Theory

Electric field

How the electric field is defined, how it is computed for point charges, for the dipole and for continuous distributions such as the wire, the ring, the disk and the plane, how field lines describe it and how a charge moves in a uniform field.

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Definition of the electric field
Field of point charges

Definition of the electric field

Throughout the topic the charges that produce the field are at rest in vacuum, in an inertial reference frame.

Electric field

Consider a set of charges at rest, called source charges, and place at the point $P$ a point charge $q_{0}$ , called the test charge, small enough not to displace the source charges. If $F$ is the force that the source charges exert on $q_{0}$ , the electric field at $P$ is

E (P) = \frac{F}{q _{0}} .

By Coulomb’s law the force is proportional to $q_{0}$ , so the ratio does not depend on the test charge and describes a property of the point $P$ . The electric field assigns a vector to each point of space, that is it is a vector field, for which we refer to the topic on vector fields in Mathematical Analysis 2. It is measured in $N/C$ , a unit that coincides with the $V/m$ , as will be seen in the topic on the electric potential.

For example, if a test charge $q_{0} = 2.0 nC$ experiences at $P$ a force of magnitude $6.0 \cdot 1 0^{- 5} N$ , the field at $P$ has magnitude $E = \frac{6.0 \cdot 1 0 ^{- 5}}{2.0 \cdot 1 0 ^{- 9}} N/C = 3.0 \cdot 1 0^{4} N/C$ and has the direction and orientation of the force.

Force on a charge placed in the field

If the field $E$ at a point is known, the force on a point charge $q$ placed at that point is

F = q E .

If $q > 0$ the force has the same orientation as the field, if $q < 0$ it has the opposite orientation. The field therefore splits the problem into two parts: first the field produced by the sources is computed, then the force on any charge placed in the field. An electron in a field of $3.0 \cdot 1 0^{4} N/C$ experiences the force of magnitude $e E = 1.60 \cdot 1 0^{- 19} \cdot 3.0 \cdot 1 0^{4} N = 4.8 \cdot 1 0^{- 15} N$ , opposite to the field.

Field of point charges

1. Field of a point charge

The field produced by a point charge $q$ at rest in vacuum, at a point $P$ at distance $r$ from the charge, is

E = k \frac{q}{r ^{2}} \overset{r}{^} = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}} \overset{r}{^},

where $\overset{r}{^}$ is the unit vector directed from the charge toward $P$ . The field has magnitude $E = k \frac{∣ q ∣}{r ^{2}}$ and is radial, outward from the charge if $q > 0$ and inward if $q < 0$ . Since the magnitude is inversely proportional to the square of the distance, when the distance doubles the field becomes four times weaker. At $10 cm$ from a charge of $5.0 nC$ we have $E = \frac{8.99 \cdot 1 0 ^{9} \cdot 5.0 \cdot 1 0 ^{- 9}}{0.1 0 ^{2}} N/C \approx 4.5 \cdot 1 0^{3} N/C$ .

Field of a positive point charge: the vectors point outward and, at twice the distance, their length becomes one quarter

Proof

Place a test charge $q_{0}$ at $P$ . By Coulomb’s law the force that $q$ exerts on $q_{0}$ is $F = k \frac{q q _{0}}{r ^{2}} \overset{r}{^}$ , with $\overset{r}{^}$ directed from $q$ toward $q_{0}$ ; the orientation is correct for every combination of signs, because the product $q q_{0}$ is positive between charges of the same sign, which repel each other, and negative between charges of opposite sign, which attract each other. Dividing by $q_{0}$ , by the definition of electric field we obtain the claim.

2. Superposition principle for the electric field

The field produced at a point $P$ by the point charges $q_{1}, \dots, q_{n}$ is the vector sum of the fields that each charge would produce alone:

E (P) = i = 1 \sum n k \frac{q _{i}}{r _{i}^{2}} \overset{r}{^}_{i},

where $r_{i}$ is the distance of $P$ from $q_{i}$ and $\overset{r}{^}_{i}$ the unit vector directed from $q_{i}$ toward $P$ . The principle follows from the superposition principle for electric forces, which is an experimental law, by dividing the total force on the test charge by $q_{0}$ . In calculations each field is written in Cartesian components, the components are added, and only at the end are the magnitude and direction of the resultant field obtained.

Field of two opposite charges

A charge $q = 4.0 nC$ is at $A = (- 3.0 cm, 0)$ and a charge $- q$ at $B = (3.0 cm, 0)$ . The point $P = (0, 4.0 cm)$ is $5.0 cm$ from both, so the two fields have the same magnitude

E_{1} = E_{2} = \frac{8.99 \cdot 1 0 ^{9} \cdot 4.0 \cdot 1 0 ^{- 9}}{0.05 0 ^{2}} N/C \approx 1.44 \cdot 1 0^{4} N/C .

The field $E_{1}$ points outward from $A$ and has the direction of the unit vector $(\frac{3}{5}, \frac{4}{5})$ ; the field $E_{2}$ is directed toward $B$ , with unit vector $(\frac{3}{5}, - \frac{4}{5})$ . The $y$ components cancel and the $x$ components add up, therefore

E = 2 E_{1} \cdot \frac{3}{5} \overset{}{^} \approx 1.7 \cdot 1 0^{4} N/C \overset{}{^},

that is the field at $P$ is parallel to the line $A B$ joining the charges and points toward the negative charge.

The field

E_{1}

of the positive charge points outward, the field

E_{2}

of the negative charge points inward; the sum

E

is parallel to the line joining the charges

Warning

In general the magnitudes of the fields do not add up. In the previous example $E_{1} + E_{2} \approx 2.9 \cdot 1 0^{4} N/C$ , whereas the resultant field has magnitude $1.7 \cdot 1 0^{4} N/C$ , because the two vectors form an angle and their $y$ components cancel. The sum of the magnitudes gives the magnitude of the resultant field only when the fields are parallel and have the same orientation.

The theory continues with 6 more sections (field lines, electric dipole, continuous charge distributions…), plus all worked examples, the solutions and the PDF.

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