Coulombs Law and Electric Field Questions and Answers

Answer: C

Both forces fall as the inverse square, so the ratio is independent of distance and is set by the constants and the charge and mass of the electron. The result, of order ten to the forty-second, is why gravity is ignored entirely in atomic physics.

Answer: D

A uniform field gives a constant force and therefore a constant acceleration along the field for a positive charge. A parabola results only when the particle enters the field with a velocity across it, as in a cathode ray tube.

Answer: A

The static field is irrotational, so its line integral round any closed path vanishes and the field is conservative. This is what allows a scalar potential to be defined and makes energy calculations independent of the path taken.

Answer: B

Correct answer: an attractive force, through induced charge separation

The positive charge draws electrons in the conductor towards the near face and leaves the far face positive. The negative induced charge is closer, so its attraction exceeds the repulsion of the more distant positive charge, and the net force is attractive. The conductor is neutral throughout - it is the separation, not any net charge, that produces the force, which is why a charged rod attracts any uncharged conductor.

Answer: C

Mutual repulsion drives the charge as far apart as it can go, and Gauss's law confirms that a vanishing interior field requires no enclosed charge. On an irregular conductor it collects most densely where the curvature is greatest.

Answer: D

Correct answer: where the surface curves most sharply

Charge accumulates where the radius of curvature is smallest, since the mutual repulsion pushes it as far apart as the shape allows and a sharp region offers the greatest escape from its neighbours. The field just outside is \( \sigma/\epsilon_0 \), so the field is largest there too - which is why discharge begins at points and why a lightning conductor is pointed.

Answer: A

In a medium the force is divided by the relative permittivity, so at the same separation it would be a quarter of its free space value. To recover the free space force the separation must shrink until the inverse square makes up the factor of four: \[ \frac{1}{\epsilon_r r^{2}} = \frac{1}{\left(1\right)^{2}} \;\Longrightarrow\; r^{2} = \frac{1}{\epsilon_r} = \frac{1}{4} \;\Longrightarrow\; r = 0.5\ \text{m}. \] Whenever a permittivity is traded against a distance the square root appears, because the law is inverse square in one and inverse first power in the other.

Answer: B

Both magnitude and direction must be constant, which is why the field between large parallel plates is the standard example. Such a field does have zero divergence and curl, but so do many non-uniform fields, so that is not a sufficient description.

Answer: C

Correct answer: a linear quadrupole with zero net charge and zero dipole moment

The charges sum to \( q - 2q + q = 0 \), so there is no monopole term, and the two dipole moments formed by the outer charges against the centre are equal and opposite, so the dipole moment vanishes as well. What remains is the next term in the expansion, the quadrupole, and its field falls as \( 1/r^4 \) - faster still than a dipole’s.

Answer: D

The field within the metal must vanish, so a Gaussian surface inside the metal must enclose no net charge, which forces the induced inner charge to cancel the enclosed one. An equal charge of the original sign then appears on the outer surface.