9.2 Electric fields — Practise

Work through these on your own. Change microcoulombs to coulombs and nanometres to metres before you substitute, keep the signs of the charges, and write the unit of every answer. In the questions that ask for a reason, write your answer in full sentences.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A charge of +6.0 microcoulombs and a charge of -3.0 microcoulombs are 0.20 m apart in air. Find the magnitude of the force between them.
Find the electric field strength at a distance of 0.50 m from a point charge of +4.0 microcoulombs.
Two parallel plates are 3.0 mm apart and have a potential difference of 150 V. Find the electric field strength between them.
Find the electric potential at a distance of 0.50 m from a point charge of +4.0 microcoulombs.
A charge of +2.0 microcoulombs is placed at a point where the electric potential is 3.0e4 V. Find its electric potential energy.
Find the work done moving a charge of +5.0 microcoulombs through a potential difference of 1200 V.
An electron is accelerated from rest through a potential difference of 400 V. Find its speed. The mass of an electron is 9.11e-31 kg.
Two grids 2.0 mm apart have a potential difference of 900 V. Find the electric field strength between them and the force on a xenon ion of charge +1.60e-19 C.
Two point charges 0.10 m apart repel each other with a force of 0.40 N. Find the force between them when they are 0.20 m apart.
A calculation gives an electric force of 0.16182 N. Write the answer to two significant figures.
The potential difference across two plates is measured with an uncertainty of 0.5%, and their separation with an uncertainty of 1.0%. The field strength is 6.0e4 V/m. Find its absolute uncertainty.
A plate separation is 3.0 mm. Write it in metres.
A charged particle is released from rest in a uniform electric field. Explain how you would find its kinetic energy after it has crossed the field.
A current of 0.25 A flows for 40 s. Find the charge that passes and the number of electrons.

Answers and concise solutions

1. The magnitude of the force

Answer: $F = \dfrac{kQq}{r^{2}} = \dfrac{8.99 \times 10^{9} \times 6.0 \times 10^{-6} \times 3.0 \times 10^{-6}}{0.20^{2}} = 4.0\,\text{N}$ . The charges have opposite signs, so the force is attractive.

2. The electric field strength

Answer: $E = \dfrac{kQ}{r^{2}} = \dfrac{8.99 \times 10^{9} \times 4.0 \times 10^{-6}}{0.50^{2}} = 1.4 \times 10^{5}\,\text{N}\,\text{C}^{-1}$ , directed away from the charge.

3. The electric field strength

Answer: $E = \dfrac{V}{d} = \dfrac{150}{3.0 \times 10^{-3}} = 5.0 \times 10^{4}\,\text{V}\,\text{m}^{-1}$ .

4. The electric potential

Answer: $V = \dfrac{kQ}{r} = \dfrac{8.99 \times 10^{9} \times 4.0 \times 10^{-6}}{0.50} = 7.2 \times 10^{4}\,\text{V}$ .

5. The electric potential energy

Answer: $E_{p} = qV = 2.0 \times 10^{-6} \times 3.0 \times 10^{4} = 6.0 \times 10^{-2}\,\text{J}$ .

6. The work done

Answer: $W = q\Delta V = 5.0 \times 10^{-6} \times 1200 = 6.0 \times 10^{-3}\,\text{J}$ .

7. The speed

Answer: The energy gained is $E_{k} = eV = 6.4 \times 10^{-17}\,\text{J}$ . Then $\tfrac{1}{2}mv^{2} = E_{k}$ , so $v = \sqrt{\dfrac{2E_{k}}{m}} = 1.2 \times 10^{7}\,\text{m}\,\text{s}^{-1}$ .

8. The field strength and the force

Answer: $E = \dfrac{V}{d} = \dfrac{900}{2.0 \times 10^{-3}} = 4.5 \times 10^{5}\,\text{V}\,\text{m}^{-1}$ , and $F = QE = 1.60 \times 10^{-19} \times 4.5 \times 10^{5} = 7.2 \times 10^{-14}\,\text{N}$ .

9. The new force

Answer: The force follows an inverse-square law, so doubling the distance divides it by 4: $F = \dfrac{0.40}{4} = 0.10\,\text{N}$ .

10. The force to two significant figures

Answer: $0.16182\,\text{N}$ is $0.16\,\text{N}$ to two significant figures.

11. The absolute uncertainty

Answer: The field strength is $E = \dfrac{V}{d}$ . A quotient adds percentage uncertainties, so the percentage uncertainty is $0.5 + 1.0 = 1.5\%$ . The absolute uncertainty is $\dfrac{1.5}{100} \times 6.0 \times 10^{4} = 9.0 \times 10^{2}\,\text{V}\,\text{m}^{-1}$ . So $E = (6.0 \pm 0.09) \times 10^{4}\,\text{V}\,\text{m}^{-1}$ .

12. The separation in metres

Answer: $3.0\,\text{mm} = 3.0 \times 10^{-3}\,\text{m}$ .

13. A charged particle is released from rest in a uniform electric field. Explain how you would find its kinetic energy after it has crossed the field.

Find the force from $F = QE$ .
The work done by the field is $W = QV$ , where $V$ is the potential difference crossed.
The work done becomes kinetic energy, so $E_{k} = QV$ .
Alternatively, find the acceleration from $a = \dfrac{F}{m}$ and use the equations of motion.

14. The charge and the number of electrons

Answer: The charge is $Q = It = 0.25 \times 40 = 10\,\text{C}$ . Each electron carries $1.60 \times 10^{-19}\,\text{C}$ , so the number is $\dfrac{10}{1.60 \times 10^{-19}} = 6.3 \times 10^{19}$ .