(-q_0\), the force is opposite to the outward electric field. Therefore, it is directed toward the centre. The magnitude is \(F=q_0E=\frac{1}{4\pi\varepsilon_0}\frac{qq_0}{r^2}\), where \(q_0\) is used as a positive magnitude.
415. A dipole of moment \(p\) is placed in a uniform electric field \(E\). Its angle with the field changes slowly from \(180^\circ\) to \(0^\circ\). The change in potential energy is
ⓐ. \(0\)
ⓑ. \(+2pE\)
ⓒ. \(-pE\)
ⓓ. \(-2pE\)
Correct Answer: \(-2pE\)
Explanation: \( \textbf{Potential energy relation:} \)
\[
U=-pE\cos\theta
\]
\( \textbf{Initial angle:} \) \(\theta_i=180^\circ\).
\( \textbf{Initial energy:} \)
\[
U_i=-pE\cos180^\circ
\]
\[
U_i=+pE
\]
\( \textbf{Final angle:} \) \(\theta_f=0^\circ\).
\( \textbf{Final energy:} \)
\[
U_f=-pE\cos0^\circ
\]
\[
U_f=-pE
\]
\( \textbf{Change in energy:} \)
\[
\Delta U=U_f-U_i
\]
\( \textbf{Substitution:} \)
\[
\Delta U=(-pE)-(+pE)
\]
\( \textbf{Result:} \)
\[
\Delta U=-2pE
\]
\( \textbf{Final answer:} \) The potential energy changes by \(-2pE\).
416. A mixed proportionality record is shown below. Identify the row that is not consistent with standard electrostatic results.
| Row | Situation | Change made | Predicted effect |
|---|
| P | Point-charge field | \(r\) doubled | \(E\) becomes \(\frac{E}{4}\) |
| Q | Infinite line-charge field | \(r\) doubled | \(E\) becomes \(\frac{E}{2}\) |
| R | Infinite sheet field | distance doubled | \(E\) remains unchanged |
| S | Far dipole field | \(r\) doubled | \(E\) becomes \(\frac{E}{4}\) |
ⓐ. Row P
ⓑ. Row S
ⓒ. Row R
ⓓ. Row Q
Correct Answer: Row S
Explanation: Row S is incorrect because the far field of a dipole varies as \(\frac{1}{r^3}\). If \(r\) is doubled, a far dipole field becomes \(\frac{E}{8}\), not \(\frac{E}{4}\). Row P is correct because a point-charge field varies as \(\frac{1}{r^2}\). Row Q is correct because an infinite line-charge field varies as \(\frac{1}{r}\). Row R is correct for an ideal infinite sheet because its field is independent of distance. Comparing the powers of \(r\) is the cleanest way to separate these field patterns.
417. Assertion: Two equal and opposite infinite plane sheets produce a uniform electric field between them and zero field outside in the ideal model.
Reason: Between the sheets the field contributions are in the same direction, while outside they are in opposite directions.
ⓐ. Assertion is true, but Reason is false
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Each infinite sheet produces a field of magnitude \(\frac{\sigma}{2\varepsilon_0}\) on each side. For equal and opposite sheets, the fields between the sheets both point from the positive sheet toward the negative sheet. They add to give \(E=\frac{\sigma}{\varepsilon_0}\). Outside the two sheets, the field due to one sheet is opposite to the field due to the other sheet. Since the magnitudes are equal in the ideal model, the outside resultant is zero. The Reason correctly describes the directional addition that produces the stated field pattern.
418. A charge \(+3.0\,\mu\text{C}\) lies inside a closed surface, and a charge \(-1.0\,\mu\text{C}\) lies outside it. A separate uniform external field also passes through the surface. The net flux through the closed surface is
ⓐ. \(\frac{3.0\,\mu\text{C}}{\varepsilon_0}\)
ⓑ. \(\frac{2.0\,\mu\text{C}}{\varepsilon_0}\)
ⓒ. \(-\frac{1.0\,\mu\text{C}}{\varepsilon_0}\)
ⓓ. \(0\)
Correct Answer: \(\frac{3.0\,\mu\text{C}}{\varepsilon_0}\)
Explanation: Gauss’s law uses the net charge enclosed by the closed surface. The charge \(+3.0\,\mu\text{C}\) is inside, so it contributes to \(q_{\text{enc}}\). The charge \(-1.0\,\mu\text{C}\) is outside, so it can affect the local field on the surface but not the net closed-surface flux. A uniform external field contributes equal entering and leaving flux through a closed surface, so its net contribution is zero. Therefore, \(q_{\text{enc}}=+3.0\,\mu\text{C}\). The net flux is \(\frac{3.0\,\mu\text{C}}{\varepsilon_0}\).
419. A spherical shell, an infinite line charge, and an infinite plane sheet are used in three Gauss-law derivations. The common reason Gauss’s law becomes directly useful in all three is that
ⓐ. all three charge distributions produce the same distance dependence
ⓑ. every Gaussian surface encloses zero charge
ⓒ. the electric field is always constant everywhere in space
ⓓ. symmetry simplifies \(\vec{E}\cdot d\vec{A}\) on the Gaussian surface
Correct Answer: symmetry simplifies \(\vec{E}\cdot d\vec{A}\) on the Gaussian surface
Explanation: Gauss’s law is valid for any closed surface, but it directly gives \(E\) only when the flux integral can be simplified. For a spherical shell or point charge, spherical symmetry makes \(E\) constant on a centred sphere. For an infinite line charge, cylindrical symmetry makes \(E\) constant on the curved surface of a coaxial cylinder. For an infinite plane sheet, planar symmetry makes a pillbox surface useful. The distance dependences are different in these cases. The shared strategy is not identical formulas, but symmetry-guided choice of Gaussian surface.
420. A final comparison is made between electric force, electric field, electric flux, and Gauss’s law. The correctly paired statement is
ⓐ. \(\vec{F}=q\vec{E}\), \(\vec{E}\) is vector, \(\Phi_E\) is scalar, and \(\oint \vec{E}\cdot d\vec{A}=\frac{q_{\text{enc}}}{\varepsilon_0}\)
ⓑ. \(\vec{F}=\frac{\vec{E}}{q}\), \(\vec{E}\) is scalar, \(\Phi_E\) is vector, and \(\oint \vec{E}\cdot d\vec{A}=q_{\text{outside}}\varepsilon_0\)
ⓒ. \(\vec{F}=q+\vec{E}\), \(\vec{E}\) has no direction, \(\Phi_E\) has unit \(\text{C m}^{-2}\), and Gauss’s law uses total outside charge
ⓓ. \(\vec{F}=q\vec{E}\), \(\vec{E}\) is vector, \(\Phi_E\) is always zero, and Gauss’s law is valid only for spheres
Correct Answer: \(\vec{F}=q\vec{E}\), \(\vec{E}\) is vector, \(\Phi_E\) is scalar, and \(\oint \vec{E}\cdot d\vec{A}=\frac{q_{\text{enc}}}{\varepsilon_0}\)
Explanation: The force on a charge in an electric field is \(\vec{F}=q\vec{E}\). Electric field \(\vec{E}\) is a vector because it has magnitude and direction. Electric flux \(\Phi_E\) is a scalar obtained from the dot product \(\vec{E}\cdot\vec{A}\) or \(\int\vec{E}\cdot d\vec{A}\). Gauss’s law states that the total flux through a closed surface is \(\frac{q_{\text{enc}}}{\varepsilon_0}\). The enclosed charge, not outside charge, determines net closed-surface flux. Gauss’s law is valid for any closed surface, though symmetry decides whether it easily gives the field.
