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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Outside the outer conductor of a coaxial cable carrying equal and opposite charges, the field is
Answer: C
A Gaussian cylinder drawn beyond the outer conductor encloses both charges, whose sum is zero, so the flux and hence the field vanish. This complete external screening is why coaxial cable neither radiates nor picks up interference.
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Maxwell's first equation in differential form states that
Answer: D
Correct answer: \( \nabla \cdot \mathbf{D} = \rho_v \)
This is Gauss’s law at a point: the divergence of the flux density equals the volume charge density there. The integral form follows by the divergence theorem. Note that it is \( \mathbf{D} \) rather than \( \mathbf{E} \) that appears, which is what makes the equation hold in any medium without a permittivity in it.
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The theorem that converts Gauss's law between its integral and differential forms is
Answer: A
The divergence theorem equates the volume integral of the divergence to the closed surface integral of the flux, so equating both to the enclosed charge gives the two forms of the same law. Stokes's theorem performs the same office for Ampere's law.
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A region in which the divergence of \( \mathbf{D} \) is zero contains
Answer: B
Correct answer: no net charge density
Zero divergence means no source or sink at that point, so the charge density vanishes there. It says nothing at all about the field, which may be strong and varying - the field of charges lying outside the region has zero divergence throughout it.
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The divergence of \( \mathbf{D} \) at a point tells us
Answer: C
Correct answer: the volume charge density at that point
Divergence is a local operation, so it reports the source strength at the point itself and nothing about what lies elsewhere. The total charge enclosed by a surface is what the integral form gives, and the two are connected by the divergence theorem rather than being the same statement.
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The electric field inside the body of a conductor in electrostatic equilibrium is
Answer: D
Any field within the metal would drive a current, and in equilibrium none flows, so the field must vanish. Gauss's law then requires that no net charge exist within the metal either, so all of it resides on the surface.
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A conducting shell placed round a charged object shields the outside world because
Answer: A
Charge of opposite sign is induced on the inner surface and an equal charge of the original sign on the outer, so an earthed shell removes the latter and leaves no external field. Unearthed, the outer charge remains and the field outside is that of the original charge.
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