Gauss's law states that the flux of \( \mathbf{D} \) out of any closed surface equals
Answer: A
Written \( \oint_s \mathbf{D} \cdot d\mathbf{s} = Q_{enc} \), it involves the flux density and so needs no permittivity on the right. Expressed in terms of the field intensity instead, the permittivity appears as a divisor on the charge.
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Charge lying outside a closed surface contributes to the flux through it
Answer: B
Every line of flux from an external charge that enters the surface must also leave it, so the net contribution cancels. This is why only the enclosed charge appears on the right hand side, however much charge lies just outside.
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The size and shape of the closed surface used in Gauss's law
Answer: C
The law holds for any closed surface whatever, which is what makes it so general. A surface chosen to match the symmetry of the problem lets the field be taken outside the integral, and that is the only reason for preferring one shape to another.
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For Gauss's law to yield the field directly, the Gaussian surface must be chosen so that
Answer: D
On such a surface the dot product becomes a constant magnitude times an area, and the field can be solved for. Where the field is tangential the contribution is zero, which disposes of the end faces of a cylinder round a line charge.
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The Gaussian surface appropriate to a long straight line charge is
Answer: A
The field is radial and of constant magnitude at a given radius, so a coaxial cylinder makes the integral trivial and its flat ends contribute nothing because the field is tangential there. A sphere would suit only a point-symmetric source.
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For an infinite sheet of charge the surface of choice is
Answer: B
Correct answer: a rectangular box straddling the sheet
The field of an infinite sheet is uniform, perpendicular to the sheet and outward on both sides. A box with two faces parallel to the sheet therefore has flux through those two faces only - the sides being parallel to the field - and the enclosed charge is the sheet area they span. That gives \( E = \sigma/2\varepsilon_0 \) immediately.
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Applied to a point charge with a concentric spherical surface, Gauss's law gives a field proportional to
Answer: C
The enclosed charge is fixed while the area of the sphere grows as the square of the radius, so their ratio falls as the inverse square. The same reasoning gives the inverse first power for a cylinder and a constant for a box straddling a sheet.
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Inside a uniformly charged spherical shell, the field is
Answer: D
A concentric sphere drawn inside the shell encloses no charge, so the flux through it is zero, and the spherical symmetry then forces the field itself to vanish. Outside the shell the field is the same as that of a point charge at the centre.
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At a radius \( r \) inside a solid sphere carrying uniform volume charge density, the enclosed charge grows as
Answer: A
The volume within radius r grows as the cube, and with the density constant so does the charge. Dividing by an area growing as the square leaves a field rising linearly, which is the standard result for the interior of such a sphere.
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Between the conductors of a charged coaxial cable, the field varies as
Answer: B
A coaxial Gaussian cylinder encloses the fixed charge on the inner conductor and has an area growing linearly with radius, so the field falls as the first power. Outside the outer conductor the enclosed charge is zero and so is the field.
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