## Electric Field and Electric Field Lines

The interaction between two charges is determined by Coulomb’s law. How does the interaction itself occur? Consider a point charge kept at a point in space. If another point charge is placed at some distance from the first point charge, it experiences either an attractive force or repulsive force.

This is called ‘action at a distance’. But how does the second charge know about existence of the first charge which is located at some distance away from it? To answer this question, Michael Faraday introduced the concept of field.

According to Faraday, every charge in the universe creates an electric field in the surrounding space, and if another charge is brought into its field, it will interact with the electric field at that point and will experience a force. It may be recalled that the interaction of two masses is similarly explained using the concept of gravitational field (Refer unit 6, volume 2, XI physics).

Both the electric and gravitational forces are non-contact forces, hence the field concept is required to explain action at a distance. Consider a source point charge q located at a point in space. Another point charge q_{0} (test charge) is placed at some point P which is at a distance r from the charge q. The electrostatic force experienced by the charge q_{0} due to q is given by Coulomb’s law.

The charge q creates an electric field in the surrounding space within which its effect can be felt by another charge. It is measured in terms of a quantity called electric field intensity or simply called electric field \(\vec{E}\). The electric field at the point P at a distance r from the point charge q is defined as the force that would be experienced by a unit positive charge placed at that point P and is given by

Here \(\hat{r}\) is the unit vector pointing from q to the point of interest P. The electric field is a vector quantity and its SI unit is newton per coulomb (NC^{-1}).

**Important aspects of Electric field**

(i) If the charge q is positive then the electric field points away from the source charge and if q is negative, the electric field points towards the source charge q. This is shown in the Figure 1.4

(ii) If the electric field at a point P is \(\vec{E}\), then the force experienced by the test charge q_{0} placed at the point P is

\(\vec{F}\) = q_{0}\(\vec{E}\) ……….. (1.5)

This is Coulomb’s law in terms of electric field. This is shown in Figure 1.5

(iii) The equation (1.4) implies that the electric field is independent of the test charge q_{0} and it depends only on the source charge q.

(iv) Since the electric field is a vector quantity, at every point in space, this field has unique direction and magnitude as shown in Figures 1.6(a) and (b). From equation (1.4), we can infer that as distance increases, the electric field decreases in magnitude.

Note that in Figures 1.6 (a) and (b) the length of the electric field vector is shown for three different points. The strength or magnitude of the electric field at point P is stronger than at the points Q and R because the point P is closer to the source charge.

(v) In the definition of electric field, it is assumed that the test charge q_{0} is taken sufficiently small, so that bringing this test charge will not move the source charge. In other words, the test charge is made sufficiently small such that it will not modify the electric field of the source charge.

(vi) The expression (1.4) is valid only for point charges. For continuous and finite size charge distributions, integration techniques must be used (Refer Appendix A1.1). However, this expression can be used as an approximation for a finite-sized charge if the test point is very far away from the finite sized source charge. Note that we similarly treat the Earth as a point mass when we calculate the gravitational field of the Sun on the Earth (Refer unit 6, volume 2, XI physics).

(vii) There are two kinds of the electric field: uniform (constant) electric field and non-uniform electric field. Uniform electric field will have the same direction and constant magnitude at all points in space.

Non-uniform electric field will have different directions or different magnitudes or both at different points in space. The electric field created by a point charge is basically a non uniform electric field. This non-uniformity arises, both in direction and magnitude, with the direction being radially outward (or inward) and the magnitude changes as distance increases. These are shown in Figure 1.7.

Calculate the electric field at points P, Q for the following two cases, as shown in the figure

(a) A positive point charge +1 µC is placed at the origin

(b) A negative point charge -2 µC is placed at the origin

Solution

Case (a)

The magnitude of the electric field at point P is

Since the source charge is positive, the electric field points away from the charge. So the electric field at the point P is given by

For the point Q

Hence \(\vec{E}_{Q}\) = 0.56 × 10^{3} \(\hat{j}\) NC^{-1}

Case (b)

The magnitude of the electric field at point P

Since the source charge is negative, the electric field points towards the charge. So the electric field at the point P is given by

\(\vec{E}_{Q}\) = 0.5 × 10^{3} \(\hat{i}\) NC^{-1}

At the point Q the electric field is directed along the positive x-axis.

**Electric field due to the system of point charges**

Suppose a number of point charges are distributed in space. To find the electric field at some point P due to this collection of point charges, superposition principle is used. The electric field at an arbitrary point due to a collection of point charges is simply equal to the vector sum of the electric fields created by the individual point charges. This is called superposition of electric fields.

Consider a collection of point charges q_{1}, q_{2}, q_{3} ……………. q_{n} located at various points in space. The total electric field at some point P due to all these n charges is given by

where r_{1p}, r_{2p}, r_{3p} ………….. r_{nP} are the distance of the the charges q_{1}, q_{2}, q_{3} …………. q_{n} from the point P respectively. Also \(\hat{r}_{1 P}, \hat{r}_{2 P}, \hat{r}_{3 P} \ldots \ldots \ldots . \hat{r}_{n P}\) are the corresponding unit vectors directed from q_{1}, q_{2}, q_{3} ………. q_{n} to P. Equation (1.7) can be re-written as

For example in Figure 1.8, the resultant electric field due to three point charges q_{1}, q_{2}, q_{3} at point P is shown.

Note that the relative lengths of the electric field vectors for the charges depend on relative distances of the charges to the point P.

**Example 1.7**

Consider the charge configuration as shown in the figure. Calculate the electric field at point A. If an electron is placed at points A, what is the acceleration experienced by this electron? (mass of the electron = 9.1 × 10^{-31} kg and charge of electron = – 1.6 × 10^{-19} C)

Solution

By using superposition principle, the net electric field at point A is

where r_{1A} and r_{2A} are the distances of point A from the two charges respectively

The direction of \(\vec{E}\)_{A} is given by image 21 which is the unit vector along OA as shown in the figure.

The acceleration experienced by an electron placed at point A is

The electron is accelerated in a direction exactly opposite to \(\vec{E}\)_{A}.

Electric field due to continuous charge distribution

The electric charge is quantized microscopically. The expressions (1.2), (1.3), (1.4) are applicable to only point charges. While dealing with the electric field due to a charged sphere or a charged wire etc., it is very difficult to look at individual charges in these charged bodies.

Therefore, it is assumed that charge is distributed continuously on the charged bodies and the discrete nature of charges is not considered here. The electric field due to such continuous charge distributions is found by invoking the method of calculus. (For further reading, refer Appendix A1.1).

**Example 1.8**

A block of mass m carrying a positive charge q is placed on an insulated frictionless inclined plane as shown in the figure. A uniform electric field E is applied parallel to the inclined surface such that the block is at rest. Calculate the magnitude of the electric field E.

Solution

Note: A similar problem is solved in XI^{th} Physics volume I, unit 3 section 3.3.2. There are three forces that acts on the mass m:

(i) The downward gravitational force exerted by the Earth (mg)

(ii) The normal force exerted by the inclined surface (N)

(iii) The Coulomb force given by uniform electric field (qE)

The free body diagram for the mass m is drawn below.

A convenient inertial coordinate system is located in the inclined surface as shown in the figure. The mass m has zero net acceleration both in x and y-direction. Along x-direction, applying Newton’s second law, we have

mg sinθ\(\hat {i} \) – qE\(\hat {i} \) = 0

mg sinθ – qE = 0

E = \(\frac{mgsinθ}{q}\)

Note that the magnitude of the electric field is directly proportional to the mass m and inversely proportional to the charge q. It implies that, if the mass is increased by keeping the charge constant, then a strong electric field is required to stop the object from sliding. If the charge is increased by keeping the mass constant, then a weak electric field is sufficient to stop the mass from sliding down the plane.

The electric field also can be expressed in terms of height and the length of the inclined surface of the plane.

E = \(\frac{mg h}{qL}\)

**Electric Field Lines**

Electric field vectors are visualized by the concept of electric field lines. They form a set of continuous lines which are the visual representation of the electric field in some region of space. The following rules are followed while drawing electric field lines for charges.

The electric field lines start from a positive charge and end at negative charges or at infinity. For a positive point charge the electric field lines point radially outward and for a negative point charge, the electric field lines point radially inward. These are shown in Figure 1.9 (a) and (b).

Note that for an isolated positive point charge the electric field line starts from the charge and ends only at infinity. For an isolated negative point charge the electric field lines start at infinity and end at the negative charge.

The electric field vector at a point in space is tangential to the electric field line at that point. This is shown in Figure 1.10

The electric field lines are denser (more closer) in a region where the electric field has larger magnitude and less dense in a region where the electric field is of smaller magnitude. In other words, the number of lines passing through a given surface area perpendicular to the lines is proportional to the magnitude of the electric field in that region. This is shown in Figure 1.11

Figure 1.11 shows electric field lines from a positive point charge. The magnitude of the electric field for a point charge decreases as the distance increases (|\(\vec{E}\)| ∝ \(\frac{1}{r^{2}}\)). So the electric field has greater magnitude at the surface A than at B. Therefore, the number of lines crossing the surface A is greater than the number of lines crossing the surface B. Note that at surface B the electric field lines are farther apart compared to the electric field lines at the surface A.

No two electric field lines intersect each other. If two lines cross at a point, then there will be two different electric field vectors at the same point, as shown in Figure 1.12.

As a consequence, if some charge is placed in the intersection point, then it has to move in two different directions at the same time, which is physically impossible. Hence, electric field lines do not intersect.

The number of electric field lines that emanate from the positive charge or end at a negative charge is directly proportional to the magnitude of the charges.

For example in the Figure 1.13, the electric field lines are drawn for charges +q and -2q. Note that the number of field lines emanating from +q is 8 and the number of field lines ending at -2q is 16. Since the magnitude of the second charge is twice that of the first charge, the number of field lines drawn for – 2q is twice in number than that for charge +q.

**Example 1.9**

The following pictures depict electric field lines for various charge configurations.

(i) In figure (a) identify the signs of two charges and find the ratio |\(\frac{q_{1}}{q_{2}}\)|

(ii) In figure (b), calculate the ratio of two positive charges and identify the strength of the electric field at three points A, B, and C

(iii) Figure (c) represents the electric field lines for three charges. If q_{2} = -20 nC, then calculate the values of q_{1} and q_{3}

Solution

(i) The electric field lines start at q_{2} and end at q_{1}. In figure (a), q_{2} is positive and q_{1} is negative. The number of lines starting from q_{2} is 18 and number of the lines ending at q_{1} is 6. So q_{2} has greater magnitude. The ratio of |\(\frac{q_{1}}{q_{2}}\)| = \(\frac{N_{1}}{N_{2}}\) = \(\frac{6}{18}\) = \(\frac{1}{3}\). It implies that |q_{2}| = 3|q_{1}|

(ii) In figure (b), the number of field lines emanating from both positive charges are equal (N=18). So the charges are equal. At point A, the electric field lines are denser compared to the lines at point B. So the electric field at point A is greater in magnitude compared to the field at point B. Further, no electric field line passes through C, which implies that the resultant electric field at C due to these two charges is zero.

(iii) In the figure (c), the electric field lines start at q_{1} and q_{3} and end at q_{2}. This implies that q_{1} and q_{3} are positive charges. The ratio of the number of field lines is |\(\frac{q_{1}}{q_{2}}\)| = \(\frac{1}{2}\), implying that q_{1} and q_{3} are half of the magnitude of q_{2}. So q_{1} = q_{3} = +10 nC.

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