301. What is the main reason Gauss's law is especially useful for calculating electric field only in highly symmetric charge distributions?
ⓐ. Gauss's law is false for non-symmetric charge distributions
ⓑ. Symmetry can make \(E\) constant on parts of a Gaussian surface
ⓒ. Symmetry removes the need for enclosed charge
ⓓ. Non-symmetric distributions always have zero electric field
Correct Answer: Symmetry can make \(E\) constant on parts of a Gaussian surface
Explanation: Gauss's law is valid for any closed surface and any charge distribution. However, using it to calculate electric field directly requires simplifying the flux integral. In highly symmetric cases, a suitable Gaussian surface can be chosen so that \(E\) has the same magnitude on important parts of the surface. Also, \(\vec{E}\) may be parallel to \(d\vec{A}\) on some parts and perpendicular on others. This allows \(E\) to be taken outside the integral. Without symmetry, Gauss's law still gives net flux, but it may not give the field at each point easily. The distinction is between validity and practical usefulness.
302. Which Gaussian surface is most suitable for finding the field due to an isolated point charge?
ⓐ. An arbitrary open surface
ⓑ. A cylinder with the charge on its curved surface
ⓒ. A sphere centred on the point charge
ⓓ. A pillbox with the charge on one flat face
Correct Answer: A sphere centred on the point charge
Explanation: The electric field due to an isolated point charge is radial and has the same magnitude at all points at the same distance from the charge. A sphere centred on the point charge has every point on its surface at the same distance from the charge. On this spherical surface, \(\vec{E}\) is parallel to \(d\vec{A}\) everywhere. Therefore, the flux integral simplifies to \(EA\). A cylinder or pillbox does not match the spherical symmetry of a point charge. The Gaussian surface should be chosen to match the symmetry of the field.
303. Which Gaussian surface is usually chosen for an infinitely long uniformly charged straight wire?
ⓐ. A sphere centred at one point of the wire
ⓑ. A coaxial cylinder around the wire
ⓒ. A cube with one edge touching the wire
ⓓ. A flat circular disk perpendicular to the wire only
Correct Answer: A coaxial cylinder around the wire
Explanation: An infinitely long uniformly charged straight wire has cylindrical symmetry. The electric field points radially outward from the wire for positive charge and radially inward for negative charge. A coaxial cylindrical Gaussian surface matches this symmetry. On the curved surface, \(E\) has the same magnitude and is parallel to \(d\vec{A}\). On the flat end caps, the electric field is perpendicular to \(d\vec{A}\), so their flux is zero. The symmetry match is the reason a cylinder is preferred over a sphere or cube.
304. Which row correctly matches charge distribution with the usual Gaussian surface?
| Row | Charge distribution | Suitable Gaussian surface |
|---|
| P | Point charge | Sphere centred on charge |
| Q | Infinite line charge | Coaxial cylinder |
| R | Infinite plane sheet | Pillbox crossing the sheet |
| S | Spherical shell | Spherical surface concentric with shell |
Which rows are correct?
ⓐ. P and Q only
ⓑ. P, Q and R only
ⓒ. Q, R and S only
ⓓ. P, Q, R and S
Correct Answer: P, Q, R and S
Explanation: A point charge has spherical symmetry, so a sphere centred on the charge is suitable. An infinite line charge has cylindrical symmetry, so a coaxial cylinder is suitable. An infinite plane sheet has planar symmetry, so a pillbox crossing the sheet is used. A spherical shell has spherical symmetry, so a concentric spherical Gaussian surface is chosen. In each case, the surface is selected so that the flux integral becomes simple. The key strategy is to match the Gaussian surface to the symmetry of the charge distribution, not to choose a surface randomly.
305. Why is a spherical Gaussian surface suitable for a point charge placed at its centre?
ⓐ. Because the electric field is tangential to the spherical surface everywhere
ⓑ. Because \(E\) is constant on the sphere and parallel to \(d\vec{A}\)
ⓒ. Because the enclosed charge becomes zero on a sphere
ⓓ. Because a sphere always makes electric flux zero
Correct Answer: Because \(E\) is constant on the sphere and parallel to \(d\vec{A}\)
Explanation: A point charge produces a radial electric field. If a spherical Gaussian surface is centred on the point charge, every point on the sphere is at the same distance from the charge. Therefore, the magnitude \(E\) is the same everywhere on the spherical surface. The outward area element \(d\vec{A}\) is also radial, so \(\vec{E}\parallel d\vec{A}\) everywhere. This makes \(\vec{E}\cdot d\vec{A}=E\,dA\), allowing \(E\) to be taken outside the integral. The symmetry mistake is choosing a surface that encloses the charge but does not simplify the flux integral.
306. For an infinitely long uniformly charged wire, why is the flux through the flat end caps of a coaxial cylindrical Gaussian surface zero?
ⓐ. Because the electric field is zero at the end caps
ⓑ. Because the electric field is parallel to the end-cap area vectors
ⓒ. Because the field is perpendicular to the end-cap area vectors
ⓓ. Because the end caps do not have area
Correct Answer: Because the field is perpendicular to the end-cap area vectors
Explanation: The electric field due to an infinitely long uniformly charged wire is radial. For a coaxial cylindrical Gaussian surface, the area vectors of the flat end caps are along the cylinder axis. The radial electric field is perpendicular to the cylinder axis. Therefore, on the end caps, \(\vec{E}\perp d\vec{A}\). Since \(\vec{E}\cdot d\vec{A}=E\,dA\cos90^\circ=0\), the flux through the end caps is zero. The field itself is not zero there; only its normal component through the end caps is zero.
307. Which condition allows \(E\) to be taken outside the flux integral \(\oint \vec{E}\cdot d\vec{A}\)?
ⓐ. \(E\) must be zero everywhere on the surface
ⓑ. Same \(E\) on that surface part
ⓒ. The Gaussian surface must be open
ⓓ. The enclosed charge must be negative
Correct Answer: Same \(E\) on that surface part
Explanation: In Gauss's law, the flux integral is \(\oint \vec{E}\cdot d\vec{A}\). To take \(E\) outside an integral over a surface, the magnitude of \(E\) must be constant over that part of the surface. Also, the angle between \(\vec{E}\) and \(d\vec{A}\) should be simple, usually \(0^\circ\), \(180^\circ\), or \(90^\circ\). This is why symmetry is essential for direct field calculation using Gauss's law. If \(E\) varies from point to point, the integral cannot be simplified as \(E\oint dA\). The common mistake is to think enclosing charge is enough to calculate \(E\), but symmetry is what makes the calculation easy.
308. Which Gaussian surface is most suitable for finding the electric field due to an infinite uniformly charged plane sheet?
ⓐ. A sphere centred on a point of the sheet
ⓑ. A coaxial cylinder along a line charge
ⓒ. A pillbox crossing the sheet
ⓓ. A cone with vertex on the sheet
Correct Answer: A pillbox crossing the sheet
Explanation: An infinite uniformly charged plane sheet has planar symmetry. The electric field is perpendicular to the sheet and has equal magnitude on both sides. A pillbox Gaussian surface is chosen with its flat faces parallel to the sheet. The electric field is parallel to the area vectors of the two flat faces, so those faces contribute flux. The curved side of the pillbox has area vector parallel to the sheet, while the field is perpendicular to the sheet, so the curved side contributes zero flux. The Gaussian surface should match the symmetry of the charge distribution, not merely enclose some charge.
309. In using a pillbox Gaussian surface for an infinite plane sheet, why is the flux through the curved side zero?
ⓐ. Because the curved side has no area
ⓑ. Because the electric field is parallel to the sheet and perpendicular to \(d\vec{A}\)
ⓒ. Because the field is tangent to the curved side of the pillbox
ⓓ. Because the enclosed charge on the sheet is zero
Correct Answer: Because the field is tangent to the curved side of the pillbox
Explanation: For an infinite uniformly charged sheet, the electric field is normal to the sheet. In a pillbox Gaussian surface, the curved side has area vectors parallel to the plane of the sheet, radially outward from the side surface. The electric field is perpendicular to those side area vectors. Therefore, \(\vec{E}\cdot d\vec{A}=0\) on the curved side. The flux comes only through the two flat faces of the pillbox. The field is not zero on the curved side; its normal component through that side is zero.
310. Which statement correctly explains why Gauss's law may not directly give the electric field for an irregular charge distribution?
ⓐ. Gauss's law becomes invalid for irregular charge distributions
ⓑ. The enclosed charge becomes meaningless for irregular distributions
ⓒ. The field may not be constant on a convenient closed surface
ⓓ. Electric flux cannot be defined for irregular surfaces
Correct Answer: The field may not be constant on a convenient closed surface
Explanation: Gauss's law is valid for any closed surface and any charge distribution. However, direct calculation of electric field requires simplifying the flux integral. For irregular charge distributions, the field may vary in magnitude and direction over the surface. Then \(E\) cannot be taken outside the integral easily. The law still gives the total flux if the enclosed charge is known. The limitation is not validity, but practical usefulness for finding \(E\). The important distinction is that Gauss's law is always true, while field extraction from it requires symmetry.
311. Which row correctly identifies the zero-flux part for the stated Gaussian surface?
| Row | Charge distribution and Gaussian surface | Zero-flux part |
|---|
| P | Infinite line charge with coaxial cylinder | Flat end caps |
| Q | Infinite plane sheet with pillbox | Two flat faces |
| R | Point charge with centred sphere | Entire spherical surface |
| S | Spherical shell with concentric sphere outside | Entire spherical surface |
ⓐ. Row P only
ⓑ. Rows P and Q only
ⓒ. Rows Q and R only
ⓓ. Rows R and S only
Correct Answer: Row P only
Explanation: For an infinite line charge with a coaxial cylinder, the electric field is radial and perpendicular to the area vectors of the flat end caps, so row P is correct. For an infinite plane sheet with a pillbox, the two flat faces contribute flux; the curved side has zero flux, so row Q is wrong. For a centred sphere around a point charge, the field is radial and parallel to \(d\vec{A}\), so the spherical surface has non-zero flux, making row R wrong. For a spherical shell with a concentric outside sphere, the outside field is radial and gives non-zero flux, so row S is wrong. The zero-flux part is found from \(\vec{E}\perp d\vec{A}\), not from visual surface shape alone.
312. A Gaussian surface encloses charge \(+q\), but the chosen surface is irregular so that \(E\) is not constant on it. Which statement is correct?
ⓐ. Gauss's law cannot be applied
ⓑ. \(\Phi=\frac{q}{\varepsilon_0}\), but \(E\ne\Phi/A\) in general
ⓒ. The net flux becomes zero because the surface is irregular
ⓓ. The electric field must be constant because the enclosed charge is fixed
Correct Answer: \(\Phi=\frac{q}{\varepsilon_0}\), but \(E\ne\Phi/A\) in general
Explanation: Gauss's law gives \(\Phi_E=\frac{q_{\text{enc}}}{\varepsilon_0}\) for any closed surface. If the surface encloses \(+q\), the net flux is \(\frac{q}{\varepsilon_0}\), regardless of whether the surface is regular or irregular. However, if \(E\) varies over the surface, the flux is not simply \(EA\). In that case, \(E\) cannot be obtained by dividing total flux by total area. The law gives total flux, but not automatically the field at each point. The common mistake is to treat all closed surfaces as if \(E\) were uniform on them.
313. A spherical Gaussian surface is chosen for a point charge not located at the centre of the sphere. What remains true?
ⓐ. \(\vec{E}\) is parallel to \(d\vec{A}\) everywhere
ⓑ. The flux must be zero
ⓒ. \(E\) is constant everywhere on the sphere despite the off-centre charge
ⓓ. Net flux remains \(q_{\text{enc}}/\varepsilon_0\)
Correct Answer: Net flux remains \(q_{\text{enc}}/\varepsilon_0\)
Explanation: Gauss's law depends only on the charge enclosed by the closed surface. If the point charge remains inside the sphere, the net flux is still \(\frac{q_{\text{enc}}}{\varepsilon_0}\). But if the charge is not at the centre, points on the spherical surface are at different distances from the charge. Therefore, \(E\) is not constant everywhere on the sphere. Also, \(\vec{E}\) is not generally parallel to \(d\vec{A}\) at every point. The total flux remains simple, but the local field distribution becomes less symmetric. The centre condition is needed for easy field calculation, not for Gauss's law itself.
314. Which statement is correct about choosing a Gaussian surface?
ⓐ. It should always be the smallest possible closed surface
ⓑ. Choose it so symmetry simplifies the flux integral
ⓒ. It must always be made of a real physical material
ⓓ. It must always pass through all charges in the system
Correct Answer: Choose it so symmetry simplifies the flux integral
Explanation: A Gaussian surface is an imaginary closed surface used to apply Gauss's law. It is chosen for mathematical convenience, not because it is a physical boundary. The best choice is one that matches the symmetry of the charge distribution. Then \(E\) may be constant on useful parts of the surface, and \(\vec{E}\cdot d\vec{A}\) becomes easy to evaluate. The surface need not be the smallest and need not pass through charges. In fact, passing through point charges can create difficulties because the field becomes singular there. The guiding idea is symmetry-based simplification.
315. Which statement correctly compares \(\vec{E}\parallel d\vec{A}\) and \(\vec{E}\perp d\vec{A}\) in Gaussian-surface calculations?
ⓐ. \(\vec{E}\parallel d\vec{A}\): zero; \(\vec{E}\perp d\vec{A}\): maximum
ⓑ. \(\vec{E}\parallel d\vec{A}\): zero; \(\vec{E}\perp d\vec{A}\): zero
ⓒ. \(\vec{E}\parallel d\vec{A}\): maximum; \(\vec{E}\perp d\vec{A}\): maximum
ⓓ. \(\vec{E}\parallel d\vec{A}\): maximum; \(\vec{E}\perp d\vec{A}\): zero
Correct Answer: \(\vec{E}\parallel d\vec{A}\): maximum; \(\vec{E}\perp d\vec{A}\): zero
Explanation: Flux through an area element is \(d\Phi_E=\vec{E}\cdot d\vec{A}=E\,dA\cos\theta\). If \(\vec{E}\parallel d\vec{A}\), then \(\theta=0^\circ\) or \(180^\circ\), so the contribution has maximum magnitude. The sign depends on whether the directions are same or opposite. If \(\vec{E}\perp d\vec{A}\), then \(\theta=90^\circ\), and the contribution is zero. This is why end caps or curved sides are often ignored in symmetric Gauss-law derivations. The angle with the area vector is the deciding factor.
316. For an infinitely long uniformly charged straight wire with linear charge density \(\lambda\), which expression gives the electric field magnitude at distance \(r\)?
ⓐ. \(E=\frac{\sigma}{2\varepsilon_0}\)
ⓑ. \(E=\frac{\lambda r}{2\pi\varepsilon_0}\)
ⓒ. \(E=\frac{\lambda}{4\pi\varepsilon_0 r^2}\)
ⓓ. \(E=\frac{\lambda}{2\pi\varepsilon_0 r}\)
Correct Answer: \(E=\frac{\lambda}{2\pi\varepsilon_0 r}\)
Explanation: An infinitely long uniformly charged wire has cylindrical symmetry. Using a coaxial cylindrical Gaussian surface of radius \(r\) and length \(L\), the field is radial and constant on the curved surface. The flux through the curved surface is \(E(2\pi rL)\). The enclosed charge is \(\lambda L\). Gauss's law gives \(E(2\pi rL)=\frac{\lambda L}{\varepsilon_0}\). Cancelling \(L\) gives \(E=\frac{\lambda}{2\pi\varepsilon_0 r}\). The important distance dependence is \(E\propto r^{-1}\), not \(r^{-2}\).
317. In deriving the electric field due to an infinitely long uniformly charged wire, what is the charge enclosed by a coaxial cylindrical Gaussian surface of length \(L\)?
ⓐ. \(\lambda L\)
ⓑ. \(\lambda 2\pi rL\)
ⓒ. \(\frac{\lambda}{L}\)
ⓓ. \(\lambda \pi r^2L\)
Correct Answer: \(\lambda L\)
Explanation: Linear charge density \(\lambda\) means charge per unit length. If a cylindrical Gaussian surface of length \(L\) encloses a segment of the wire of length \(L\), the enclosed charge is \(q_{\text{enc}}=\lambda L\). The factor \(2\pi rL\) is the curved surface area of the cylinder, not the enclosed charge. The factor \(\pi r^2L\) is a volume and would be relevant for volume density, not line charge density. The unit check confirms the result: \((\text{C m}^{-1})(\text{m})=\text{C}\). In Gauss-law derivations, confusing enclosed charge with Gaussian surface area is a common error.
318. A long uniformly charged wire has \(\lambda=4.0\times10^{-6}\,\text{C m}^{-1}\). What is the electric field magnitude at \(r=0.20\,\text{m}\)? Take \(\frac{1}{2\pi\varepsilon_0}=1.8\times10^{10}\,\text{N m}^2\text{C}^{-2}\).
ⓐ. \(1.8\times10^5\,\text{N C}^{-1}\)
ⓑ. \(7.2\times10^4\,\text{N C}^{-1}\)
ⓒ. \(9.0\times10^4\,\text{N C}^{-1}\)
ⓓ. \(3.6\times10^5\,\text{N C}^{-1}\)
Correct Answer: \(3.6\times10^5\,\text{N C}^{-1}\)
Explanation: \( \textbf{Given:} \) \(\lambda=4.0\times10^{-6}\,\text{C m}^{-1}\), \(r=0.20\,\text{m}\), and \(\frac{1}{2\pi\varepsilon_0}=1.8\times10^{10}\,\text{N m}^2\text{C}^{-2}\).
\( \textbf{Required:} \) Electric field magnitude \(E\).
\( \textbf{Formula for long line charge:} \)
\[
E=\frac{\lambda}{2\pi\varepsilon_0 r}
\]
\( \textbf{Rewrite using given constant:} \)
\[
E=\left(\frac{1}{2\pi\varepsilon_0}\right)\frac{\lambda}{r}
\]
\( \textbf{Substitution:} \)
\[
E=(1.8\times10^{10})\frac{4.0\times10^{-6}}{0.20}
\]
\( \textbf{Divide by radius:} \)
\[
\frac{4.0\times10^{-6}}{0.20}=20.0\times10^{-6}=2.0\times10^{-5}
\]
\( \textbf{Final calculation:} \)
\[
E=(1.8\times10^{10})(2.0\times10^{-5})=3.6\times10^5\,\text{N C}^{-1}
\]
\( \textbf{Direction note:} \) If \(\lambda\) is positive, the field is radially outward.
\( \textbf{Final answer:} \) The electric field magnitude is \(3.6\times10^5\,\text{N C}^{-1}\).
319. If the distance from an infinitely long uniformly charged wire is doubled, how does the electric field magnitude change?
ⓐ. It becomes twice
ⓑ. It becomes four times
ⓒ. It becomes one-fourth
ⓓ. It becomes one-half
Correct Answer: It becomes one-half
Explanation: For an infinitely long uniformly charged wire, the electric field magnitude is \(E=\frac{\lambda}{2\pi\varepsilon_0 r}\). For fixed \(\lambda\), this gives \(E\propto\frac{1}{r}\). If the distance \(r\) is doubled, the denominator doubles. Therefore, the field becomes half of its original value. This is different from a point charge, for which doubling distance gives one-fourth field. The distance-dependence mistake is mixing up line-charge \(r^{-1}\) behaviour with point-charge \(r^{-2}\) behaviour.
320. What is the direction of electric field around a positively charged infinitely long straight wire?
ⓐ. Circular around the wire
ⓑ. Radially inward toward the wire
ⓒ. Along the length of the wire
ⓓ. Radially outward from the wire
Correct Answer: Radially outward from the wire
Explanation: A uniformly charged infinitely long straight wire has cylindrical symmetry. For a positive line charge density \(\lambda\), a positive test charge near the wire is repelled. Therefore, the electric field points radially outward from the wire. The field is perpendicular to the wire, not along its length. Circular field lines around a wire are associated with magnetic fields around currents, not electrostatic field of a charged wire. If \(\lambda\) were negative, the electric field would be radially inward. The sign of \(\lambda\) decides inward or outward radial direction.
