Electric Field: Definition, Formula, Explanation, and Solved Problems

 17.2.1 Electric Field

1. Definition of electric field

An electric field is a region in which an electric force will act on a charge that is placed in the region.

2. ‘Production’ of electric field

A charge or an electric potential difference across two conductors can produce an electric field. Hence an electric field exists in the space surrounding the charge.

17.2.2 Electric Field Pattern

1. Electric field lines

Figure 17.9

Consider a positive test charge placed within an electric field produced by a positive point charge. A repulsive electric force will act on the test charge along a certain direction. Suppose we allow the test charge to move without any restriction within the electric field. Then the electric force will drive the test charge to move along a certain path away from the point charge, as shown in Figure 17.9(a). This path is known as the electric field line.

(Note: The word ‘line’ does not necessarily mean a straight line. It may be a curve.)

Electric field lines have the following features:

(a) They do not intersect each other.

(b) Each electric field line has a direction.

The direction is the same as the direction of motion of the positive test charge that is allowed to move freely in the field. Because of this, the line points away from a positive charge that produces the field, as shown in Figure 17.9(b). This also means that the line ‘flows’ out of a positive charge.

Conversely, the line points towards a negative charge that produces the field. This is because of the fact that a positive test charge is attracted by the negative charge and so it will move towards the negative charge. This also means that the line ‘flows’ into a negative charge.

(c) The electric field strength vector E at a point in an electric field is tangential to the field line that passes through that point. Both the vector and the field line point in the same direction.

2. Direction of electric force in electric field

There is always an electric field line that passes through any point charge found in an electric field. Since the charge is in an electric field, an electric force acts on it. The direction of this force acting on the charge is along the tangent to the field line at the point where the charge is found.

The direction of the electric force acting on a positive charge in an electric field is the same as the direction of the field line, as shown in Figure 17.10. For a negative charge, the electric force points in a direction that is opposite to the direction of the field line.

3. Electric field pattern

Suppose we draw many of these electric field lines around the charge that produces the electric field. This will show a pattern of the electric field. Figure 17.11 shows examples of some electric field patterns.

Figure 17.11: (a) Point charges +q and –q, (b) Point charges +q and +q, (c) Point charges +3q and –q

4. Electric field produced by a charged electric conductor in electrostatic equilibrium

Consider a charged conductor of any shape that is in electrostatic equilibrium. This means that no current flows in any part of the conductor because the charges carried by it is stationary. The electric field produced by the charges on the conductor has the following properties:

(a) The electric field exists outside the conductor but there is no electric field inside the charged conductor.

(b) In electrostatic equilibrium, the electric field outside the charged conductor must be perpendicular to the surface of the conductor.

Figure 17.12

Suppose the electric field line is not perpendicular to the surface, as shown in Figure 17.12(a). The electric field strength vector E would have a horizontal component E₁ parallel to the conductor surface and a vertical component E₂ perpendicular to the surface.

The horizontal component E₁ would cause a free electron on the surface of a negatively charged conductor to experience an electric force, which would cause the electron to move on the conductor surface. However, no such current could be observed on the surface of a charged conductor in electrostatic equilibrium. Hence, we conclude that E₁ does not exist when the free electrons on the conductor are in “static” condition and only the vertical component exists. Figure 17.12(b) shows the electric field lines that are drawn perpendicular to the surface of the conductor. 

17.2.3 Uniform Electric Field

1. Field pattern of uniform field

A uniform electric field is represented by a set of electric field lines that are
(a) straight,
(b) parallel to each other, and
(c) equally spaced, as shown in Figure 17.13(a).

2. Formation of uniform electric field

Consider a pair of identical parallel plates connected to a constant voltage, as shown in Figure 17.13(b). A uniform electric field will exist in the region between the two plates. However, a non-uniform field is found at the edges of the plates. 

17.2.4 Electric Field Strength

1. Definition

The electric field strength at a point in an electric field is defined as the electric force per unit charge experienced by a charge placed at that point.

2. Equation from definition

Suppose a charge q is placed at a point X in an electric field.
If q is positive, q experiences an electric force of F N.

If q C of charge at X will experience a force of F/q N.

By definition, F/q is the electric field strength at X, which is represented by E. Hence the electric field strength at a point in an electric field is given by

$E=\frac{F}{q}$

3. Electric force

Using the defining equation above we have

$F=qE$

4. Unit of E

E has a unit of N C⁻¹ or voltmetre (V m⁻¹).

5. Vector quantity

E is a vector quantity. Its direction is the same as
(a) the direction of the electric field lines
(b) the direction of the electric force F acting on a positive charge q in the electric field.

We have

$E=\left(\frac{1}{q}\right)F$

6. Electric field lines and field strength

The strength of an electric field can be inferred from the spacing of electric field lines, as follows:

(a) In a region where the field lines are drawn very close together, the electric field strength is strong.

(b) Conversely, in another region of the same electric field, where the lines are drawn far apart, the field strength there is comparatively weaker.

Example 17.4

Charge (μC)	Electric force (N)	Direction of force	Electric field strength (kN C⁻¹)	Direction of electric field
(a) +10		...	20	to the left
(b) -500	10	to the right	...	...
(c) ...	30	to the left	150	to the right
(d) ...	20	to the left	100	to the left

$$\mathrm{<span\; style="font-family:\; "Times\; New\; Roman";\; font-size:\; medium;">A\; point\; charge\; is\; placed\; in\; a\; horizontal\; uniform\; electric\; field\; that\; points\; either\; to\; the\; right\; or\; left.\; Complete\; the\; table\; above.</span>}$$

$$Answer(a) $F = qE = (10 \ \mu C)(20 \ kN C^{-1})=0.20 N$Since the charge is positive, the direction of F is opposite to that of the field. E points to the left.(b) $E=\frac{F}{q}=\frac{10}{500 \times 10^{-6}}=20 \ kN \ C^3$Since the charge is positive, the direction of F is opposite to that of the field. E points to the left.(c) $|q|=\frac{F}{E}=\frac{10}{500 \times 10^3}=200 \ \mu C$Since the direction of F is opposite to that of E, q is negative.(d) $|q|=\frac{F}{E}=\frac{20}{100 \times 10^3}=200 \ \mu C$Since the direction of F is the same as that of E, q is positive.$$

$$\mathrm{<span\; style="font-size:\; medium;"><br></span>}\mathrm{<br>}$$Example 17.5

A particle of mass 0.50 g and carrying a charge of +5.0 $\mu C$ is held in a uniform electric field of sterngth 2.0 kN $C^{-1}$. If the particle is related from rest, determine

(a) the acceleration of the particle

(b) the speed of the particle 3.0 s after release.

$$$$Answer

(a) $F=ma$, also $F=qE$

$a=\frac{qE}{m}$

$a=\frac{(5 \times 10^{-6})(2 \times 10^3)}{0.5 \times 10^{-3}}=20 \ m \ s^{-2}$

(b) Use $v=v_0 +at$

$v=0+(20)(3)=60 \ m \ s^{-1}$

17.2.5 Electric Field Strength of an Electric Field Produced by an Isolated Point Charge

1. Equation for E

(a) An isolated positive point charge Q in free space produces a non-uniform radial electric field.

(b) Suppose another positive point charge qis placed at a point X in the electric field produced by charge Q. Point X is at distance rfrom Q, as shown in Figure 17.14(a). We wish to determine an expression for the electric field strength E at X.

(c) According to Coulomb’s law, the electric force Fthat acts on qis given by

$F=\frac{Qq}{4 \pi \epsilon_0r^2}$

By definition, the electric field strength at X given by

$E=\frac{F}{q}$

$E=\frac{Q}{4 \pi \epsilon_0 r^2}$ (in free space)

Notice that $E \propto \frac{1}{r^2}$

Hence, the electric field produced by an isolated point charge obeys the inverse square law. 

2. Graph of E against r

Figure 17.14(b) shows how the electric field strength E varies with distance r from the isolated point charge Q. 

EXAMPLE 17.6

(a) Estimate the electric field strength at a point X that is 10 nm from a gold nucleus, which is assumed to be a point charge. 

(b) If an alpha particle arrives at point X, what would be the magnitude of the electric force that acts on the particle? (Atomic number: gold, 79; helium, 2)

Answer

(a) $E=\frac{Q}{4 \pi \epsilon_0 r^2}$

$Q= 79e$ with $e=1.6 \times 10^{-19} \ C$

$E=\frac{9 \times 10^9 \times 79 \times 1.6 \times 10^{-19}}{(10 \times 10^{-9})^2}$

$E=1.1 \times 10^9 \ N \ C^{-1}$

(b) $F=qE=(2e)E$

$F=2(1.6 \times 10^{-19})(1.1 \times 10^9)=3.5 \times 10^{-10}$ N

EXAMPLE 17.7

The magnitude of the electric field strength at a point 10 cm from a point charge is 200 $kN C^{-1}$. Determine the field strength at a point 20 cm from the point charge.

Answer 

$E \propto \frac{1}{r^2}$

$\frac{E}{200}=\left(\frac{10 \ cm}{20 \ cm} \right)^2$

$E=50 \ kN \ C^{-1}$

17.2.6 Resultant Field Strength at a Point Due to Several Point Charges

Consider a point X in a resultant electric field produced by two point charges, $Q_1$ and $Q_2$, as shown in Figure 17.15. $Q_1$ and $Q_2$ produce electric field strengths $E_1$ and $E_2$ respectively at X. The resultant field strength at X is the resultant of the vector sum of the two field vectors $E_1$ and $E_2$.

EXAMPLE 17.8

A point X is 10 cm away from a point charge $q_1$ of charge -2.0μC. Another point charge $q_2$ has a charge of magnitude 5μC. $q_2$  is placed on the line joining point X and $q_1$ such that the resultant electric field strength at X is zero. Determine the position of $q_2$. Is the charge of $q_2$ positive or negative? 

Answer

Let charge $q_2$ be placed at distance x from $q_1$, as shown in Figure 17.16. Since charge $q_1$ is negative, field strength vector $E_1$ produced by $q_1$ points towards $q_1$ (to the right). The resultant field strength at X is zero. Hence the field strength vector $E_2$ produced by $q_2$ must point to the left, away from $q_2$.

This means that $q_2$ is positively charged. 

Distance x

Magnitude of $E_1$ = magnitude of $E_2$

$k\frac{q_1}{r_1^2}=k\frac{q_2}{r_2^2}$

$\frac{2 \mu C}{(10 \ cm)^2}=\frac{5 \mu C}{(10 \ cm+x \ cm)^2}$

Solving, x = 5.8 cm

Hence, $q_2$ is 15.8 cm from point X

EXAMPLE 17.9

Point charges of +2.0 μC and −2.0 μC are placed at corners A and B respectively of a right-angled triangle ABC, as shown in Figure 17.17. Determine the resultant electric field strength at corner C. 

Answer

Let

$E_1$= electric field strength produced by charge -2.0μC

$E_2$= electric field strength produced by charge +2.0μC 

$E_1=\frac{(9 \times 10^9)(2 \times 10^{-6})}{(0.05)^2}=7.19 MN \ C^{-1}$

$E_2=\frac{(9 \times 10^9)(2 \times 10^{-6})}{(0.13)^2}=1.06 MN \ C^{-1}$

Magnitued of resultant E

Use cosine rule:

$E^2=E_1^2+E_2^2-2E_1E_2 \ cos \ \theta$

$E^2=(7.19)^2+(1.06)^2+2(7.19)(1.06) \frac{5}{13}=46.96$

$E=6.85 \ MN \ C^{-1}$

Direction of resultant E

Use sin rule:

$\frac{sin \ \phi}{E_2}=\frac{sin \ \phi}{E_1}$

$sin \ \phi=\frac{1.06}{6.85} \left (\frac{12}{13} \right)=0.1428$

$\phi=8.21^o$ with BC

17.2.7 Similarities and Differences between Electric and Gravitational Fields

1. Similarities

(a) The field strength of each kind of field obeys the inverse square law.

(b) Both kinds of fields are radial if the charge is a point charge and the mass is a point mass.

(c) The force associated with each kind of field is a conservative force. Because of this each kind of field is associated with a potential at a point in the field. 

2. Differences

Gravitational field	Electric field
Gravitational field is produced by an object which has mass.	Electric field is produced by a charge.
The gravitational force in the field is always attractive.	The electric force in the field can be attractive or repulsive.

17.2.8 Applications

An electric field is a region around a charged object where other charges experience an electric force. Electric fields are widely used in many technologies and everyday applications.

1. Photocopiers and Laser Printers

Photocopiers and laser printers use electrostatic forces produced by electric fields.

Working principle:

1. A rotating drum is given an electric charge.

2. Light from the document removes charges from certain areas.

3. Toner particles (charged ink powder) are attracted to the charged areas by the electric field.

4. The toner is transferred to paper and heated to fix the image.

Physics concept used:

$$F = qE$$

Charged toner particles move due to the electric force in an electric field.

2. Electrostatic Precipitator (Air Pollution Control)

An electrostatic precipitator is used in factories to remove dust and smoke particles from exhaust gases.

Working principle:

1. Smoke particles are given an electric charge.

2. A strong electric field is created between metal plates.

3. Charged particles are attracted to the plates.

4. The particles stick to the plates, and cleaner air is released.

Used in:

• Power plants

• Cement factories

• Steel industries

3. Electrostatic Painting

Electrostatic painting is commonly used in car manufacturing and industrial coating.

Working principle:

1. Paint droplets are given an electric charge.

2. The object being painted has an opposite charge.

3. The electric field attracts the paint droplets toward the object.

Advantages:

• Even coating

• Less paint waste

• Higher efficiency

4. Cathode Ray Tube (CRT)

Electric fields are used to control electron beams in CRT devices such as:

• Old television sets

• Oscilloscopes

• Computer monitors (older models)

Working principle:

Electrons moving through an electric field experience a force:

$$F = qE$$

This force changes the direction of the electron beam so it can form images on the screen.

5. Air Purifiers

Some electrostatic air purifiers use electric fields to remove dust from the air.

Working principle:

1. Dust particles are electrically charged.

2. An electric field attracts the charged particles.

3. The particles stick to collector plates.

This helps clean the air more effectively.

17.2.9 Conclusion of Electric Field

The electric field is a region around a charged object where other charged particles experience an electric force. It explains how electric forces act without direct contact between charges. The strength of an electric field depends on the magnitude of the charge and the distance from the charge, following the inverse square law.

Electric fields are important in understanding many physical phenomena and are widely used in modern technology. Applications include photocopiers, laser printers, electrostatic precipitators, electrostatic painting, air purifiers, and electronic display devices.

In conclusion, the concept of the electric field helps scientists and engineers analyze and control the behavior of charged particles, making it essential in both physics and technological applications.

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