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 = r, 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 is given as
E(r) = $\frac{1}{4πε_0}\frac{Q}{r^2}\hat{\mathbf{r}}$ (1)
where $\hat{\mathbf{r}} = \frac{\mathbf{r}}{r}$, is a unit vector from the origin to the point r. Thus, Eq.(1.6) specifies the value of the electric field for each value of the position vector r. 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 exerted by a charge Q on a charge q, as
F = $\frac{1}{4πε_0}\frac{Qq}{r^2}\hat{\mathbf{r}}$ (2)
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, it experiences a force F equal to the charge q multiplied by the electric field E at the location of q. Thus,
F(r) = qE(r) (3)
Equation (3) defines the SI unit of electric field as N/C. Some important remarks may be made here:
(i) From Eq. (3), 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 is then negligibly small but the ratio F/q is finite and defines the electric field:
E(r) = $\lim_{q \to 0}\frac{\mathbf{F}}{q}$ (4)
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, 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 due to Q, though defined operationally in terms of some test charge q, is independent of q. This is because F is proportional to q, so the ratio F/q does not depend on q. The force F 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 due to Q is also dependent on the space coordinate r. For different positions of the charge q all over the space, we get different values of electric field E. 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.
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| Fig.1 |
(iv) Since the magnitude of the force F 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 will also depend only on the distance r. Thus at equal distances from the charge Q, the magnitude of its electric field E is same. The magnitude of electric field E 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.
Electric field due to a system of charges
Consider a system of charges $q_1$ , $q_2$, ..., $q_n$ with position vectors $\mathbf{r_1}$, $\mathbf{r_2}$, ..., $\mathbf{r_n}$ 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 $q_1$, $q_2$, ..., $q_n$. We can use Coulomb’s law and the superposition principle to determine this field at a point P denoted by position vector r.
Electric field $\mathbf{E_1}$ at r due to $q_1$ at $r_{1}$ is given by
$\mathbf{E_1}=\frac{1}{4πε_0}\frac{q_1}{r^2_{1P}}\hat{\mathbf{r}_{1P}}$
where $\hat{\mathbf{r}_{1P}}$ is a unit vector in the direction from $q_1$ to P, and $r_{1P}$ is the distance between $q_1$ and P. In the same manner, electric field $\mathbf{E_2}$ at r due to $q_2$ at $\mathbf{r_2}$ is
$\mathbf{E_2}=\frac{1}{4πε_0}\frac{q_2}{r^2_{2P}}\hat{\mathbf{r}_{2P}}$
where $\hat{\mathbf{r}_{2P}}$ is a unit vector in the direction from $q_2$ to P, and $r_{2P}$ is the distance between $q_2$ and P. Similar expressions hold good for fields $\mathbf{E_3}$, $\mathbf{E_4}$, ..., $\mathbf{E_n}$ due to charges $q_3$, $q_4$, ..., $q_N$.
By the superposition principle, the electric field E at r due to the system of charges is (as shown in Fig. 2)
$\mathbf{E(r)}=\mathbf{E_1(r)}+\mathbf{E_2(r)}+...+\mathbf{E_n(r)}$
$\mathbf{E(r)} = \frac{1}{4πε_0}\frac{q_1}{r^2_{1P}}\hat{\mathbf{r}_{1P}} + \frac{1}{4πε_0}\frac{q_2}{r^2_{2P}}\hat{\mathbf{r}_{2P}} + ...+\frac{1}{4πε_0}\frac{q_n}{r^2_{nP}}\hat{\mathbf{r}_{nP}}$
$\mathbf{E(r)}=\frac{1}{4πε_0}\sum_{i=2}^{n}\frac{q_i}{r^2_{iP}}\hat{\mathbf{r}}_{iP}$ (5)
E is a vector quantity that varies from one point to another point in space and is determined from the positions of the source charges.
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 Eq.
$\mathbf{F_1}=\frac{q_1}{4πε_0}\sum_{i=2}^{n}\frac{q_i}{r^2_{1i}}\hat{\mathbf{r}}_{1i}$.
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 $q_1$, $q_2$ 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 $q_1$ on $q_2$ cannot arise instantaneously. There will be some time delay between the effect (force on $q_2$) and the cause (motion of $q_1$). 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 $q_2$ and cause a force on $q_2$. 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
What is the magnitude of the electric field at a field E point 2.0 m from a point charge q = 4.0 nC?
Solution:
This problem concerns the electric field due to a point charge. We are given the magnitude of the charge and the distance from the charge to the field point, so we use Eq. (1) to calculate the field magnitude.
From (1):
E(r) = $\frac{1}{4πε_0}\frac{Q}{r^2}\hat{\mathbf{r}}$
E = $\frac{1}{4πε_0}\frac{|Q|}{r^2}=(8.987 \times 10^9 \ Nm^2/C^2) \frac{4.0 \times 10^{-9} \ C}{(2.0 \ m)^2}$
E = 9.0 N/C
Our result means that if we placed a 1.0 C charge at a point 2.0 m from q, it would experience a 9.0 N force. The force on a 2.0 C charge at that point would be (2.0 C)(9.0 N/C) = 18 N, and so on.
Example 2
A point charge $q=-8.0$ nC is located at the origin. Find the electric-field vector at the field point x = 1.2 m, y = -1.6 m.
Solution
Our sketch for this problem.
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| Fig.2 |
We must find the electric-field vector due to a point charge. Figure 21.19 shows the situation. We use Eq. (21.7); to do this, we must find the distance from the source point S (the position of the charge which in this example is at the origin) to the field point and we must obtain an expression for the unit vector $\hat{\mathbf{r}}=\frac{\mathbf{r}}{r}$ that points from S to P.
The distance from to is
$r=\sqrt{x^2 +y^2}=\sqrt{(1.2 \ m)^2 + (-1.6 \ m)^2}=2.0 \ m$
The unit vector $\hat{\mathbf{r}}$ is then
$\hat{\mathbf{r}}=\frac{\mathbf{r}}{r}=\frac{x \hat{\mathbf{i}}+y\hat{\mathbf{j}}}{r}$
$\hat{\mathbf{r}}=\frac{(1.2 \ m) \hat{\mathbf{i}}+(-1.6 \ m)\hat{\mathbf{j}}}{2.0 \ m}$
$\hat{\mathbf{r}}=(0.60 \ m) \hat{\mathbf{i}}-(0.80 \ m)\hat{\mathbf{j}}$
Then, from Eq. (1),
E(r) = $\frac{1}{4πε_0}\frac{Q}{r^2}\hat{\mathbf{r}}$
E(r) = $(8.987 \times 10^9 \ Nm^2/C^2)\frac{(-8.0 \times 10^{-9} \ C)}{(2.00 \ m)^2}(0.60 \ m) \hat{\mathbf{i}}-(0.80 \ m)\hat{\mathbf{j}}$
E(r) = $(-11 \hat{\mathbf{i}}+14\hat{\mathbf{j}}) \ N/C$
Example 3
When the terminals of a battery are connected to two parallel conducting plates with a small gap between them, the resulting charges on the plates produce a nearly uniform electric field between the plates. (In the next section we’ll see why this is.) If the plates are 1.0 cm apart and are connected to a 100-volt battery as shown in Fig. 21.20, the field is vertically upward and has magnitude E = 1.00 $\times 10^4$ N/C. (a) If an electron (charge, $-e = -1.60 \times 10^{-19} \ C$, mass $9.11 \times 10^{-31} kg$) is released from rest at the upper plate, what is its acceleration? (b) What speed and kinetic energy does it acquire while traveling 1.0 cm to the lower plate? (c) How long does it take to travel this distance?
Solution:
A uniform electric field between two parallel conducting plates connected to a 100-volt battery. (The separation of the plates is exaggerated in this figure relative to the dimensions of the plates.)
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| Fig.3 |
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