1.electric Charges and Fields
1.electric Charges and Fields
1.electric Charges and Fields
Electricity:
It is the branch of physics which deals with the study of electric charges. It is divided in to two
branches. They are
1. Electrostatics (or) static electricity: which deals with the study of charges at rest.
2. Current electricity: which deals with the study of charges in motion.
Historical development of electricity:
One of the Greek scientist Thales of Miletus (600 B.C) discovered that, when the amber rubbed with
fur, the amber acquires the property of attracting light objects like dust, light feather, lint etc.
William Gilbert (1544-1603) made a systematic study on this phenomenon. His study showed
that, many other substances behaved in this manner when they rubbed suitably. He named this phenomenon
as electrification. He introduced the word “electricity” which
is derived from Greek word ‘elektron’, meaning is amber.
Amber is a fossilized resin found near the seashore and fur is a
kind of wool.
Modern concept of matter:
1. A matter is made up of atoms. The atom consists of
positively charged nucleus at the centre and negatively
charged electrons revolving around the nucleus in
different orbits.
2. The nucleus consists of protons and neutrons. They together called as nucleons.
3. Mass of proton mp=1.673×10-27kg, mass of neutron, mn=1.675×10-27 kg and mass of the electron
me= 9.1×10-31kg.
4. The smallest charge available in the nature is charge on the electron. It is equal to -1.6×10-19C. This
was experimentally determined by Millikan.
5. The magnitude of charge on proton is same as that on the electron but opposite in sign.
6. In an atom number of electrons is equal to the number of protons and hence an atom is said to be
electrically neural.
7. The neutrons have no charge.
8. A body is said to be charged if it has less (deficiency or shortage) electrons or more (surplus or excess)
electrons.
9. A body is said to be positively charged if it has less number of electrons.
10. A body is said to be negatively charged if it has more number of electrons.
11. Two identical bodies, one with positively charged and the other with negatively charged have different
masses.
12. Mass of the body with positive charge is slightly less than the mass of a body with negative charge.
Showing two kinds of charges:
Experiment -1:
1. When two glass rods rubbed with a piece of
silk cloth are brought close to each other,
they repel each other.
2. When two ebonite rods (plastic rods) rubbed
with a fur (wool) are brought close to each
other, they repel each other.
3. When a glass rod rubbed with a piece of silk cloth and a ebonite rod rubbed with a fur are brought
close to each other, they attract each other.
Experiment -2:
1. When a glass rod rubbed with a piece of silk
is made to touch with two pith balls suspended
by silk threads, the balls repel each other.
Coulomb's law is valid for any sign of 𝑞1 and 𝑞2 . If 𝑞1 and 𝑞2 are of same sign (either both positive or
both negative), then the product 𝑞1 𝑞2 > 0 and the force is repulsive force. If 𝑞1 and 𝑞2 are of opposite sign,
then the product 𝑞1 𝑞2 < 0 and the force is attractive force.
NP3. Coulomb’s law for electrostatic force between two point charges and Newton’s law for gravitational
force between two stationary point masses, both have inverse-square dependence on the distance between the
charges and masses respectively. (a) Compare the strength of these forces by determining the ratio of their
magnitudes (i) for an electron and a proton and (ii) for two protons. (b) Estimate the accelerations of electron
and proton due to the electrical force of their mutual attraction when they are 1 Ao (= 10-10 m) apart? (mp =
1.67×10–27 kg, me = 9.11× 10–31 kg)
(a) (i) The electric force between an electron and a proton at a distance r apart is:
1 𝑒2
𝐹𝑒 = −
4𝜋𝜖𝑜 𝑟 2
where the negative sign indicates that the force is attractive. The corresponding gravitational force (always
attractive) is:
𝑚𝑝 𝑚𝑒
𝐹𝐺 = −𝐺
𝑟2
where mp and me are the masses of a proton and an electron respectively.
𝐹 𝑒2
|𝐹𝑒 |= 𝐺 4𝜋𝜖 = 2.4× 1039
𝐺 𝑜 𝐺𝑚𝑝 𝑚𝑒
(ii) On similar lines, the ratio of the magnitudes of electric force to the gravitational force between two
protons at a distance r apart is:
𝐹 𝑒2
|𝐹𝑒 |= 𝐺 4𝜋𝜖 = 1.3 × 1036
𝐺 𝑜 𝐺𝑚𝑝 𝑚𝑒
However, it may be mentioned here that the signs of the two forces are different. For two protons, the
gravitational force is attractive in nature and the Coulomb force is repulsive. The actual values of these forces
between two protons inside a nucleus (distance between two protons is ~ 10-15 m inside a nucleus) are Fe ~
230 N, whereas, FG ~ 1.9×10–34 N.
The (dimensionless) ratio of the two forces shows that electrical forces are enormously stronger than the
gravitational forces.
(b) The electric force F exerted by a proton on an electron is same in magnitude to the force exerted by an
electron on a proton; however, the masses of an electron and a proton are different. Thus, the magnitude of
force is
1 𝑒2
|F| = 4𝜋𝜖 = 8.987 ×109 Nm2/C2 × (1.6×10–19C)2 / (10–10m)2 = 2.3 ×10–8 N
𝑜 𝑟2
Using Newton’s second law of motion, F = ma, the acceleration that an electron will undergo is
a = 2.3×10–8 N / 9.11×10–31 kg = 2.5× 1022 m/s2
Dr. Sankara Rao Gattu, (9949435575) Page 5
Comparing this with the value of acceleration due to gravity, we can conclude that the effect of gravitational
field is negligible on the motion of electron and it undergoes very large accelerations under the action of
Coulomb force due to a proton.
The value for acceleration of the proton is
2.3 × 10–8 N / 1.67×10–27 kg = 1.4 ×1019 m/s2
NP.4 A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by
an insulating handle is brought close to A such that the distance between their centres is 10 cm, as shown in
Fig. (a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the
deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively,
as shown in Fig. (b). C and D are then removed and B is brought closer to A to a distance of 5.0 cm between
their centres, as shown in Fig.(c). What is the expected repulsion of A on the basis of Coulomb’s law? Spheres
A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation
between their centres.
Solution
Let the original charge on sphere A be q and that on B be q’. At a distance r between their centres, the
magnitude of the electrostatic force on each is given by
1 𝑞𝑞′
𝐹=
4𝜋𝜖𝑜 𝑟 2
neglecting the sizes of spheres A and B in comparison to r. When an identical but uncharged sphere C touches
A, the charges redistribute. on A and C and, by symmetry, each sphere carries a charge q/2. Similarly, after D
touches B, the redistributed charge on each is q’/2. Now, if the separation between A and B is halved, the
magnitude of the electrostatic force on each is
𝑞𝑞′
′ 1 22 1 𝑞𝑞′
𝐹 = 4𝜋𝜖 𝑟 2
= 4𝜋𝜖 =F
𝑜( ) 𝑜 𝑟2
2
Thus the electrostatic force on A, due to B, remains unaltered.
TP1: 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?
TP2: 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?
TP7: 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?
TP8: 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?
TP 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?
Coulomb's law in vector form in terms of position vectors of two charges.(or) Coulomb's law agrees
with the Newton’s third law of motion:
Consider two point charges q1 and q2 located at two points in vacuum. Let r1 and r2 be the position
vectors of q1 and q2 respectively with reference to the origin O of the three-dimensional co-ordinate system.
From triangle law of vector addition, r21 = (r2 − r1 ) and r12 = (r1 − r2 )
From coulomb's law,
q n
q
F = 1 2i rˆ1i where r̂12 , r̂13 , r̂14 ....... rˆ1n are the unit vectors.
4 o i = 2 r1i
Note: The vector sum is obtained by the parallelogram law of vector addition for two forces and polygon law
of vector addition for more than two forces.
NP.5 Consider three charges q1, q2, q3 each equal to q at the vertices of an
equilateral triangle of side l. What is the force on a charge Q (with the same
sign as q) placed at the centroid of the triangle, as shown in Fig.
The forces acting on charge q at A due to charges q at B and –q at C are 𝐹⃑12 along BA and 𝐹⃑13 along AC
respectively, as shown in Fig. By the parallelogram law, the total force 𝐹⃑1 on the charge q at A is given by
𝐹⃑1 = 𝐹 𝑟̂1 where 𝑟̂1 is a unit vector along BC.
1 𝑞2
The force of attraction or repulsion for each pair of charges has the same magnitude F = 4𝜋𝜖
𝑜 𝑙2
The total force 𝐹⃑2 on charge q at B is thus 𝐹⃑2 = 𝐹 𝑟̂2, where 𝑟̂2 is a unit vector along AC.
Similarly the total force on charge –q at C is 𝐹⃑3 = √3𝐹 𝑛̂, where 𝑛̂ is the unit vector along the direction
bisecting the ∠𝐵𝐶𝐴.
It is interesting to see that the sum of the forces on the three charges is zero, i.e.,
𝐹⃑1 + 𝐹⃑2 + 𝐹⃑3 = 0
The result is not at all surprising. It follows straight from the fact that Coulomb’s law is consistent with
Newton’s third law. The proof is left to you as an exercise.
Electric Field:
It is the region in which the charged particles experience electrostatic force.
Source charge: It is the charge which produces the electric field that we define the source charge can be a
single charge or a group of charges or any continuous distribution of charge.
The limit q o → 0 indicates that the test charge is infinitesimally small so that it does not change
original field configuration (arrangement). But, minimum value of charge is the charge on the proton or
electron (1.6×10-19C). Therefore, in classical and microscopic situations, we can safely ignore this problem.
The test charge is merely introduced for measurement of electric field in convenient manner.
Note: The magnitude of electric field depends only on the source charge and the distance of the test charge
from source charge. It does not depend on the magnitude of test charge.
If the source charge is positive, the direction of electric field is directed radially outward from the
source charge. If the source charge is negative, the direction of electric filed is directed radially inwards
(towards the source charge)
TP 13: Figure 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?
Superposition principle of electric fields and electric field due to system of n point charges:
Statement:
The resultant electric field at a point due to number of charges is the
vector sum of all the electric fields due to number of charges.
Let 𝑞1 , 𝑞2 , 𝑞3 … … … , 𝑞𝑛 be the point charges at distances
𝑟1 , 𝑟2 , 𝑟3 … … , 𝑟𝑛 respectively from a point P.
1 q1
Electric field at P due to charge 𝑞1 , E1 = rˆ1
4 o r12
1 q2
Electric field at P due to charge 𝑞2 , E 2 = rˆ2
4 o r22
1 qn
Similarly, Electric field at P due to charge 𝑞𝑛 , E n = rˆn
4 o rn2
From superposition principle,
The resultant electric field, E = E1 + E 2 + ....... + E n
1 q1 1 q2 1 qn
E= rˆ + rˆ +……. + rˆn
4 o r12 1
4 o r2 2 2
4 o rn2
1 n qi
E= rˆi where r̂1 , r̂2 , r̂3 ....... r̂n are the unit vectors.
4 o i =1 ri 2
The vector sum is obtained by the parallelogram law of vector addition for two electric fields and
polygon law of vector addition for more than two electric fields.
In Fig.(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
ae = 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
2ℎ 2ℎ𝑚𝑒
𝑡𝑒 = √ 𝑎 = √ = 2.9 ×10–9s
𝑒 𝑒𝐸
For e = 1.6 ×10–19C, me = 9.11×10–31 kg, E = 2.0 ×104 N C–1, h = 1.5 ×10–2 m, te = 2.9 ×10–9s
In Fig. (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
2ℎ 2ℎ𝑚𝑝
𝑡𝑝 = √𝑎 = √ 𝑒𝐸
= 1.3 ×10–7s
𝑝
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
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:
ap = eE/mp = 1.9 ×1012 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.
The electric field vector 𝐸⃗⃑1𝐴 at A due to the positive charge q1 points
towards the right and has a magnitude
𝐸⃗⃑1𝐴 = 3.6 ×104 N C–1
The electric field vector 𝐸⃗⃑2𝐴 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 at A is
EA = E1A + E2A = 7.2 ×104 N C–1
𝐸⃗⃑𝐴 is directed toward the right.
The electric field vector 𝐸⃗⃑1𝐵 at B due to the positive charge q1 points towards the left and has a magnitude
𝐸⃗⃑1𝐵 = 3.6 ×104 N C–1
The electric field vector 𝐸⃗⃑2𝐵 at B due to the negative charge q2 points towards the right and has a magnitude
𝐸⃗⃑2𝐵 = 4 ×103 N C–1
The magnitude of the total electric field at B is. EB = E1B – E2B = 3.2 ×104 N C–1
1 q
Similarly, electric field at P due to +q, E 2 =
4 o r + a 2
2
(
along BPQ
)
From resolution of vectors,
E1 cos is the component of E1 along PR
E2 cos is the component of E 2 along PR
Resultant electric field at P, E = E1 cos + E2 cos
But E1 = E2, E = 2 E1 cos -------- (3)
1 q
Eq.(2) in Eq (3) E = 2 cos ---- (4)
4 o (r + a 2 )
2
a a
From triangle POA, cos = =
AP (r 2
+ a2 )
1 q a
Eq (4) E = 2
4 o (r + a2 2
) (r 2
+ a2 )
1 2qa 1 p
E= E=− q×2a = p = electric dipole moment
(
4 o r 2 + a 2 2 ) 3
4 o r 2 + a 2 3 2 ( )
where negative sign indicates that the electric field and dipole moment are in opposite directions
1 p
In vector form E = −
4 o r 2 + a 2 3 2 ( )
p 1
Electric field at the midpoint of the dipole is E = ( at mid-point r = 0)
4 o a 3
1 p 1 p
For short dipole r >>a, a can be neglected. E =
2 In vector form E = −
4 o r 3 4 o r 3
Point dipole:
When the distance between the two charges of a dipole approaches zero, then the dipole is said to be
point dipole.
Dipole in a uniform external electric field:
When an electric dipole is placed in uniform electric field, net force on the dipole is zero. When an
electric dipole is placed in non-uniform electric field, net force on the dipole is not zero. A torque acts on the
dipole placed in a uniform or non-uniform electric field till the dipole is completely aligned along the direction
of electric field.
Physical significance of dipoles:
Most of the molecules have zero dipole moment. However they develop dipole moment when they are
placed in an external electric field. But in some molecules there is permanent electric dipole moment even in
the absence of electric field.
The substances having this type of molecules are used in important applications in the presence or
absence of external electric field.
NP 9. Two charges ±10 𝜇 C are placed 5.0 mm apart. Determine the
electric field at (a) a point P on the axis of the dipole 15 cm away
from its centre O on the side of the positive charge, as shown in
Fig.(a), and (b) a point Q, 15 cm away from O on a line passing
through O and normal to the axis of the dipole, as shown in Fig. (b).
1 q q
(1) = 4r 2 = This is Gauss's theorem.
4 o r 2
o
Verification of gauss's law:
To verify Gauss's law, let us calculate electric flux through a closed
cylindrical surface containing no charge. Let the cylinder be held in an external
uniform electric field E along the axis of the cylinder. Suppose 1 and 2
represents the electric flux passing through the surfaces 1 and 2 each of area A
of the cylinder and 3 represent the flux passing through the curved surface 3
of the cylinder.
Total electric flux through the cylindrical surface is given by
= 1 + 2 + 3 ----- (1)
We have, = EA cos
For surface-1, = 180 o cos180o = -1 1 = − EA ----- (2)
For surface-2, = 0 cos 0 = 1 2 = EA ------- (3)
o o
(a) Since the electric field has only an x component, for faces
perpendicular to x direction, the angle between 𝐸⃗ and ∆𝑆 is ± 𝜋/2.
Therefore, the flux 𝜙 = 𝐸⃗ ∙ ∆𝑆 is separately zero for each face of the
cube except the two shaded ones. Now the magnitude of the electric
field at the left face is
EL = 𝛼x1/2 = 𝛼a1/2
(x = 𝛼 at the left face).
The magnitude of electric field at the right face is
ER = 𝛼 x1/2 = 𝛼 (2a)1/2 (x = 2a at the right face).
The corresponding fluxes are
𝜙𝐿 = 𝐸⃗𝐿 ∙ ∆𝑆 = ∆𝑆 𝐸⃗𝐿 ∙ 𝑛̂𝐿 = ∆𝑆𝐸𝐿 cos 𝜃= – 𝐸𝐿 ∆𝑆, since 𝜃 = 180°
= – 𝐸𝐿 𝑎2
𝜙𝑅 = 𝐸⃗𝑅 ∙ ∆𝑆 = ∆𝑆𝐸𝑅 cos 𝜃= 𝐸𝑅 ∆𝑆, since 𝜃 = 0°
= 𝐸𝐿 𝑎2
Net flux through the cube 𝜙𝐿 + 𝜙𝑅 = – 𝐸𝐿 ∆𝑆 + 𝐸𝑅 ∆𝑆 = a2 (𝐸𝑅 – 𝐸𝐿 ) = 𝛼a2 [(2a)1/2 – a1/2]
= 𝛼a5/2 ( √2 – 1) = 1.05 N m2 C–1
(b) We can use Gauss’s law to find the total charge q inside the cube. We have 𝜙 = q/ 𝜖𝑜 or q = 𝜙 𝜖𝑜 .
Therefore,
q = 1.05× 8.854× 10–12 C = 9.27× 10–12 C.
(a) We can see from the figure that on the left face 𝐸⃗ and ∆𝑆 are parallel. Therefore, the outward flux is
𝜙𝐿 = 𝐸⃗𝐿 ∙ ∆𝑆= -200 𝑖̂ ∙ ∆𝑆 = + 200 ∆S, since 𝑖̂ ∙ ∆𝑆 = – ∆S
= + 200× 𝜋 (0.05)2 = + 1.57 N m2 C–1
4) when r = , then E = 0
Electric field due to an infinitely long straight uniformly charged wire by using Gauss’s law:
Let AB is the infinitely long wire, E is the electric
field, P is a point at a distance r from the wire, r be the radius
of Gaussian cylinder, l be the length of the Gaussian cylinder.
Let q be the charge enclosed by the Gaussian cylinder.
Let be the linear charge density on the wire. Flux
through the end faces is zero because there are no components
of electric field along the normal to the end faces.
= flux through curved surface
= E area of curved surface [ = E area]
= E 2rl ----- (1)
q
From Gauss's theorem, = ------- (2)
o
q l
But = q = l (2) = ------ (3)
l o
l
On comparing (l) and (3), we get, E 2rl = E=
o 2 o r
The direction of E is perpendicular to the wire and directed away from the wire.