Electric Potential and Energy Questions and Answers

Answer: A

The two contributions are equal in magnitude and opposite in sign at the midpoint, so they cancel. The field there is not zero but twice that of either charge alone, which shows the two quantities behave quite differently.

Answer: B

The energy is the product of the charges over \( 4\pi\epsilon_0 \) times the separation, \( 9 \times 10^{9} \times 4 \times 10^{-18}/1 = 3.6 \times 10^{-8} \) J. It is positive because work was needed to bring like charges together.

Answer: C

Correct answer: 4.43 \( \mu \)J/m3

\[ u = \tfrac{1}{2}\varepsilon_0 E^2 = 0.5 \times 8.854\times10^{-12} \times 10^6 = 4.43\ \mu\mathrm{J/m^3}. \] Omitting the factor of a half gives 8.85, which is offered here - the half comes from integrating the energy as the field is built up rather than from any property of the medium.

Answer: D

Correct answer: four times its former value

Energy density goes as \( E^2 \), so doubling the field quadruples it. That square is why the energy in a capacitor rises so steeply with voltage, and why the field concentration at a sharp point matters so much for breakdown.

Answer: A

Correct answer: 29 V

\[ V = \frac{\lambda}{2\pi\varepsilon_0}\ln\frac{b}{a} = 17.98 \times \ln 5 = 17.98 \times 1.609 = 28.9\ \mathrm{V}. \] The logarithm is the signature of cylindrical geometry: the potential of a line charge has no finite reference at infinity, so only differences between two radii can be quoted.

Answer: B

Correct answer: 450 V

\[ V = \frac{Q}{4\pi\varepsilon_0 r} = \frac{5\times10^{-9}}{4\pi \times 8.854\times10^{-12} \times 0.1} = 449\ \mathrm{V}. \] Outside a conducting sphere the field and potential are those of a point charge at its centre, which is why the same formula serves.

Answer: C

A static field is conservative, so the potential returns to its starting value and no net work is done however long or convoluted the path. The same is not true when a changing magnetic flux threads the path, which is Faraday's law.

Answer: D

Correct answer: 100 eV, that is \( 1.6 \times 10^{-17} \) J

A charge falling through a potential difference gains \( qV \) of energy, and for an electron through one volt that defines the electronvolt. Through 100 volts it gains 100 eV, which is \( 100 \times 1.6\times10^{-19} = 1.6\times10^{-17} \) joules - the two are the same quantity in different units.

Answer: A

Correct answer: \( -6x\,\mathbf{a}_x \)

\[ \mathbf{E} = -\nabla V = -\frac{dV}{dx}\mathbf{a}_x = -6x\,\mathbf{a}_x. \] The minus sign is not decoration: the field points from high potential to low, that is downhill, while the gradient points uphill.

Answer: B

Correct answer: \( (10/r^2)\,\mathbf{a}_r \) V/m

\[ E_r = -\frac{dV}{dr} = -\frac{d}{dr}\left(\frac{10}{r}\right) = \frac{10}{r^2}. \] Two minus signs cancel here - the one in the definition and the one from differentiating \( 1/r \) - so the field is positive outward, as it should be for a positive charge at the origin.