# Physics Tutorial: Electric Field

In this Physics tutorial, you will learn:

• What is electric field?
• What are the similarities and differences between electric and gravitational field?
• What are electric field lines and why do we use them?
• What is the pattern and direction of electric field lines?
• What is the superposition principle of electric field?
• What are the two electric constants and which of them is more used in formulae?
• What is the pattern of electric field lines in a charged sphere?
• What is the electric field inside and outside a conducing material?
• What happens with conductors placed inside an electric field?

## Introduction

Is electrostatic force a contact force or a force that acts in distance? How do you know this?

Can you identify any similarity between gravitational attraction and electric interaction between charged objects? What about the differences between these two concepts?

How can you convince the others for the presence of electric charges without touching the object involved?

All these questions regard to a new concept: electric field, which we will discuss in this tutorial.

## What Is Electric Field? Similarities between Gravitational and Electric Field

In the tutorial "Electric Charges. Conductors and Insulators", we have seen that one of methods used for building up charges in objects is induction. In this method, a charged object is brought near a neutral one. As a result, the neutral object is locally charged with opposite signs despite the total amount of extra charge in this object is zero.

This phenomenon is the best indicator that electric charges are generated not only when the objects are in contact, but also when there is a certain distance between them.

Since such an interaction is quantitatively represented through the concept of electrostatic force, we can conclude that electrostatic force is a force that acts in distance, like the gravitational force we have discussed earlier. Yet, there is a similarity in the formulae of these two forces as well. Look at the table below. Given the above analogies, we can elaborate further this approach, i.e. analyse the similarities between gravity and electrostatics. Thus, given that gravitational force is produced from the gravitational field generated by the massive objects (which is the space around an object in which its gravitational attraction is present), we can deduce that there must exist an electrostatic field as well, i.e. a space around a charged object in which the attractive or repulsive effect of charged objects is observed.

Let's elaborate further this new concept by using again the analogy between gravity and electrostatics.

We have seen that gravitational field is represented mathematically by means of gravitational field strength g. The simplified formula for gravitational force of an object on Earth surface is

F = m × g

Therefore, the gravitational field strength of Earth acting on the object of mass m is

g = F/m

In the same way in which we calculate the magnitude of gravitational field strength g (i.e. by finding the weight of a 1 kg object at rest on the Earth surface, which corresponds to the gravitational force exerted by the Earth on that object), we can also calculate the magnitude of electrostatic field E. This can be achieved by considering a +1C test charge Q0 brought near a charged object and then measuring the electrostatic force exerted by this charged object to the test charge. Thus, we can write

E = F/Q0

Since the formula of electrostatic force exerted by a charge Q on the test charge Q0 is

F = k × Q × Q0/r2

we can find the corresponding (long) formula for the electrostatic field generated by the charge Q and experienced by the test charge Q0 which is at distance r from the charge Q:

E = F/Q0
= k × Q × Q0/r2/Q0
= k × Q/r2

The unit of electrostatic field (for now) is [N/C]. (We will see in the next tutorial that there is another unit that is officially recognized as the unit of electrostatic field in the SI system of units).

In this way, we obtain the definition of electrostatic (or electric) field, that is

Electric field is the amount of electrostatic force exerted by a charged object on a test charge, which is at a certain distance from object.

Remark!

Despite many similarities between electric and gravitational fields, there is one important difference between them. Gravitational field has only attractive nature while electric field may have attractive or repulsive nature, in dependence of the charges involved.

### Example 1

What is the electrostatic field produced by a + 8 μC point charge at 2 m away from it? Take k = 9 × 109 N × m2 / C2.

### Solution 1

Clues:

Q = + 8 μC = + 8 × 10-6 C
r = 2 m
k = 9 × 109 N × m2 / C2

We have:

F = k × Q/r2
= 9 × 109 × 8 × 10-6/22
= 18 × 103 N/C
= 18 000 N/C

## Electric Field Lines

In order to show visually the electrostatic field produced by a charged object, we draw some imaginary arrows that show the direction of electric field at every region of the field. This is a confirmation of the fact that electric field E is a vector quantity.

By agreement, the arrows that show the direction of electrostatic field (electric field lines) start from the charge positive charge and terminate at infinity if no other charges are presents. Otherwise, the lines terminate to the negative charges around. This means electric field lines are open lines, i.e. only one thing between origin and endpoint is known. Look at the figure: On the other hand, when two like charges are near each other, the electric field lines are deformed due to the presence of individual fields. If charges have opposite signs, the electric field lines produced by them have the pattern shown below: When charges are of the same sign, their electric field lines are:  As you see, there are no electric field lines in the shortest path between two opposite charges. It means the resultant electric field at that part of the space is zero.

We can summarize the properties of electric field lines as follows:

Electric field lines begin at positive charges and terminate at negative ones

The number of lines per unit area in space is proportional to the strength of electric field in that part of space

Electric field vector at any point is tangent to the field lines passing through that point

Electric field lines never cross or touch each other

## Electrostatic Constants

In the previous tutorial, we explained that the electrostatic constant k = 8.988 × 109 N × m2/C2 found experimentally (and known as Coulomb's constant), is used to turn the proportion between the two sides of Coulomb's Law into equality. However, the Coulomb's constant k is the simplified version of electrostatic constant (as its power is positive). In fact, there is another quantity, which is recognized as the "true" electrostatic constant. It is denoted by ϵ0 and its relation with the Coulomb's constant k is

ϵ0 = 1/4π × k

When substituting the known values, we obtain for ϵ0

ϵ0 = 1/4 × 3.14 × 8.988 × 109 N × m2/C2
= 8.85 × 10-12 C2/N × m2

The constant ϵ0 is more common in electricity-related formulae. This is the reason why it is more important than the Coulomb's constant k.

We can therefore write for the formulae of electrostatic force and electrostatic field in terms of ϵ0:

F = 1/4πϵ0 × Q × Q0/r2

and

E = 1/4πϵ0 × Q/r2

## The Superposition Principle

If in a given region of space there is more than one charge that produce their own electric fields E1, E2, + En, the net electric field in that region is the sum of all individual electric field vectors, i.e.

Enet = E1 + E2 + ... + En

### Example 2

Two point charges Q1 and Q2 where Q1 = + 20 μC and Q2 = - 30 μC are placed as shown in the figure. 1. What is the magnitude of net electric field at the point P?
2. What is the angle formed by the vector of net electric field at P to the horizontal direction?

### Solution 2

a) First, we find the distance from Q1 to P. Using the Pythagorean Theorem, we have

r(Q1P) = √r(Q1 Q2 )2 - r(Q2 P)2
= √152 - 92
= √225 - 81
= √144
= 12 cm

Now, we find the individual electric fields E1P and E2P where E1P vector is only horizontal (due right, as the electric field lines start from Q2 and extend to infinity) and E2P is only vertical (upward, as the electric field lines start from infinity and end to Q2 because the charge Q2 is negative).

We have:

E1P = 1/4πϵ0 × Q1/r1P2
= 9 × 109 × 20 × 10-6/0.122
= 180 × 10-6/0.0144
= 1.8 × 10-4/1.44 × 10-2
= 1.25 × 10-2 N/C

Also,

E2P = 1/4πϵ0 × Q2/r2P2
= 9 × 109 × -30 × 10-6/0.092
= -270 × 10-6/0.0081
= -27 × 10-5/8.1 × 10-3
= -3.33 × 10-2 N/C

The magnitude of the resultant electric field at the point P therefore is

EP = √E1P2 + E2P2
= √(1.25 × 10-2 )2 + (-3.33 × 10-2 )2
= √1.625 × 10-4 + 11.111 × 10-4
= √12.736 × 10-4
= 3.569 × 10-2 N/C

The vector of resultant electric field at the point P is shown in the figure below. b) The angle θ formed by the vector of resultant electric field to the horizontal direction is obtained by calculating its tangent. Thus, we have

tan θ = |E2P|/|E1P|
= |-3.33| × 10-2/|1.25| × 10-2
= 3.33/1.25
= 2.664

Therefore, the angle θ is

θ = arctan 2.664
= 69.40

## Electric Field on a Charged Sphere

We have explained in the previous tutorials that an object remains permanently charged if it contains only one type of extra charge. Also, we have stated that all charges distribute the outer part of object, leaving the inner part neutral. This is because like charges repel each other as far as possible.

Now that we know more about electric field, it is easy to understand that this phenomenon brings concentric electric field lines as those shown in the first part of this tutorial.

Since a sphere is regular in all directions, a charged conducting sphere of radius R with a uniform charge Q (let's assume it as positive), has a regular distribution of charged throughout the outer surface, as shown in the figure below. To find the electric field of the charged sphere at any point in the space, we assume that the charge is composed by many point charges q1, q2, q3, on the upper hemisphere and symmetrically, q1, q2, q3, on the lower hemisphere of the sphere. Thus, we obtain the following figure for the resultant electric field at any point (for illustration we have taken a point A): Therefore, the resultant electric field vector at the point A is the vector E, which is the vector sum of all vectors E2 in this point, as shown in the figure above. The vector sum is in the radial direction, pointing out of the sphere. The same thing can be said for a negatively charged sphere as well. The only difference is that the electric field resultant vectors point towards the centre of sphere.

The above conclusions are proofs of the fact that electric field lines for a uniform object (a point charge or a charged sphere) have a radial direction, away from the sphere when the sphere is positively charged and towards the centre of sphere when it is negatively charged, as discussed earlier.

As the study of electric field on a charged sphere involves the same approach as for a point charge, the formula of electric field must be the same as well. Thus, the electric field produced by a charged sphere Q on a point of the space, which is at distance r from the centre of the sphere, is:

E = 1/4πϵ0 × Q/r2

The graph below shows the relationship between the electric field on a charged sphere and the distance from the centre of circle, r: The above graph, is interpreted as follows:

When distance from centre of sphere is from 0 to R, that is for all points inside the sphere, the electric field is zero. When distance from centre is R, the electric field takes the maximum value. Then, with the increase in distance, electric field decreases until it becomes zero (at the infinity).

### Example 3

Two spherical concentric shells of radii R and 3R are placed as shown in the figure. Determine the electric fields at the points O, A, B, C and D if each spherical shell carries a charge of +Q. ### Solution 3

The electric field at any point is the vector sum of all electric field vectors produced by each sphere at that point.

The points O and A are inside both spherical shells, so their electric field is zero as

EB = Elarge shell + Esmall shell
= 0 + 0
=0

The point B is inside the large spherical shell and on the surface of the small shell. Thus, the electric field at this point is

EB = Elarge shell + Esmall shell
= 0 + 1/4πϵ0 × Q/R2
= Q/4πϵ0 R2

The point C is out of the small spherical shell and on the surface of the large spherical shell, at a distance of 3R from the centre. Therefore, the net electric field at this point is

EC = Elarge shell + Esmall shell
= 1/4πϵ0 × Q/(3R)2 + 1/4πϵ0 × Q/(3R)2
= 2Q/4πϵ0 × 9R2
= Q/18πϵ0 R2

The point D is out of both spherical shells, at distance 3R + R = 4R from the centre. Therefore, the net electric field at this point is

ED = Elarge shell + Esmall shell
= 1/4πϵ0 × Q/(4R)2 + 1/4πϵ0 × Q/(4R)2
= 2Q/4πϵ0 × 16R2
= Q/32πϵ0 R2

As you see, the magnitude of electric field decreases drastically with the increase in distance from the charged objects.

## Conductors in an Electric Field

As explained in previous tutorials, conductors are materials that allow electricity flowing through them. Furthermore, their atoms have free electrons available, which can carry electricity in other parts.

Now, let's see what happens if we place a metal slab inside a uniform electric field. The free electrons of metal experience an electric force which is in the opposite direction of the electric field, because electric field lines extend from positive to negative charges and since electrons are negative, they tend to go in the opposite direction to the field lines as they are attracted by the positive charges which have produced the electric field. As a result, the electrons accumulate on the left part of the slab, leaving the right part positively charged. Hence, an electric field E' is generated inside the slab, whose direction is opposite to the electric field outside. The charge accumulation continues until the magnitude of the electric field inside the slab becomes equal to that outside the slab, i.e. until |E| = |E'|. From this moment and on, the resultant electric field inside the slab becomes zero. In few words, the electric field inside a conductor is zero. This means when we place a charge inside a cavity of a conductor, the charge does not move, as it does not experience any electric force. This property has a wide range of applications in physics. It had been discovered by Faraday, who invented the famous Faraday's Cage, a conducting spherical grid whose internal space has zero electric field despite the grid is charged. Many electric-related phenomena have been studied by inserting the apparatuses inside such a cage, as in this way they are not affected by the surrounding electric field.

### Example 4

A hollow metal box is placed inside a magnetic field produced by two plates as shown in the figure. What is the direction of electric field in the sections A, B and C of the figure?

### Solution 4

It is known that electric field lines extend from positive to negative. This means, the left plate is positively charged while the right one is negatively charged. Thus, the part A of the box contains extra negative charges while the part C contain extra positive ones (opposite charged attract each other). When the equilibrium is reached, the opposite electric field produced inside the box cancels the field E and as a result, there is no electric field at the section B.

On the other hand, sections A and C are exposed to the external field. As a result, in both sections the direction of electric field is from left to right, just like the original field. ## Summary

Electric charges are generated not only when the objects are in contact, but also when there is a certain distance between them.

Since such an interaction is quantitatively represented through the concept of electrostatic force, we can say it is a force that acts in distance, like the gravitational force. Yet, there is a similarity in the formulae of these two forces as well.

In simple words, electrostatic field is the space around a charged object in which the attractive or repulsive effect of charged objects is observed.

In the same way in which we calculate the magnitude of gravitational field strength g (i.e. by finding the weight of a 1 kg object at rest on the Earth surface, which corresponds to the gravitational force exerted by the Earth on that object), we can also calculate the magnitude of electrostatic field E. This can be achieved by considering a +1C test charge Q0 brought near a charged object and then measuring the electrostatic force exerted by this charged object to the test charge. Thus, we can write

E = F/Q0

or

E = k × Q/r2

The unit of electrostatic field (for now) is [N/C]. (We will see in the next tutorial that there is another unit that is officially recognized as the unit of electrostatic field in the SI system of units).

In this way, we obtain the definition of electrostatic (or electric) field, that is

Electric field is the amount of electrostatic force exerted by a charged object on a test charge, which is at a certain distance from object.

Unlike gravitational field which has only attractive nature, electric field may have attractive or repulsive nature, depending on the charges involved.

There are no electric field lines in the shortest path between two opposite charges. It means the resultant electric field at that part of the space is zero.

We can summarize the properties of electric field lines as follows:

• Electric field lines begin at positive charges and terminate at negative ones
• The number of lines per unit area in space is proportional to the strength of electric field in that part of space
• Electric field vector at any point is tangent to the field lines passing through that point
• Electric field lines never cross or touch each other

There is another quantity, which is recognized as the "true" electrostatic constant instead of Coulomb's constant k. It is denoted by ϵ0 and its relation with the Coulomb's constant k is

ϵ0 = 1/4π × k

The value of electric constant ϵ0 is

8.85 × 10-12 C2/N × m2

The constant ϵ0 is more common in electricity-related formulae. This is the reason why it is more important than the Coulomb's constant k.

We can therefore write for the formulae of electrostatic force and electrostatic field in terms of ϵ0:

F = 1/4πϵ0 × Q × Q0/r2

and

E = 1/4πϵ0 × Q/r2

If in a given region of space there is more than one charge that produce their own electric fields E1, E2, + En, the net electric field in that region is the sum of all individual electric field vectors, i.e.

Enet = E1 + E2 + ... + En

The electric field produced by a charged sphere Q at a point of the space that is at distance r from the centre of the sphere, is:

E = 1/4πϵ0 × Q/r2

When distance from centre of sphere is from 0 to R, that is for all points inside the sphere, the electric field is zero. When distance from centre is R, the electric field takes the maximum value. Then, with the increase in distance, electric field decreases until it becomes zero (at the infinity).

The electric field inside a conductor is zero. This means when we place a charge inside a cavity of a conductor, the charge does not move, as it does not experience any electric force.

This property has a wide range of applications in physics. It had been discovered by Faraday, who invented the famous Faraday's Cage, a conducting spherical grid whose internal space has zero electric field despite the grid is charged. Many electric-related phenomena have been studied by inserting the apparatuses inside such a cage, as in this way they are not affected by the surrounding electric field.

## Physics Revision Questions for Electric Field

1) What point charge can produce an electric field of -500 N/C at a distance of 3 m from it?

1. -5 × 10-5 C
2. -5 × 10-6 C
3. -5 × 10-7 C
4. 5 × 10-7 C

2) Two charges Q1 = 400 nC and Q2 = -1200 nC are placed at a distance of 3 m from each other. What is the magnitude and direction of the net electric field at the point P which is 1 m away from the 400 nC charge? (1 nC = 10-9 C) 1. 3600 N/C due right
2. 900 N/C due left
3. 900 N/C due right
4. 6300 N/C due right

3) Two concentric spherical shells of radii R and 2R are shown in the figure. What is the magnitude of the net electric field produced by the two shells at the point A which is at distance 3/2R from the surface of the larger shell? The large sphere has a charge of +3Q and the small sphere has a charge of +Q.

1. 5kQ/2R2
2. 12kQ/25R2
3. 12kQ/5R2
4. 6kQ/25R2