Electric Force and Electric Field

Lessons

Senior 4 · Physics

Electric Force and Electric Field

5 lessons

Lesson 2 of 5

Electric Field and Field Lines

25 min

Semolina in oil

Sprinkle semolina grains on a thin layer of castor oil in a glass dish, dip two metal electrodes into the oil, and connect them to a high-voltage supply. The grains swing round and line up in chains that run from one electrode to the other, curving out through the space between them like the threads of a map. Nothing visible joins the electrodes, yet every grain "knows" which way to point. Michael Faraday's answer was that a charge changes the space around it, setting up an electric field, and any other charge placed in that space feels a force from the field. In this lesson you measure fields, draw them with field lines and add the fields of several charges.

Field strength

To probe a field, imagine placing a small positive test charge $q$ at a point, small enough not to disturb the charges that make the field. The force on it is proportional to $q$ , so the force per unit charge describes the field itself, whatever test charge is used:

\vec{E} = \frac{\vec{F}}{q} \qquad\Leftrightarrow\qquad \vec{F} = q\vec{E}

The electric field strength $E$ is a vector, measured in newtons per coulomb ( $\mathrm{N\,C^{-1}}$ ), which is the same as volts per metre (Lesson 3). A positive charge is pushed along the field; a negative charge, such as an electron, is pushed the opposite way.

Field of a point charge. Put the test charge $q$ at distance $r$ from a point charge $Q$ . Coulomb's law gives $F = \frac{kQq}{r^{2}}$ , so

E = \frac{F}{q} = \frac{kQ}{r^{2}} = \frac{Q}{4\pi\varepsilon_0 r^{2}}

The field points radially away from a positive charge and towards a negative one, and falls with the square of the distance.

Definition

The electric field strength at a point is the force per unit positive charge on a small test charge placed there: $\vec{E} = \frac{\vec{F}}{q}$ , unit $\mathrm{N\,C^{-1}}$ . An electric field line is a line drawn so that its direction at every point is the direction of the field there. Example: a charge of $+2.0\,\mathrm{nC}$ feels a force of $6.0 \times 10^{-5}\,\mathrm{N}$ at a point, so $E = \frac{6.0 \times 10^{-5}}{2.0 \times 10^{-9}} = 3.0 \times 10^{4}\,\mathrm{N\,C^{-1}}$ there.

Field lines

Field lines are drawn by these rules (Figure 2.1):

they start on positive charges and end on negative charges (or run off to great distances);
the direction of the field at a point is along the tangent to the line, shown by an arrow;
lines are closer together where the field is stronger;
lines never cross, because the field has only one direction at each point;
lines meet the surface of a conductor at right angles.

The semolina grains in the opening experiment line up along exactly these lines, because each grain is polarised by the field and its ends are pulled in opposite directions until it points along the field.

Figure 2.1: Field lines of (a) a positive point charge, (b) a pair of equal and opposite charges, (c) two equal positive charges, and (d) two oppositely charged parallel plates. Patterns (a) to (c) are computed from Coulomb's law.

Between two large parallel plates with equal and opposite charges (Figure 2.1d) the lines are parallel and equally spaced: the field has the same size and direction everywhere between them, a uniform field. This is the field used to steer electrons and ions, and you meet it again in Lesson 3.

Worked example 1: the field of a point charge

Find the electric field 0.30 m from a charge of $+5.0\,\mathrm{nC}$ , and the force on an electron placed there.

E = \frac{kQ}{r^{2}} = \frac{8.99 \times 10^{9} \times 5.0 \times 10^{-9}}{0.30^{2}} = 5.0 \times 10^{2}\,\mathrm{N\,C^{-1}}, \text{ away from the charge}

F = eE = 1.602 \times 10^{-19} \times 499 = 8.0 \times 10^{-17}\,\mathrm{N}, \text{ towards the charge}

The electron is negative, so the force on it is opposite to the field.

Fields of several charges

Because forces add as vectors, so do fields: the field at a point due to several charges is the vector sum of the fields each would make alone,

\vec{E} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \dots

Work out each field's size with $E = \frac{kQ}{r^{2}}$ , draw its direction (away from positive charges, towards negative ones), and add by components (Figure 2.2). Where the fields cancel there is a neutral point, as midway between two equal positive charges (Figure 2.1c).

Figure 2.2: Superposition of fields at P, drawn to scale: the vertical parts cancel and the horizontal parts add.

Worked example 2: between two opposite charges

Charges of $+4.0\,\mathrm{nC}$ and $-4.0\,\mathrm{nC}$ are 0.20 m apart. Find the field at the midpoint.

E_1 = E_2 = \frac{8.99 \times 10^{9} \times 4.0 \times 10^{-9}}{0.10^{2}} = 3.6 \times 10^{3}\,\mathrm{N\,C^{-1}}

The field of the positive charge points away from it, towards the negative charge; the field of the negative charge points towards it, the same way. They add: $E = 7.2 \times 10^{3}\,\mathrm{N\,C^{-1}}$ , from the positive towards the negative charge.

Worked example 3: a field off the axis

In Figure 2.2, charges of $+2.0\,\mathrm{nC}$ and $-2.0\,\mathrm{nC}$ are 0.20 m apart, and P is 0.10 m above their midpoint. Find the field at P.

Each charge is $\sqrt{0.10^{2} + 0.10^{2}} = 0.141\,\mathrm{m}$ from P, so each field has size $\frac{8.99 \times 10^{9} \times 2.0 \times 10^{-9}}{0.020} = 899\,\mathrm{N\,C^{-1}}$ , at $45^\circ$ to the horizontal. The vertical components (up from the positive charge, down towards the negative) cancel; the horizontal components add:

E = 2 \times 899 \times \cos 45^\circ = 1.27 \times 10^{3}\,\mathrm{N\,C^{-1}}

directed parallel to the line from the positive to the negative charge.

Worked example 4: a charged oil drop held still

In a version of Millikan's experiment, an oil drop of mass $3.2 \times 10^{-15}\,\mathrm{kg}$ floats at rest between horizontal plates where the field is $9.8 \times 10^{4}\,\mathrm{N\,C^{-1}}$ upwards. Find its charge, and how many elementary charges it carries.

qE = mg \quad\Rightarrow\quad q = \frac{mg}{E} = \frac{3.2 \times 10^{-15} \times 9.8}{9.8 \times 10^{4}} = 3.2 \times 10^{-19}\,\mathrm{C} = 2e

The field is upwards and the electric force must also be upwards, so the charge is positive: the drop has lost two electrons. Measurements like this on many drops showed that charge always comes in whole multiples of $e$ .

Warning

Common mistakes. (1) Using $E = \frac{kQ}{r^{2}}$ with the test charge $q$ instead of the source charge $Q$ . (2) Forgetting that the force on a negative charge is opposite to $\vec{E}$ . (3) Adding the sizes of fields that point in different directions. (4) Thinking that field lines are the paths charges follow: a charge accelerates along the line through it, but if it is already moving, or the lines curve, its path leaves the line.

Tip

Think of a little positive charge. To find the direction of a field anywhere, imagine a tiny positive charge there and ask which way it would be pushed: away from positive charges, towards negative ones. Then add the arrows.

Electric Field Visualised with Semolina and Castor Oil (David Ferguson). Watch the grains line up between electrodes of different shapes, and compare each pattern with Figure 2.1, especially the uniform field between parallel plates.

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Practice

Find the electric field 0.10 nm ( $1.0 \times 10^{-10}\,\mathrm{m}$ ) from a proton.
Find the field 0.50 m from a charge of $+2.0\,\mu\mathrm{C}$ , and the force on a charge of $+3.0\,\mathrm{nC}$ placed there.
At what distance from a charge of $+1.0\,\mathrm{nC}$ is the field $1.0 \times 10^{3}\,\mathrm{N\,C^{-1}}$ ?
An electron is in a uniform field of $2.0 \times 10^{3}\,\mathrm{N\,C^{-1}}$ . Find the size and direction of its acceleration ( $m_e = 9.11 \times 10^{-31}\,\mathrm{kg}$ ).
Sketch the field lines around two equal negative charges, and mark the neutral point.
Charges of $+1.0\,\mathrm{nC}$ sit at two corners, A and B, of a square of side 0.30 m. Find the field at the corner D next to A (AD = 0.30 m, BD = 0.42 m).
An oil drop of mass $4.9 \times 10^{-15}\,\mathrm{kg}$ is held at rest between horizontal plates where the field is $1.0 \times 10^{5}\,\mathrm{N\,C^{-1}}$ downwards. Find the size and sign of its charge, and how many electrons it has gained or lost.
A student says: "Field lines show the paths that charged particles follow." Evaluate this statement.

Show answersHide answers

$E = \frac{8.99 \times 10^{9} \times 1.602 \times 10^{-19}}{(1.0 \times 10^{-10})^{2}} = 1.4 \times 10^{11}\,\mathrm{N\,C^{-1}}$ .
$E = \frac{8.99 \times 10^{9} \times 2.0 \times 10^{-6}}{0.50^{2}} = 7.2 \times 10^{4}\,\mathrm{N\,C^{-1}}$ ; $F = 3.0 \times 10^{-9} \times 7.19 \times 10^{4} = 2.2 \times 10^{-4}\,\mathrm{N}$ , away from the charge.
$r = \sqrt{\frac{kQ}{E}} = \sqrt{\frac{8.99 \times 10^{9} \times 1.0 \times 10^{-9}}{1.0 \times 10^{3}}} = 0.095\,\mathrm{m}$ .
$a = \frac{eE}{m} = \frac{1.602 \times 10^{-19} \times 2.0 \times 10^{3}}{9.11 \times 10^{-31}} = 3.5 \times 10^{14}\,\mathrm{m\,s^{-2}}$ , opposite to the field.
Lines come in from far away and end on each charge (pointing inwards); they bend away from the space between the charges, leaving a neutral point midway between them.
Take A at the origin, B at $(0.30, 0)$ m and D at $(0, 0.30)$ m. From A: $E_A = \frac{8.99 \times 10^{9} \times 1.0 \times 10^{-9}}{0.30^{2}} = 99.9\,\mathrm{N\,C^{-1}}$ along $+ y$ . From B: $BD = 0.424\,\mathrm{m}$ , $E_B = \frac{8.99}{0.180} = 49.9\,\mathrm{N\,C^{-1}}$ along BD, components $(- 35.3, + 35.3)$ . Total: $E_{x} = - 35.3$ , $E_{y} = 135.2$ ; $E = \sqrt{35.3^{2} + 135.2^{2}} = 140\,\mathrm{N\,C^{-1}}$ , at $14.6^\circ$ to AD, tilted away from B.
$q = \frac{mg}{E} = \frac{4.9 \times 10^{-15} \times 9.8}{1.0 \times 10^{5}} = 4.8 \times 10^{-19}\,\mathrm{C} = 3e$ . The electric force must be upwards, against a downward field, so the charge is negative: the drop has gained 3 electrons.
Not in general. A field line gives the direction of the force, and so of the acceleration, not of the velocity. A charge released from rest in a uniform field does move along a straight line; but a moving charge, or one in a field whose lines curve, follows a different path, just as a thrown ball does not follow the vertical lines of the gravitational field.

Quick check

The electric field at a point is 500 N/C pointing east. Which way is the force on an electron placed there?

Key points

Electric field strength $\vec{E} = \frac{\vec{F}}{q}$ ( $\mathrm{N\,C^{-1}}$ ); the force on a charge is $q\vec{E}$ , opposite to $\vec{E}$ for a negative charge.
Point charge: $E = \frac{kQ}{r^{2}}$ , radially outward for $+ Q$ , inward for $- Q$ .
Field lines start on $+$ , end on $-$ , never cross, and crowd together where the field is strong; between charged parallel plates the field is uniform.
Fields add as vectors; where they cancel there is a neutral point.

Sources and credits

Photo: Semolina grains on castor oil: dipole field by MikeRun, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: Electric Field Visualised with Semolina and caster oil, David Ferguson, embedded from YouTube.

Written by Ganzaa: the lesson text, derivations, computed field-line diagrams, worked examples and practice exercises. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Semolina in oil

2Field strength

3Field lines

4Fields of several charges

Semolina in oil

Field strength

Field lines

Fields of several charges