1.16 SUMMARY

1. Electric and magnetic forces determine the properties of atoms, molecules and bulk matter.

2. From simple experiments on frictional electricity, one can infer that there are two types of charges in nature; and that the like charges repel and unlike charges attract each other. By convention, the charge on a glass rod rubbed with silk is taken as positive; and that on a plastic rod rubbed with fur is then taken as negative.

3. Conductors allow movement of electric charge through them, insulators do not. In metals, the mobile charges are electrons; in electrolytes both positive and negative ions are mobile.

4. Electric charge has three basic properties: quantisation, additivity and conservation.

Quantisation of electric charge means that total charge (q) of a body is always an integral multiple of a basic quantum of charge (e) i.e., q = n e, where n = 0, ±1, ±2, ±3, ....; Proton and electron have charges +e, and −e, respectively. For macroscopic charges for which n is a very large number, quantisation of charge can be ignored.

Additivity of electric charges means that the total charge of a system is the algebraic sum (i.e., the sum taking into account proper signs) of all individual charges in the system.

Conservation of electric charges means that the total charge of an isolated system remains unchanged with time. This means that when bodies are charged through friction, there is a transfer of electric charge
from one body to another, but no creation or destruction of charge.

5. Coulomb’s Law: The mutual electrostatic force between two point charges q1 and q2 is proportional to the product q1q2 and inversely proportional to the square of the distance r21 separating them.

Mathematically,

F21 vector = force on q2 due to q1 = k(q1q2)/r21^2 = r21 cap

where r21 cap is a unit vector in the direction from q1 to q2 and k = 1/(4πε0) is the constant of proportionality.

In SI units, the unit of charge is coulomb. The experimental value of the constant ε0 is

ε0 = 8.854 × 10^−12 C^2 N^−1 m^−2

The approximate value of k is

k = 9 × 10^9 N m^2 C^−2

6. The ratio of electric force and gravitational force between a proton and an electron is

K e^2/G me mp ≅ 2.4 × 10^39

7. Superposition Principle: The principle is based on the property that the forces with which two charges attract or repel each other are not affected by the presence of a third (or more) additional charge(s). For an assembly of charges q1, q2, q3, ..., the force on any charge, say q1, is the vector sum of the force on q1 due to q2, the force on q1 due to q3, and so on. For each pair, the force is given by the Coulomb’s law for two charges stated earlier.

8. The electric field E vector at a point due to a charge configuration is the force on a small positive test charge q placed at the point divided by the magnitude of the charge. Electric field due to a point charge q has a magnitude |q|/4πε0 r^2; it is radially outwards from q, if q is positive, and radially inwards if q is negative.
Like Coulomb force, electric field also satisfies superposition principle.

9. An electric field line is a curve drawn in such a way that the tangent at each point on the curve gives the direction of electric field at that point.


The relative closeness of field lines indicates the relative strength of electric field at different points; they crowd near each other in regions of strong electric field and are far apart where the electric field is weak.


In regions of constant electric field, the field lines are uniformly spaced parallel straight lines.

10. Some of the important properties of field lines are:


(i) Field lines are continuous curves without any breaks.


(ii) Two field lines cannot cross each other.


(iii) Electrostatic field lines start at positive charges and end at negative charges; they cannot form closed loops.

11. An electric dipole is a pair of equal and opposite charges q and −q separated by some distance 2a. Its dipole moment vector p vector has magnitude 2q a and is in the direction of the dipole axis from −q to +q.

12. Field of an electric dipole in its equatorial plane (i.e., the plane perpendicular to its axis and passing through its centre) at a distance r from the centre:


E vector = −p vector /4πε0 × 1/(a^2 + r^2)^3/2


≅ −p vector /4πε0r^3, for r >> a

Dipole electric field on the axis at a distance r from the centre:


E vector = 2p vector r/4πε0(r^2 − a^2)^2


≅ 2p vector /4πε0r^3 for r >> a

The 1/r^3 dependence of dipole electric fields should be noted in contrast to the 1/r^2 dependence of electric field due to a point charge.

13. In a uniform electric field E vector, a dipole experiences a torque τ vector given by


τ vector = p vector × E vector


but experiences no net force.

14. The flux Δφ of electric field E vector through a small area element ΔS vector is given by


Δφ = E vector • ΔS vector

The vector area element ΔS vector is


ΔS vector = ΔS n cap


where ΔS is the magnitude of the area element and n cap is normal to the area element, which can be considered planar for sufficiently small ΔS.

For an area element of a closed surface, n cap is taken to be the direction of outward normal, by convention.

15. Gauss’s law: The flux of electric field through any closed surface S is 1/ε0 times the total charge enclosed by S. The law is especially useful in determining electric field E vector, when the source distribution has simple symmetry:

(i) Thin infinitely long straight wire of uniform linear charge density λ


E vector = λ/2πε0r × n cap


where r is the perpendicular distance of the point from the wire and n cap is the radial unit vector in the plane normal to the wire passing through the point.

(ii) Infinite thin plane sheet of uniform surface charge density σ


E vector = σ/2ε0 × n cap


where n cap is a unit vector normal to the plane, outward on either side.

(iii) Thin spherical shell of uniform surface charge density σ


E vector = q/4πε0r^2 × r cap (r ≥ R)


E vector = 0 (r < R)


where r is the distance of the point from the centre of the shell and R the radius of the shell. q is the total charge of the shell: q = 4πR^2 σ.

The electric field outside the shell is the same as though the total charge of the shell is concentrated at the centre of the shell. The same result is true for a solid sphere of uniform volume charge density.

The field is zero at all points inside the shell.

Physical quantity	Symbol	Dimensions	Unit	Remarks
Vector area element	ΔS vector	[L^2]	m^2	ΔS vector = ΔS n cap
Electric field	E vector	[MLT^−3A^−1]	V m^−1
Electric flux	φ	[ML^3 T^−3A^−1]	V m	Δφ = E vector. ΔS vector
Dipole moment	p vector	[LTA]	C m	Vector directed from negative to positive charge
Charge density:
Linear	λ	[L^−1 TA]	C m^−1	Charge/length
Surface	σ	[L^−2 TA]	C m^−2	Charge/area
Volume	ρ	[L^−3 TA]	C m^−3	Charge/volume

POINTS TO PONDER

1. You might wonder why the protons, all carrying positive charges, are compactly residing inside the nucleus. Why do they not fly away? You will learn that there is a third kind of a fundamental force, called the strong force which holds them together. The range of distance where this force is effective is, however, very small ~10^−14 m. This is precisely the size of the nucleus. Also the electrons are not allowed to sit on top of the protons, i.e. inside the nucleus, due to the laws of quantum mechanics. This gives the atoms their structure as they exist in nature.

2. Coulomb force and gravitational force follow the same inverse-square law. But gravitational force has only one sign (always attractive), while Coulomb force can be of both signs (attractive and repulsive), allowing possibility of cancellation of electric forces. This is how gravity, despite being a much weaker force, can be a dominating and more pervasive force in nature.

3. The constant of proportionality k in Coulomb’s law is a matter of choice if the unit of charge is to be defined using Coulomb’s law. In SI units, however, what is defined is the unit of current (A) via its magnetic effect (Ampere’s law) and the unit of charge (coulomb) is simply defined by (1C = 1 A s). In this case, the value of k is no longer arbitrary; it is approximately 9 × 10^9 N m^2 C^−2.

4. The rather large value of k, i.e., the large size of the unit of charge (1C) from the point of view of electric effects arises because (as mentioned in point 3 already) the unit of charge is defined in terms of magnetic forces (forces on current-carrying wires) which are generally much weaker than the electric forces. Thus while 1 ampere is a unit of reasonable size for magnetic effects, 1 C = 1 A s, is too big a unit for electric effects.

5. The additive property of charge is not an ‘obvious’ property. It is related to the fact that electric charge has no direction associated with it; charge is a scalar.

6. Charge is not only a scalar (or invariant) under rotation; it is also invariant for frames of reference in relative motion. This is not always true for every scalar. For example, kinetic energy is a scalar under rotation, but is not invariant for frames of reference in relative motion.

7. Conservation of total charge of an isolated system is a property independent of the scalar nature of charge noted in point 6. Conservation refers to inVariance in time in a given frame of reference. A quantity may be scalar but not conserved (like kinetic energy in an inelastic collision). On the other hand, one can have conserved vector quantity (e.g., angular momentum of an isolated system).

8. Quantisation of electric charge is a basic (unexplained) law of nature; interestingly, there is no analogous law on quantisation of mass.

9. Superposition principle should not be regarded as ‘obvious’, or equated with the law of addition of vectors. It says two things: force on one charge due to another charge is unaffected by the presence of other charges, and there are no additional three-body, four-body, etc., forces which arise only when there are more than two charges.

10. The electric field due to a discrete charge configuration is not defined at the locations of the discrete charges. For continuous volume charge distribution, it is defined at any point in the distribution. For a surface charge distribution, electric field is discontinuous across the surface.

11. The electric field due to a charge configuration with total charge zero is not zero; but for distances large compared to the size of the configuration, its field falls off faster than 1/r^2, typical of field due to a single charge. An electric dipole is the simplest example of this fact.

EXERCISES

Q1.1 What is the force between two small charged spheres having charges of 2 × 10^−7C and 3 × 10^−7C placed 30 cm apart in air?

A1.1 6 × 10^−3 N (repulsive)

Q1.2 The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge −0.8 μC in air is 0.2 N.


(a) What is the distance between the two spheres?


(b) What is the force on the second sphere due to the first?

A1.2 (a) 12 cm


(b) 0.2 N (attractive)

Q1.3 Check that the ratio k e^2/G me mp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

A1.3 2.4 × 10^39. This is the ratio of electric force to the gravitational force (at the same distance) between an electron and a proton.

Q1.4


(a) Explain the meaning of the statement ‘electric charge of a body is quantised’.


(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Q1.5 When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

A1.5 Charge is not created or destroyed. It is merely transferred from one body to another.

Q1.6 Four point charges qA = 2 μC, qB = −5 μC, qC = 2 μC, and qD = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

A1.6 Zero N

Q1.7


(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?


(b) Explain why two field lines never cross each other at any point?

Q1.8 Two point charges qA = 3 μC and qB = −3 μC are located 20 cm apart in vacuum.


(a) What is the electric field at the midpoint O of the line AB joining the two charges?


(b) If a negative test charge of magnitude 1.5 × 10^−9 C is placed at this point, what is the force experienced by the test charge?

A1.8 (a) 5.4 × 10^6 N C^−1 along OB


(b) 8.1 × 10^−3 N along OA

Q1.9 A system has two charges qA = 2.5 × 10^−7 C and qB = −2.5 × 10^−7 C located at points A: (0, 0, −15 cm) and B: (0, 0, +15 cm), respectively.


What are the total charge and electric dipole moment of the system?

A1.9 Total charge is zero. Dipole moment = 7.5 × 10^−8 C m along z-axis.

Q1.10 An electric dipole with dipole moment 4 × 10^−9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10^4 NC^−1.


Calculate the magnitude of the torque acting on the dipole.

A1.10 10^−4 N m

Q1.11 A polythene piece rubbed with wool is found to have a negative charge of 3 × 10^−7 C.


(a) Estimate the number of electrons transferred (from which to which?)


(b) Is there a transfer of mass from wool to polythene?

A1.11 (a) 2 × 10^12, from wool to polythene.


(b) Yes, but of a negligible amount ( = 2 × 10^−18 kg in the example).

Q1.12 (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10^−7 C? The radii of A and B are negligible compared to the distance of separation.


(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

A1.12 (a) 1.5 × 10^−2 N


(b) 0.24 N

Q1.13 Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

A1.13 5.7 × 10^−3 N

Q1.14 Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Figure 1.33

A1.14 Charges 1 and 2 are negative, charge 3 is positive. Particle 3 has the highest charge to mass ratio.

Q1.15 Consider a uniform electric field E vector = 3 × 10^3 i cap N/C.


(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?


(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

A1.15 (a) 30Nm^2/C,


(b) 15 Nm^2/C

Q1.16 What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

A1.16 Zero. The number of lines entering the cube is the same as the number of lines leaving the cube.

Q1.17 Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10^3 Nm^2/C.


(a) What is the net charge inside the box?


(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

A1.17 (a) 0.07 μC


(b) No, only that the net charge inside is zero.

Q1.18 A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

Figure 1.34

A1.18 2.2 × 10^5 N m^2/C

Q1.19 A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

A1.19 1.9 × 10^5 N m^2/C

Q1.20 A point charge causes an electric flux of −1.0 × 10^3 Nm^2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.


(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?


(b) What is the value of the point charge?

A1.20 (a) −10^3 N m^2/C; because the charge enclosed is the same in the two cases.


(b) −8.8 nC

Q1.21 A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10^3 N/C and points radially inward, what is the net charge on the sphere?

A1.21 −6.67 nC

Q1.22 A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m^2.


(a) Find the charge on the sphere.


(b) What is the total electric flux leaving the surface of the sphere?

A1.22 (a) 1.45 × 10^−3 C


(b) 1.6 × 10^8 Nm^2/C

Q1.23 An infinite line charge produces a field of 9 × 10^4 N/C at a distance of 2 cm. Calculate the linear charge density.

A1.23 10 μC/m

Q1.24 Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10^−22 C/m^2. What is E vector:


(a) in the outer region of the first plate,


(b) in the outer region of the second plate, and


(c) between the plates?

A1.24 (a) Zero,


(b) Zero,


(c) 1.9 N/C

ADDITIONAL EXERCISES

Q1.25 An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10^4 NC^−1 (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm^−3. Estimate the radius of the drop. (g = 9.81 m s^−2; e = 1.60 × 10^−19 C).

A1.25 9.81 × 10^−4 mm.

Q1.26 Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

Fig 1.35 (a)

Fig 1.35 (b)

Fig 1.35 (c)

Fig 1.35 (d)

Fig 1.35 (e)

A1.26 Only (c) is right; the rest cannot represent electrostatic field lines,


(a) is wrong because field lines must be normal to a conductor,


(b) is wrong because field lines cannot start from a negative charge,


(d) is wrong because field lines cannot intersect each other,


(e) is wrong because electrostatic field lines cannot form closed loops.

Q1.27 In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10^5 NC^−1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10^−7 Cm in the negative z-direction ?

A1.27 The force is 10^−2 N in the negative z-direction, that is, in the direction of decreasing electric field. You can check that this is also the direction of decreasing potential energy of the dipole; torque is zero.

Q1.28


(a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.

Fig. 1.36 (a)

(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. 1.36(b)].

Fig. 1.36(b)

(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

A1.28 (a) Hint: Choose a Gaussian surface lying wholly within the conductor and enclosing the cavity.


(b) Gauss’s law on the same surface as in


(a) shows that q must induce −q on the inner surface of the conductor.


(c) Enclose the instrument fully by a metallic surface.

Q1.29 A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε0) n cap, where n cap is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.

A1.29 Hint: Consider the conductor with the hole filled up. Then the field just outside is (σ/ε0) n cap and is zero inside. View this field as a superposition of the field due to the filled up hole plus the field due to the rest of the charged conductor. Inside the conductor, these fields are equal and opposite. Outside they are equal both in magnitude and direction. Hence, the field due to the rest of the conductor is (σ/2ε0) n cap.

Q1.30 Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Q1.31 It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + (2/3) e, and the ‘down’ quark (denoted by d) of charge (−1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

A1.31 p;uud; n;udd.

Q1.32 (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E vector = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.


(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

A1.32 (a) Hint: Prove it by contradiction. Suppose the equilibrium is stable; then the test charge displaced slightly in any direction will experience a restoring force towards the null-point. That is, all field lines near the null point should be directed inwards towards the null-point. That is, there is a net inward flux of
electric field through a closed surface around the null-point. But by Gauss’s law, the flux of electric field through a surface, not enclosing any charge, must be zero. Hence, the equilibrium cannot be stable.


(b) The mid-point of the line joining the two charges is a null-point. Displace a test charge from the null-point slightly along the line. There is a restoring force. But displace it, say, normal to the line. You will see that the net force takes it away from the null-point. Remember, stability of equilibrium needs restoring
force in all directions.

Q1.33 A particle of mass m and charge (−q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL^2/(2m vx^2).

Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class 11 Textbook of Physics.

Q1.34 Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx = 2.0 × 10^6 m s^−1. If E between the plates separated by 0.5 cm is 9.1 × 10^2 N/C, where will the electron strike the upper plate? (|e|=1.6 × 10^−19 C, me = 9.1 × 10^−31 kg.)

A1.34 1.6 cm

Congratulations! You have completed this chapter. I hope you enjoyed studying this chapter. In case you found any difficulties in this chapter or have any suggestions to improve it, please write to us at ‘blind2Visionary@gmail.com’.

End of Chapter 1 ELECTRIC CHARGES AND FIELDS