9.2 Electric fields — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Change microcoulombs and nanocoulombs to coulombs, and millimetres to metres, before you substitute.

Warm-up: remember

A current of 0.50 A flows through a lamp for 20 s. Find the charge that passes through it.
An electron is accelerated through a potential difference of 200 V. Find the energy it gains, in joules.
A $5.0\,\mu\text{F}$ capacitor is charged to 12 V. Find the energy stored in it.

Answers

$Q = It = 0.50 \times 20 = 10\,\text{C}$ .
$E = eV = 1.60 \times 10^{-19} \times 200 = 3.2 \times 10^{-17}\,\text{J}$ .
$E = \tfrac{1}{2}CV^{2} = 0.5 \times 5.0 \times 10^{-6} \times 12^{2} = 3.6 \times 10^{-4}\,\text{J}$ .

If question 1 goes wrong, revisit charge and current in Section 5.1. If question 2 goes wrong, remember that the energy gained is the charge multiplied by the potential difference. If question 3 goes wrong, revisit the energy stored in a capacitor in Section 5.4.

Practise with help

1. A charge of $+3.0\,\mu\text{C}$ and a charge of $-2.0\,\mu\text{C}$ are 0.50 m apart in air. Find the magnitude of the force between them, and state its direction.

Hint

The two charges have opposite signs, so the force is attractive.

Next step

Change both charges to coulombs and use $F = \dfrac{kQq}{r^{2}}$ . The product $Qq$ is negative, which tells you the force is attractive; use the size of the answer.

Full working

\begin{aligned} F &= \frac{kQq}{r^{2}} = \frac{8.99 \times 10^{9} \times 3.0 \times 10^{-6} \times 2.0 \times 10^{-6}}{0.50^{2}} \\ &= \frac{5.39 \times 10^{-2}}{0.25} = 0.22\,\text{N} \end{aligned}

The force is $0.22\,\text{N}$ , and it pulls the two charges together.

2. Find the electric field strength at a distance of 0.20 m from a point charge of $+5.0\,\text{nC}$ .

Hint

A nanocoulomb is $10^{-9}\,\text{C}$ , and the field of a point charge follows an inverse-square law.

Next step

Use $E = \dfrac{kQ}{r^{2}}$ , and square $0.20$ before you divide.

Full working

\begin{aligned} E &= \frac{kQ}{r^{2}} = \frac{8.99 \times 10^{9} \times 5.0 \times 10^{-9}}{0.20^{2}} \\ &= \frac{44.95}{0.040} = 1.1 \times 10^{3}\,\text{N}\,\text{C}^{-1} \end{aligned}

The field strength is $1.1 \times 10^{3}\,\text{N}\,\text{C}^{-1}$ , directed away from the charge.

3. Two parallel plates are 4.0 mm apart and have a potential difference of 240 V between them. Find the electric field strength and the force on an electron that is between the plates.

Hint

The field between parallel plates is uniform. Change the separation to metres.

Next step

Use $E = \dfrac{V}{d}$ , then multiply by the charge on an electron, $1.60 \times 10^{-19}\,\text{C}$ .

Full working

E = \frac{V}{d} = \frac{240}{4.0 \times 10^{-3}} = 6.0 \times 10^{4}\,\text{V}\,\text{m}^{-1}

F = QE = 1.60 \times 10^{-19} \times 6.0 \times 10^{4} = 9.6 \times 10^{-15}\,\text{N}

The field strength is $6.0 \times 10^{4}\,\text{V}\,\text{m}^{-1}$ , and the force on the electron is $9.6 \times 10^{-15}\,\text{N}$ towards the positive plate.

4. Find the electric potential at a distance of 0.20 m from the $+5.0\,\text{nC}$ charge in question 2.

Hint

The potential of a point charge is proportional to $\dfrac{1}{r}$ , not $\dfrac{1}{r^{2}}$ .

Next step

Use $V = \dfrac{kQ}{r}$ with the same charge and distance, and do not square the distance.

Full working

V = \frac{kQ}{r} = \frac{8.99 \times 10^{9} \times 5.0 \times 10^{-9}}{0.20} = 2.2 \times 10^{2}\,\text{V}

The potential is $2.2 \times 10^{2}\,\text{V}$ , or $2.2 \times 10^{2}\,\text{J}\,\text{C}^{-1}$ .

5. An electron is accelerated from rest through a potential difference of 250 V. Find the kinetic energy it gains and its final speed.

Hint

The energy gained is the electron charge multiplied by the potential difference.

Next step

Find $E_{k} = eV$ first, then use $\tfrac{1}{2}mv^{2} = E_{k}$ with the mass of an electron.

Full working

E_{k} = eV = 1.60 \times 10^{-19} \times 250 = 4.0 \times 10^{-17}\,\text{J}

\begin{aligned} v &= \sqrt{\frac{2E_{k}}{m}} = \sqrt{\frac{2 \times 4.0 \times 10^{-17}}{9.11 \times 10^{-31}}} \\ &= \sqrt{8.78 \times 10^{13}} = 9.4 \times 10^{6}\,\text{m}\,\text{s}^{-1} \end{aligned}

The electron gains $4.0 \times 10^{-17}\,\text{J}$ and reaches $9.4 \times 10^{6}\,\text{m}\,\text{s}^{-1}$ .

6. A charge of $+4.0\,\mu\text{C}$ moves through a potential difference of 500 V. Find the work done on the charge.

Hint

The work done depends on the charge and the potential difference, not on the distance travelled.

Next step

Use $W = q\,\Delta V$ with the charge in coulombs.

Full working

W = q\,\Delta V = 4.0 \times 10^{-6} \times 500 = 2.0 \times 10^{-3}\,\text{J}

The work done is $2.0 \times 10^{-3}\,\text{J}$ , which is 2.0 mJ.

7. Explain why the electric force between two electrons is much stronger than the gravitational force between them.

Hint

Both forces follow an inverse-square law with the same distance, so compare the constants and the charges.

Next step

Write both force laws for the same separation $r$ , then compare $k e^{2}$ with $G m_{e}^{2}$ .

Full working

Both forces follow an inverse-square law and use the same distance $r$ , so the ratio of the electric force to the gravitational force is $\dfrac{ke^{2}}{Gm_{e}^{2}}$ . The Coulomb constant $k$ is about $10^{20}$ times larger than $G$ . The charge and the mass of the electron make up the rest of the ratio, which works out at about $4 \times 10^{42}$ . The electric force is enormously stronger. Gravity becomes noticeable only when the mass is very large, as it is for a planet.

8. A positively charged particle enters a uniform electric field at right angles to the field lines. Describe the shape of its path and explain how the force on it changes during the motion.

Hint

Compare the situation with projectile motion under gravity.

Next step

Say which component of the velocity changes and which stays constant, and state whether the force changes.

Full working

The force is $F = QE$ , and the field is uniform, so the force is constant and points along the field lines. The component of the velocity along the field lines changes at a steady rate, so the particle accelerates in that direction. The component at right angles stays constant. The path is a parabola, the same shape as the path of a projectile under gravity.