Let us consider a point charge Q placed in vacuum, at the origin O. If we place another point charge q at a point P, where oP vector = r vector, then the charge Q will exert a force on q as per Coulomb’s law. We may ask the question: If charge q is removed, then what is left in the surrounding? Is there nothing? If there is nothing at the point P, then how does a force act when we place the charge q at P. In order to answer such questions, the early scientists introduced the concept of field. According to this, we say that the charge Q produces an electric field everywhere in the surrounding. When another charge q is brought at some point P, the field there acts on it and produces a force. The electric field produced by the charge Q at a point r vector is given as

E vector (r vector) = 1/4πε0 × Q/r^2 × r cap (1.6)

where r cap = r vector /r, is a unit vector from the origin to the point r vector. Thus, Equation(1.6) specifies the value of the electric field for each value of the position vector r vector.
The word “field” signifies how some distributed quantity (which could be a scalar or a vector) varies with position. The effect of the charge has been incorporated in the existence of the electric field. We obtain the force F vector exerted by a charge Q on a charge q, as

F vector = 1/4πε0 × (Q q)/r^2 × r cap (1.7)

Note that the charge q also exerts an equal and opposite force on the charge Q. The electrostatic force between the charges Q and q can be looked upon as an interaction between charge q and the electric field of Q and vice versa. If we denote the position of charge q by the vector r vector, it experiences a force F vector equal to the charge q multiplied by the electric field E at the location of q. Thus,

F vector (r vector) = q E vector (r vector)

Equation (1.8) defines the SI unit of electric field as N/C*.
(* An alternate unit V/m will be introduced in the next chapter.)

FIGURE 1.11 Electric field (a) due to a charge Q, (b) due to a charge −Q.

Some important remarks may be made here:
(i) From Equation (1.8), we can infer that if q is unity, the electric field due to a charge Q is numerically equal to the force exerted by it. Thus, the electric field due to a charge Q at a point in space may be defined as the force that a unit positive charge would experience if placed at that point. The charge Q, which is producing the electric field, is called a source charge and the charge q, which tests the effect of a source charge, is called a test charge. Note that the source charge Q must remain at its original location. However, if a charge q is brought at any point around Q, Q itself is bound to experience an electrical force due to q and will tend to move. A way out of this difficulty is to make q negligibly small. The force F vector is then negligibly small but the ratio F vector /q is finite and defines the electric field:
E vector = lim q → 0 (F vector /q) (1.9)

A practical way to get around the problem (of keeping Q undisturbed in the presence of q) is to hold Q to its location by unspecified forces! This may look strange but actually this is what happens in practice. When we are considering the electric force on a test charge q due to a charged planar sheet (Section 1.15), the charges on the sheet are held to their locations by the forces due to the unspecified charged constituents inside the sheet.

(ii) Note that the electric field E vector due to Q, though defined operationally in terms of some test charge q, is independent of q. This is because F vector is proportional to q, so the ratio F vector /q does not depend on q. The force F vector on the charge q due to the charge Q depends on the particular location of charge q which may take any value in the space around the charge Q. Thus, the electric field E vector due to Q is also dependent on the space coordinate r vector. For different positions of the charge q all over the space, we get different values of electric field E vector. The field exists at every point in three-dimensional space.

(iii) For a positive charge, the electric field will be directed radially outwards from the charge. On the other hand, if the source charge is negative, the electric field vector, at each point, points radially inwards.

(iv) Since the magnitude of the force F vector on charge q due to charge Q depends only on the distance r of the charge q from charge Q, the magnitude of the electric field E vector will also depend only on the distance r. Thus at equal distances from the charge Q, the magnitude of its electric field E vector is same. The magnitude of electric field E vector due to a point charge is thus same on a sphere with the point charge at its centre; in other words, it has a spherical symmetry.

1.8.1 Electric field due to a system of charges

Consider a system of charges q1, q2, ..., qn with position vectors r1 vector, r2 vector, ..., rn vector relative to some origin O. Like the electric field at a point in space due to a single charge, electric field at a point in space due to the system of charges is defined to be the force experienced by a unit test charge placed at that point, without disturbing the original positions of charges q1, q2, ..., qn. We can use Coulomb’s law and the superposition principle to determine this field at a point P denoted by position vector r vector.

Electric field E1 vector at r vector due to q1 at r1 vector is given by
E1 vector = 1/4πε0 × q1/r1P^2 × r1P cap
where r1P cap is a unit vector in the direction from q1 to P, and r1P is the distance between q1 and P.

In the same manner, electric field E2 vector at r vector due to q2 at r2 vector is
E2 vector = 1/4πε0 × q2/r2P^2 × r2P cap
where r2P cap is a unit vector in the direction from q2 to P and r2P is the distance between q2 and P.

Similar expressions hold good for fields E3 vector, E4 vector, ..., En vector due to charges q3, q4, ..., qn.
By the superposition principle, the electric field E vector at r vector due to the system of charges is (as shown in Fig. 1.12)

FIGURE 1.12 Electric field at a point due to a system of charges is the vector sum of the electric fields at the point due to individual charges.

E vector (r vector) = E1 vector (r vector) + E2 vector (r vector) + … + En vector (r vector)
= 1/(4πε0) × q1/r1P^2 × r1P cap + 1/(4πε0) × q2/r2P^2 × r2P cap + …… + 1/(4πε0) × qn/rnP^2 × rnP cap
E vector (r vector) = 1(4πε0) × Σ I from 1 to n (q i/riP^2 × riP cap) (1.10)

E vector is a vector quantity that varies from one point to another point in space and is determined from the positions of the source charges.

1.8.2 Physical significance of electric field

You may wonder why the notion of electric field has been introduced here at all. After all, for any system of charges, the measurable quantity is the force on a charge which can be directly determined using Coulomb’s law and the superposition principle [Equation (1.5)]. Why then introduce this intermediate quantity called the electric field?

For electrostatics, the concept of electric field is convenient, but not really necessary. Electric field is an elegant way of characterising the electrical environment of a system of charges. Electric field at a point in the space around a system of charges tells you the force a unit positive test charge would experience if placed at that point (without disturbing the system). Electric field is a characteristic of the system of charges and is independent of the test charge that you place at a point to determine the field. The term field in physics generally refers to a quantity that is defined at every point in space and may vary from point to point. Electric field is a vector field, since force is a vector quantity.

The true physical significance of the concept of electric field, however, emerges only when we go beyond electrostatics and deal with time dependent electromagnetic phenomena. Suppose we consider the force between two distant charges q1, q2 in accelerated motion. Now the greatest speed with which a signal or information can go from one point to another is c, the speed of light. Thus, the effect of any motion of q1 on q2 cannot arise instantaneously. There will be some time delay between the effect (force on q2) and the cause (motion of q1). It is precisely here that the notion of electric field (strictly, electromagnetic field) is natural and very useful. The field picture is this: the accelerated motion of charge q1 produces electromagnetic waves, which then propagate with the speed c, reach q2 and cause a force on q2. The notion of field elegantly accounts for the time delay. Thus, even though electric and magnetic fields can be detected only by their effects (forces) on charges, they are regarded as physical entities, not merely mathematical constructs. They have an independent dynamics of their own, i.e., they evolve according to laws of their own. They can also transport energy. Thus, a source of time dependent electromagnetic fields, turned on for a short interval of time and then switched off, leaves behind propagating electromagnetic fields transporting energy. The concept of field was first introduced by Faraday and is now among the central concepts in physics.

Example 1.8

An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 10^4 N C^−1 [Fig. 1.13(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig. 1.13(b)]. Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’.

Fig. 1.13

Solution:
In Fig. 1.13(a) the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is
A e = eE/mE
where mE is the mass of the electron.
Starting from rest, the time required by the electron to fall through a distance h is given by
t e = √(2h/a e) = √(2h m e/e E)
For e = 1.6 × 10^−19C, m e = 9.11 × 10^−31 kg,
E = 2.0 × 10^4 N C^−1, h = 1.5 × 10^−2 m,
t e = 2.9 × 10^−9s

In Fig. 1.13 (b), the field is downward, and the positively charged proton experiences a downward force of magnitude eE. The acceleration of the proton is
ap = eE/mp
where mp is the mass of the proton; mp = 1.67 × 10^−27 kg.
The time of fall for the proton is
tp = √(2h/ap) = √(2hmp/eE) = 1.3 × 10^−7

Thus, the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of ‘free fall under gravity’ where the time of fall is independent of the mass of the falling body.

Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. To see if this is justified, let us calculate the acceleration of the proton in the given electric field:
a p = e E/m p
= (1.6 × 10^−19 C) × (2.0 × 10^4 NC^−1)/1.67 × 10^−27 kg
= 1.9 × 10^12 ms^−2
which is enormous compared to the value of g (9.8 m s^−2), the acceleration due to gravity. The acceleration of the electron is even greater. Thus, the effect of acceleration due to gravity can be ignored in this example.

Example 1.9

Two point charges q1 and q2, of magnitude +10^−8 C and −10^−8 C, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Fig. 1.14.

Fig. 1.14

Solution:
The electric field vector E1A vector at A due to the positive charge q1 points towards the right and has a magnitude
E1A vector = (9 × 10^9 Nm^2C^−2) × (10^−8C)/(0.05m)^2
= 3.6 × 10^4 NC^−1

The electric field vector E2A vector at A due to the negative charge q2 points towards the right and has the same magnitude. Hence the magnitude of the total electric field EA vector at A is
EA vector = E1A vector + E2A vector = 7.2 × 10^4 N C^−1
EA vector is directed toward the right.
The electric field vector E1B vector at B due to the positive charge q1 points towards the left and has a magnitude
E1B vector = (9 × 10^9 Nm^2C^−2) × (10^−8C)/(0.05m)^2 = 3.6 × 10^4 NC^−1
The electric field vector E2B vector at B due to the negative charge q2 points towards the right and has a magnitude
E2B vector = (9 × 10^9 Nm^2C^−2) × (10^−8C)/(0.15m)^2 = 4 × 10^3 NC^−1

The magnitude of the total electric field at B is
E B vector = E1B vector − E2B vector
= 3.2 × 10^4 N C^−1
EB vector is directed towards the left.
The magnitude of each electric field vector at point C, due to charge q1 and q2 is
E1C vector = E2C vector
= (9 × 10^9 Nm^2C^−2) × (10^−8C)/(0.10m)^2
= 9 × 10^3 NC^−1

The directions in which these two vectors point are indicated in Fig. 1.14. The resultant of these two vectors is
EC vector = E1 vector cos π/3 + E2 vector cos π/3
= 9 × 10^3 NC^−1
EC points towards the right.