Electric Potential and Energy Questions and Answers

Answer: A

Correct answer: 1 and 2 only are correct

Statements 1 and 2 are correct: potential is a scalar, which is what makes it easier to compute than the field, and \( \mathbf{E} = -\nabla V \). Statement 3 confuses a quantity with its gradient - a zero field means the potential is constant, and the constant may be anything at all.

Answer: B

Correct answer: 1 and 2 only are correct

Statements 1 and 2 are correct: the density goes as \( E^2 \), and assembling like charges requires work against their repulsion. Statement 3 has the sign wrong - a bound pair of opposite charges has negative energy, since work would have to be done to separate them, and that is precisely what being bound means.

Answer: C

Correct answer: the field is tangential to it

The field is normal to an equipotential, not tangential - a tangential component would do work along the surface, contradicting the definition. The other three statements are all correct, including that two equipotentials of different value cannot intersect, since a point cannot hold two potentials at once.

Answer: D

Energy density is an algebraic function of the field, being half the permittivity times its square, with no differentiation involved. The other three pairs are connected by a gradient, a divergence and a curl respectively.

Answer: A

For a point charge the potential is \( Q/4\pi\epsilon_0 r \) and the field is that again divided by r, so at one metre they take the same numerical value. At two metres the potential would be 250 V and the field 125 V/m.

Answer: B

Correct answer: it is a scalar and needs no resolution into components

The potential of a distribution is a single integral of a scalar, whereas the field requires the vector contributions to be resolved and integrated component by component. The usual practice is therefore to compute the potential and then take its gradient, which is far less work than integrating the field directly.

Answer: C

Above that stress the field strips electrons from air molecules and an avalanche follows, so the charge leaks away and the field cannot be maintained. Every high voltage design is ultimately constrained by this figure or its equivalent in another medium.

Answer: D

Near the surface the disc behaves as an infinite sheet, whose field is uniform, and the potential of a uniform field falls linearly with distance. Far away the same disc behaves as a point charge and the potential falls as the inverse first power, so the two regimes are quite distinct and the crossover occurs at distances comparable with the radius.

Answer: A

Connecting them makes one conductor, and a conductor in equilibrium is at a single potential throughout. Since the potential of an isolated sphere is its charge divided by its radius times the constant, equal potential means the charges divide in proportion to the radii, and the smaller sphere ends with the greater surface density and the stronger field.

Answer: B

Equal potential means the charge divides in proportion to radius, so the surface density, which is charge over area, varies inversely with radius. Since the surface field is proportional to that density, the smaller sphere carries the larger field, which is why a sharp point on a charged body is where a discharge begins.