Chapter 1 Electric Charges and Fields (Electrostatics Part 1) – Physics free study material by TEACHING CARE online tuition and coaching classes
Electric Charge
Definition : Charge is the property associated with matter due to which it produces and experiences electrical and magnetic effects.
Origin of electric charge : It is known that every atom is electrically neutral, containing as many electrons as the number of protons in the nucleus.
Charged particles can be created by disturbing neutrality of an atom. Loss of electrons gives positive charge (as then n_{p} > n_{e}) and gain of electrons gives negative charge (as then n_{e} > n_{p}) to a particle. When an object is negatively charged it gains electrons and therefore its mass increases negligibly. Similarly, on charging a body with positive electricity its mass decreases. Change in mass of object is equal to n ´ m_{e}. Where, n is the
number of electrons transferred and me is the mass of electron = 9.1 ´ 10 ^{31} Kg .
Type : There exists two types of charges in nature (i) Positive charge (ii) Negative charge
Charges with the same electrical sign repel each other, and charges with opposite electrical sign attract each other.
Unit and dimensional formula : Rate of flow of electric charge is called electric current i.e.,
i = dQ
dt
Þ dQ = idt , hence S.I. unit of charge is – Ampere ´ sec = coulomb (C), smaller S.I. units are mC, mC, nC
(1mC = 10^{3} C, 1m C = 10^{6} C, 1nC = 10^{9} C) . C.G.S. unit of charge is – Stat coulomb or e.s.u. Electromagnetic
unit of charge is – ab coulomb 1C = 3 ´ 10^{9} stat coulomb = 1 ab coulomb . Dimensional formula [Q] = [AT]
10
Note : @Benjamin Franklin was the first to assign positive and negative sign of charge.
@ The existence of two type of charges was discovered by Dufog.
@ Franklin (i.e., e.s.u. of charge) is the smallest unit of charge while faraday is largest (1 Faraday = 96500 C).
@ The e.s.u. of charge is also called stat coulomb or Franklin (Fr) and is related to e.m.u.
of charge through the relation
emu of charge = 3 ´ 1010 esu of charge
Point charge : A finite size body may behave like a point charge if it produces an inverse square electric field. For example an isolated charged sphere behave like a point charge at very large distance as well as very small distance close to it’s surface.
Properties of charge
 Charge is transferable : If a charged body is put in contact with an uncharged body, uncharged body becomes charged due to transfer of electrons from one body to the
 Charge is always associated with mass, e., charge can not exist without mass though mass can exist without charge.
 Charge is conserved : Charge can neither be created nor be destroyed. g. In radioactive decay the uranium nucleus (charge = + 92e ) is converted into a thorium nucleus (charge = +90e ) and emits an a – particle (charge = +2e )
_{92}U ^{238} ®90 Th^{234} +2 He^{4} . Thus the total charge is + 92e
both before and after the decay.
 Invariance of charge : The numerical value of an elementary charge is independent of It is proved by the fact that an atom is neutral. The difference in masses on an electron and a proton suggests that electrons move much faster in an atom than protons. If the charges were dependent on velocity, the neutrality of atoms would be violated.
 Charge produces electric field and magnetic field : A charged particle at rest produces only electric field in the space surrounding it. However, if the charged particle is in unaccelerated motion it produces both electric and magnetic fields. And if the motion of charged particle is accelerated it not only produces electric and magnetic fields but also radiates energy in the space surrounding the charge in the form of electromagnetic
 Charge resides on the surface of conductor : Charge resides on the outer surface of a conductor because like charges repel and try to get as far away as possible from one another and stay at the farthest distance from each other which is outer surface of the This is why a solid and hollow conducting sphere of same outer radius will hold maximum equal charge and a soap bubble expands on charging.
 Charge leaks from sharp points : In case of conducting body no doubt charge resides on its outer surface, if surface is uniform the charge distributes uniformly on the surface and for irregular surface the
distribution of charge, i.e., charge density is not uniform. It is maximum where the radius of curvature is
minimum and vice versa. i.e., σ µ (1/R). This is why charge leaks from sharp points.
 Quantization of charge : When a physical quantity can have only discrete values rather than any value, the quantity is said to be The smallest charge that can exist in nature is the charge of an
electron. If the charge of an electron ( = 1.6 ´ 10 ^{19} C ) is taken as elementary unit i.e. quanta of charge the charge on any body will be some integral multiple of e i.e.,

with n = 1, 2, 3 ….
Charge on a body can never be ± 2 e , ± 17.2e or
3
± 10 ^{5} e etc.
Note : @Recently it has been discovered that elementary particles such as proton or neutron are composed
of quarks having charge (± 1 / 3) e and (± 2 / 3)e. However, as quarks do not exist in free state, the quanta of charge is still e.
@ Quantization of charge implies that there is a maximum permissible magnitude of charge.
Comparison of Charge and Mass.
We are familiar with role of mass in gravitation, and we have just studied some features of electric charge.
We can compare the two as shown below
Methods of Charging.
A body can be charged by following methods :
 By friction : In friction when two bodies are rubbed together, electrons are transferred from one body to the As a result of this one body becomes positively charged while the other negatively charged,
e.g., when a glass rod is rubbed with silk, the rod becomes positively charged while the silk negatively. However, ebonite on rubbing with wool becomes negatively charged making the wool positively charged. Clouds also become charged by friction. In charging by friction in accordance with conservation of charge, both positive and negative charges in equal amounts appear simultaneously due to transfer of electrons from one body to the other.
 By electrostatic induction : If a charged body is brought near an uncharged body, the charged body will attract opposite charge and repel similar charge present in the uncharged body. As a result of this one side of neutral body (closer to charged body) becomes oppositely charged while the other is similarly charged. This process is called electrostatic
Note : @Inducting body neither gains nor loses charge.
@ Induced charge can be lesser or equal to inducing charge (but never greater) and its
maximum value is given by
Q’ = –Qé1 – 1 ù
where Q is the inducing charge and K is the dielectric
ëê K úû
constant of the material of the uncharged body. Dielectric constant of different media are shown below
@Dielectric constant of an insulator can not be ¥
@For metals in electrostatics opposite to inducing charge.
K = ¥
and so
Q’ = – Q;
i.e. in metals induced charge is equal and
 Charging by conduction : Take two conductors, one charged and other Bring the
conductors in contact with each other. The charge (whether
 ve or
 ve ) under its own repulsion will spread
over both the conductors. Thus the conductors will be charged with the same sign. This is called as charging by conduction (through contact).
+
+ 
+
+ + 
+ + 
Note : @A truck carrying explosives has a metal chain touching the ground, to conduct away the charge produced by friction.
Electroscope.
It is a simple apparatus with which the presence of electric charge on a body is detected (see figure). When metal knob is touched with a charged body, some charge is transferred to the gold leaves, which then diverges due to repulsion. The separation gives a rough idea of the amount of charge on the body. If a charged body brought near a charged electroscope the leaves will further diverge. If the charge on body is similar to that on electroscope and will usually converge if opposite. If the induction effect is strong enough leaves after converging may again diverge.
(1)
Uncharged electroscope
 Charged electroscope
Example: 1 A soap bubble is given negative charge. Its radius will [DCE 2000; RPMT 1997; CPMT 1997; MNR 1988]
 Increase (b) Decrease (c) Remain unchanged (d) Fluctuate Solution: (a) Due to repulsive
Example: 2 Which of the following charge is not possible
(a)
1.6 ´ 10^{–}^{18} C
1.6 ´ 10^{–}^{19} C
1.6 ´ 10^{–}^{20} C
 None of these
Solution: (c)
1.6 ´ 10^{–}^{20} C, because this is
1 of electronic charge and hence not an integral multiple.
10
Example: 3 Five balls numbered 1 to 5 balls suspended using separate threads. Pair (1,2), (2,4) and (4,1) show electrostatic attraction, while pair (2,3) and (4,5) show repulsion. Therefore ball 1 must be [NCERT 1980]
 Positively charged (b) Negatively charged (c) Neutral (d) Made of metal
Solution: (c) Since 1 does not enter the list of repulsion, it is just possible that it may not be having any charge. Moreover, since ball no. 1 is being attracted by 2 and 4 both. So 2 and 4 must be similarly charged, but it is also given that 2 and 4 also attract each other. So 2 and 4 are certainly oppositely charged.
Since 1 is attracting 2, either 1 or 2 must be neutral but since 2 is already in the list of balls repelling each other, it necessarily has some charge, similarly 4 must have some charge. It means that though 1 is attracting 2 and 4 it does not have any charge.
Example: 4 If the radius of a solid and hollow copper spheres are same which one can hold greater charge
[BHU 1999; KCET 1994; IITJEE 1974]
(a) Solid sphere (b)Hollow sphere
(c) Both will hold equal charge (d) None of these
Solution: (c) Charge resides on the surface of conductor, since both the sphere having similar surface area so they will hold equal charge.
Example: 5 Number of electrons in one coulomb of charge will be
(a)
5.46 ´ 10^{29}
(b)
6.25 ´ 10^{18}
(c)
1.6 ´ 10^{19}
(d)
9 ´ 10^{11}
Solution: (b) By using Q = ne
Þ n = Q
e
Þ n =
1
1.6 ´ 10^{–}^{19}
= 6.25 ´ 10^{18}
Example: 6 The current produced in wire when 10^{7} electron/sec are flowing in it [CPMT 1994]
(a) 1.6 ´ 10^{–26} amp (b) 1.6 ´ 10^{12} amp (c) 1.6 ´ 10^{26} amp (d) 1.6 ´ 10^{–12} amp
Solution: (d)
i = Q = ne = 10^{7} ´ 1.6 ´ 10^{–}^{19} = 1.6 ´ 10^{–}^{12} amp
t t
Example: 7 A tabletennis ball which has been covered with a conducting paint is suspended by a silk thread so that it hangs between two metal plates. One plate is earthed. When the other plate is connected to a high voltage generator, the ball
 Is attracted to the high voltage plate and stays there
 Hangs without moving
 Swings backward and forward hitting each plate in turn
 None of these
Solution: (c) The table tennis ball when slightly displaced say towards the positive plate gets attracted towards the positive plate due to induced negative charge on its near surface.
–
–
+ –
+ –
–
The ball touches the positive plate and itself gets positively charged by the process of conduction from the plate connected to high voltage generator. On getting positively charged it is repelled by the positive plate and therefore the ball touches the other plate (earthed), which has negative charge due to induction. On touching this plate, the positive charge of the ball gets neutralized and in turn the ball shares negative charge of the earthed plate and is again repelled from this plate also, and this process is repeated again and again.
Here it should be understood that since the positive plate is connected to high voltage generator, its potential and hence its charge will always remain same, as soon as this plate gives some of its charge to ball, excess charge flows from generator to the plate, and an equal negative charge is always induced on the other plate.
Coulomb’s Law.
If two stationary and point charges attraction
Q1 and
Q _{2} are kept at a distance r, then it is found that force of
or repulsion between them is
Q1 Q2
F µ r 2
i.e.,
F = kQ1 Q2
r 2
; (k = Proportionality constant)
 Dependence of k : Constant k depends upon system of units and medium between the two
(i) Effect of units
 In G.S. for air k = 1,
F = Q1 Q2
r 2
Dyne
 In I. for air k = 1 = 9 ´ 10^{9} N –m2 , F = 1 . Q1Q2 Newton (1 Newton = 10^{5} Dyne)


4pe 0 C 2 4pe 0 r 2
Note : @ e
= Absolute permittivity of air or free space =
8.85 ´ 10 ^{12} C 2
æ= Farad ö . It’s
0
Dimension is [ML^{3}T ^{4} A^{2} ]
N – m^{2} ç m ÷
@ e_{0} Relates with absolute magnetic permeability ( m_{0} ) and velocity of light (c) according
to the following relation c = 1
(ii) Effect of medium
 When a dielectric medium is completely filled in between charges rearrangement of the charges inside the dielectric medium takes place and the force between the same two charges decreases by a factor of K
known as dielectric constant or specific inductive capacity (SIC) of the medium, K is also called relative permittivity e_{r} of the medium (relative means with respect to free space).
Hence in the presence of medium
F = Fair
m K
= 1
4pe _{0} K
. Q1 Q2
r 2
Here e _{0} K = e _{0} e _{r} = e
(permittivity of medium)
 If a dielectric medium (dielectric constant K, thickness t) is partially filled between the charges then
effective air separation between the charges becomes (r – t + t K )
Hence force F = 1 Q1Q2
4pe 0 (r – t + t K )^{2}
 Vector form of coulomb’s law : Vector form of Coulomb’s law is
F = K. q1q2 r = K. q1q2 rˆ , where rˆ
is the unit vector from first charge to second charge along the line
12 r 3 12 r 2 12 12
joining the two charges.
(3) A comparative study of fundamental forces of nature
also.

Note : @Coulombs law is not valid for moving charges because moving charges produces magnetic field
@ Coulombs law is valid at a distance greater than 10 ^{15} m.
@ A charge
Q_{1} exert some force on a second charge
Q_{2} . If third charge
Q_{3} is brought near, the
force of Q1 exerted on Q2
remains unchanged.
@ Ratio of gravitational force and electrostatic force between (i) Two electrons is 10^{–}^{43}/1. (ii) Two protons is 10^{–36}/1 (iii) One proton and one electron 10^{–39}/1.
@ Decreasing order to fundamental forces FNuclear
 FElectromagnetic
 FWeak
 FGravitational
 Principle of superposition : According to the principle of super position, total force acting on a given charge due to number of charges is the vector sum of the individual
forces acting on that charge due to all the charges.
Consider number of charge charge Q
Net force on Q will be
Q1 , Q2 , Q3 …are applying force on a
Fnet
= F1 + F2 + ………. + Fn1 + Fn
Example: 8 Two point charges
+3mC and
+8mC
repel each other with a force of 40 N. If a charge
of 5mC is added to each of them, then the force between them will become
(a)
10 N
(b)
+10 N
+20 N
20 N
Solution: (a) Initially
F = k ´ 3 ´ 8 ´ 10^{–}^{12}
and Finally
F ‘ = –k 2 ´ 3 ´ 10^{–}^{12}
so F ‘ = – 1
Þ F ‘ = 10N
r ^{2} r ^{2} F 4
Example: 9 Two small balls having equal positive charge Q (coulomb) on each are suspended by two insulated string of equal length L meter, from a hook fixed to a stand. The whole set up is taken in satellite into space where there is no gravity (state of weight less ness). Then the angle between the string and tension in the string is
(a)
180^{o}, 1 .
4pe_{0}
Q2
(2L)^{2}
[IITJEE 1986]
o 1 Q^{2}
(b)
90 , 4pe . L2

(c)
180o, 1 . Q2
4pe_{0} 2L^{2}
(d)
180^{o},
1 . QL
4pe_{0} 4 L^{2}
Solution: (a) In case to weight less ness following situation arises
+Q 180^{o} +Q
So angle q = 180o
and force F = 1 .
Q2
4pe_{0}
(2L)2 L L
Example: 10 Two point charges 1 mC
& 5mC
are separated by a certain distance. What will be ratio of
forces acting on these two
(a)
1 : 5
(b)
5 : 1
(c)
1 : 1
(d) 0
Solution: (c) Both the charges will experience same force so ratio is 1:1
Example: 11 Two charges of
40mC
and 20mC
are placed at a certain distance apart. They are touched
and kept at the same distance. The ratio of the initial to the final force between them is
(a)
8 : 1
(b)
4 : 1
(c) 1 : 8 (d) 1 : 1
Solution: (a) Since only magnitude of charges are changes that’s why
F µ q q Þ F1 =
q1q2
= 40 ´ 20 = 8
1 2 F2
q’1 q’2
10 ´ 10 1
Example: 12 A total charge Q is broken in two parts Q_{1} and Q_{2} and they are placed at a distance R from
each other. The maximum force of repulsion between them will occur, when
(a)
Q2 = Q , Q1 = Q – Q (b) Q2 = Q , Q1 = Q – 2Q (c) Q2 = Q , Q1 = 3Q
(d)
R R 4 3 4 4
Q1 = Q , Q2 = Q
2 2
Solution: (d) Force between charges Q and Q
F = k Q1Q2 = k Q1 (Q – Q1 )
1 2
For F to be maximum, dF = 0
R 2

i.e., d ìï
R 2
(Q1Q – Q2 )üï = 0 or Q – 2Q
= 0, Q = Q
Hence Q1 = Q2 = Q
2
dQ1
dQ íïk
1 ý

R2 þ
1 1 2
Example: 13 The force between two charges 0.06m apart is 5 N. If each charge is moved towards the other by 0.01m, then the force between them will become
(a) 7.20 N (b) 11.25 N (c) 22.50 N (d) 45.00 N
Solution: (b) Initial separation between the charges = 0.06m Final separation between the charges = 0.04m
1 F æ r ö 2 5 æ 0.04 ö^{2} 4
Since F µ Þ ^{ } ^{1} = ç 2 ÷ Þ = ç ÷ = Þ F_{2} = 11.25N
r ^{2} F_{2}
ç r1 ÷
F2 è 0.06 ø 9
è ø
Example: 14 Two charges equal in magnitude and opposite in polarity are placed at a certain distance apart and force acting between them is F. If 75% charge of one is transferred to another, then the force between the charges becomes
(a)
F (b)
16
9F (c) F (d)
16
15 F
16
Solution: (a)
r
+Q/4
r
æ Q ö^{2}
– Q/4
Q2 k.ç 4 ÷ F
Initially
F = k
r 2
Finally
F ‘ =
è ø =
r ^{2} 16
Example: 15 Three equal charges each +Q, placed at the corners of on equilateral triangle of side a what
æ 1 ö
will be the force on any charge ç k = ÷
è 4pe_{0} ø
(a)
kQ^{2}
a2
(b)
2kQ^{2}
a2
(c)
2kQ^{2}
a2
(d)
3kQ^{2}
a2
Solution: (d) Suppose net force is to be calculated on the charge which is kept at A. Two charges kept at B and C are applying force on that particular charge, with direction as shown in the figure.
Since F = F = F = k Q2
b c a2
So, Fnet =
3kQ^{2}
Fnet =
3F =
a2
+Q +Q
B C
Example: 16 Equal charges Q are placed at the four corners A, B, C, D of a square of length a. The magnitude of the force on the charge at B will be
3Q^{2}
4Q^{2}
æ 1 + 1 2 ö Q ^{2}
 2
 2
ç ÷



2
æ 1 ö
4pe_{0}a
Q 2
4pe_{0}a
ç ÷ 4pe _{0} a
ç 2 +
è
÷
2 ø 4pe
0 a 2
Solution: (c) After following the guidelines mentioned above F_{C}
Fnet
= FAC
 FD
=
kQ^{2}
 FD
kQ^{2}
+Q +Q
A
F_{D}
FAC
B F_{A}
Since FA = FC =
a2
kQ^{2}
and FD =
kQ^{2} æ 1 ö
Q^{2} æ 1 + 2 2 ö D C
Fnet =
2 + 2 = 2 ç
+ ÷ =
ç ÷


2 +Q
a 2a a è
2 ø 4pe_{0}a ç ÷

Example: 17 Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance, will experience maximum coulomb force when
 x = d
2
x = d
2
 x = d
 x = d
Solution: (c) Suppose third charge is similar to Q and it is q F F
So net force on it F_{net} = 2F cosq q q
Where
F = 1 .
4pe _{0} æ


2 d 2 ö
and cosq = x
q
q q

ç ÷

ç ÷ x ^{2} + d ^{2} / 4
x x ^{2} + d ^{2} / 4
\ Fnet
= 2 ´ 1 . Qq ´ x = 2Qqx

4pe _{0}
æ x + 2 ö
æ


ç x +
2 ö1 / 2



÷
4pe
æ

ç x +
2 ö 3 / 2 Q Q

÷ B d d C




ç ÷
è ø è ø
é


ê
0 ç ÷

è ø 2 2
ù
ú
for F
to be maximum
dFnet
= 0 i.e. d ê 2Qqx ú = 0
net
dx dx ê
æ 2 d 2
3 / 2 ú
ö



ê 4pe _{0} ç x + ÷ ú
éæ d 2 ö3 / 2
êë ç


æ d 2 ö5 / 2 ù
÷ úû
d



or êç x ^{2} + ÷
ëè ø
– 3x ^{2} ç x ^{2} + ÷

è ø
ú = 0
ú
û
i.e.
x = ±
Example: 18 ABC is a right angle triangle in which AB = 3 cm, BC = 4 cm and
ÐABC = p . The three
2
charges +15, +12 and – 20 e.s.u. are placed respectively on A, B and C. The force acting on B is
 125 dynes (b) 35 dynes (c) 25 dynes (d) Zero
Solution: (c) Net force on B
Fnet =
A +15 esu

FA = 15 ´ 12 = 20 dyne
3
12 ´ 20
3 cm
B FC C
FC =
(4)2 = 15 dyne
+12 esu
4 cm
– 20 esu
F F Fnet =
F ^{2} + F ^{2}
_{net} = 25 dyne
A A C
Example: 19 Five point charges each of value +Q are placed on five vertices of a regular hexagon of side
 L. What is the magnitude of the force on a point charge of value – q placed at the centre of the hexagon [IITJEE 1992]
(a)
k Q2 L2
k Q2
4 L^{2}
 Zero (d) Information is insufficient
Solution: (a) Four charges cancels the effect of each other, so the net force on the charge placed at centre due to
remaining fifth charge is
Q2
+Q +Q
F = k
L2 +Q
Example: 20 Two small, identical spheres having +Q and – Q charge are kept at a certain distance. F
force acts between the two. If in the middle of two spheres, another similar sphere having
+Q charge is kept, then it experience a force in magnitude and direction as
(a) Zero having no direction (b) 8F towards +Q
charge
charge
 8F towards – Q charge (d) 4F towards +Q
Solution: (c) Initially, force between A and C
F = k Q 2
r 2
When a similar sphere B having charge +Q is kept at the mid point of line joining A and C, then Net force on B is
+Q +Q – Q
Fnet
= FA
 FC
Q2
^{=} ^{k} (r 2)2
kQ^{2}
^{+} (r 2)2
= 8 kQ2 = 8F . (Direction is shown
r 2
r/2
r/2
r
in figure)
Electrical Field.
A positive charge or a negative charge is said to create its field around itself. If a charge
Q_{1} exerts a force
on charge
Q2 placed near it, it may be stated that since Q2
is in the field of Q_{1} , it experiences some force, or it
may also be said that since charge
Q_{1} is inside the field of
Q_{2} , it experience some force. Thus space around a
charge in which another charged particle experiences a force is said to have electrical field in it.
 Electric field intensity (E) : The electric field intensity at any point is defined as the force
r F
experienced by a unit positive charge placed at that point. E = q
0
Where q_{0} ® 0 so that presence of this charge may not affect the source charge Q and its electric field is
not changed, therefore expression for electric field intensity can be better written as
r F
E = Lim
 Unit and Dimensional formula : It’s I. unit –
Newton =
volt =
q0 ®0 q 0
Joule
and
C.G.S. unit – Dyne/stat coulomb. Dimension : [ E ] =[ MLT ^{3} A^{1} ]
coulomb
r
meter
coulomb ´ meter
 Direction of electric field : Electric field (intensity) E is a vector Electric field due to a
positive charge is always away from the charge and that due to a negative charge is always towards the charge
 Relation between electric force and electric field : In an electric field E a charge (Q)
experiences a force F = QE . If charge is positive then force is directed in the direction of field while if charge is
negative force acts on it in the opposite direction of field
 Super position of electric field (electric field at a point due to various charges) : The resultant electric field at any point is equal to the vector sum of electric fields at that point due to various
E = E1 + E2 + E3 + …
The magnitude of the resultant of two electric fields is given by
E = and the direction is given by
tana =
E2 sinq
E1 + E2 cosq
 Electric field due to continuous distribution of charge : A system of closely spaced electric charges forms a continuous charge distribution
To find the field of a continuous charge distribution, we divide the charge into infinitesimal charge
elements. Each infinitesimal charge element is then considered, as a point charge and electric field dE is determined due to this charge at given point. The Net field at the given point is the summation of fields of all
the elements. i.e.,
E = ò dE
Electric Potential.
 Definition : Potential at a point in a field is defined as the amount of work done in bringing a unit positive test charge, from infinity to that point along any arbitrary path (infinity is point of zero potential).
Electric potential is a scalar quantity, it is denoted by V;
V = W
q0
 Unit and dimensional formula : I. unit –
Joule = volt Coulomb
C.G.S. unit – Stat volt (e.s.u.); 1 volt
= 1
300
Stat volt Dimension – [V] = [ML^{2}T ^{3} A^{1} ]
 Types of electric potential : According to the nature of charge potential is of two types
 Positive potential : Due to positive charge. (ii) Negative potential : Due to negative
 Potential of a system of point charges : Consider P is a point at which net electric potential is to be determined due to several charges. So net potential at P
V = k Q1 + k Q2 + k Q3 + k ( Q4 ) + … In general V = åX kQi
r1 r2 r3 r4
i=1 ri
Note : @At the centre of two equal and opposite charge V = 0 but
E ¹ 0
@ At the centre of the line joining two equal and similar charge V ¹ 0, E = 0
 Electric potential due to a continuous charge distribution : The potential due to a continuous charge distribution is the sum of potentials of all the infinitesimal charge elements in which the distribution may
be divided i.e.,
V = ò dV,
= dQ

4πε 0 r
 Graphical representation of potential : When we move from a positive charge towards an equal negative charge along the line joining the two then initially potential decreases in magnitude and at centre become zero, but this potential is throughout positive because when we are nearer to positive charge, overall potential must be positive. When we move from centre towards the negative charge then though potential remain always negative but increases in magnitude fig. (A). As one move from one charge to other when both charges are like, the potential first decreases, at centre become minimum and then increases (B).
 Potential difference : In an electric field potential difference between two points A and B is defined as equal to the amount of work done (by external agent) in moving a unit positive charge from point A to point

i.e., VB – VA = q in general W = Q. DV ; DV = Potential difference through which charge Q moves.
0
Electric Field and Potential Due to Various Charge Distribution.
 Point charge : Electric field and potential at point P due to a point charge Q is
Q r Q æ 1 ö Q
E = k r 2
or E = k r 2 rˆ
ç k = ÷ ,
è 4pe _{0} ø
V = k r
Note : @ Electric field intensity and electric potential due to a point charge q, at a distance t_{1} + t_{2} where t_{1} is thickness of medium of dielectric constant K_{1} and t_{2} is thickness of medium of dielectric constant K_{2} are :
(2) Line charge
E = 1 Q ;
4πε 0 (t1
V = 1 Q
4πε 0 (t1
 Straight conductor : Electric field and potential due to a charged straight conducting wire of length l
and charge density l
(a) Electric field :
E = kl (sina + sin b ) and E
x r y
= kl (cos b – cosa)
r
If a = b;
Ex =
2kl
r
sina
and E_{y} = 0
If l ® ¥ i.e. a = b = p ; E
2 x
= 2kl
r
and E_{y}
= 0 so Enet
= l
2pe_{0}r
If a = 0, b = p ;  E = E = kl
so E = =
2 x y
l
r net r
é r ^{2} + l ^{2} – 1 ù – l
(b) Potential : V =
log e ê ú
for infinitely long conductor V = log e r + c
2pe _{0}
ëê r ^{2} + l ^{2} + 1 úû
2pe _{0}