GyaanSankalp
ELECTRIC CHARGES AND FIELDS
C H A P T E R
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LEARNING OBJECTIVES
(i) Electric and magnetic forces determine the properties of atoms, molecules and bulk matter. 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.
(ii) 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. Conservation of total charge of an isolated system is a property independent of the scalar nature of charge. 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). Quantisation of electric charge is a basic (unexplained) law of nature; interestingly, there is no analogous law on quantisation of mass.
(iii) 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 = force on q2 due to q1 = 1 22 21 21 k (q q )
ˆr
r where ˆr is a unit vector in the direction from q21 1 to q2 and 0 1 k
4
is the constant of proportionality. In SI units, the unit of charge is coulomb. The experimental value of the constant is0
12 2 1 2 0 8.854 10 C N m
. The approximate value of k is k = 9 × 109 Nm2C–2.
(iv) 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.
(v) The electric field E 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 |/40r2 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. 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.
(vi) 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. 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.
(vii) An electric dipole is a pair of equal and opposite charges q and –q separated by some distance 2a. Its dipole moment vector p
has magnitude 2qa and is in the direction of the dipole axis from –q to q.
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: 2 2 3/2
0 p 1 E 4 (a r ) 3 0 p 4 r , for r >> a
Dipole electric field on the axis at a distance r from the centre:
2 2 2 3 0 0 2pr 2p E 4 (r a ) 4 r for r >> a
The 1/r3 dependence of dipole electric fields should be noted in contrast to the 1/r2 dependence of electric field due to a point charge. In a uniform electric field E , a dipole experiences a torque given by p E but experiences no net force.
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(viii) The flux of electric field E through a small area element S is given by E. S The vector area element S is S S nˆ
where S is the magnitude of the area element and ˆn is normal to the area element, which can be considered planar for sufficiently small S. For an area element of a closed surface, ˆn is taken to be the direction of outward normal, by convention.
(ix) Gauss’s law: The flux of electric field through any closed surface S is 1/0 times the total charge enclosed by S. The law is especially useful in determining electric field E, when the source distribution has simple symmetry:
(a) Thin infinitely long straight wire of uniform linear charge density
0 ˆ E n 2 r
where r is the perpendicular distance of the point from the wire and ˆn is the radial unit vector in the plane normal to the wire passing through the point.
(b) Infinite thin plane sheet of uniform surface charge density
0 ˆ E n 2
(c) Thin spherical shell of uniform surface charge density 2 0 ˆ E r 4 r (rR) E 0 (r < R) INTRODUCTION
Many students feel that in electrostatic charge remains at rest & hence formulae derive in electrostatic are applicable when charge is at rest but fact is formulae are applicable when charge is in motion only the difference is when charge is in motion we will consider additional effect called magnetic effect. Electrostatics deals with the study of forces, fields and potentials arising from static charges. Like mass, electric charge is an intrinsic property of protons and electrons.
In nature, atoms are normally found with equal numbers of protons and electrons, i.e. atom is electrically neutral.
The charge on an electron or a proton is the smallest amount of free charge that has been discovered. Charges of larger magnitude are built up on an object by adding or removing electrons. If in a body there is excess of electrons over its neutral configuration, conventionally the body is said to be negatively charged and if there is deficiency of electron it is said to be positively charged.
– ve charged body Body has gained electrons + ve charged boy Body has lost some electrons + ve & – ve charge named by benjamin Franklin. INTERESTING EXPERIMENT
Take any two materials from the following list and then rubbed with each other. We can always find that the former one is positively charged and the later one is negatively charged.
Fur glass paper metal silk plastic amber rubber sulfur
When a charged body is close enough to a neutral body, they attract each other. One of the applications of this effect is to use tiny paint droplets to paint the automobiles uniformly.
CONDUCTOR AND INSULATORS
Suppose you charge a rubber rod and then touch it to a neutral object. Some charge, repelled by the negative charge on the rod, will be transferred to the originally-neutral object. What happens to that charge then depends on the material of which the originally-neutral object consists. In the case of some materials, the charge will stay on the spot where the originally neutral object is touched by the charged rod. Such materials are referred to as insulators, materials through which charge cannot move, or, through which the movement of charge is very limited. Examples of good insulators are quartz, glass, and air. In the case of other materials, the charge, almost instantly spreads out all over the material, in response to the force of repulsion (recalling that force causes acceleration which leads to the movement) that each elementary particle of the charge exerts on every other elementary particle of charge. Materials in which the charge is free to move about are referred to as conductors. Examples of good conductors are metals and saltwater.
When you put some charge on a conductor, it immediately spreads out all over the conductor. The larger the conductor, the more it spreads out. In the case of a very large object, the charge can spread out so much that any chunk of the object has a negligible amount of charge and hence, behaves as if were neutral. Near the surface of the earth, the earth itself is large enough to play such a role. If we bury a good conductor such as a long copper rod or pipe, in the earth, and connect to it another good conductor such
GyaanSankalp
as a copper wire, which we might connect to another metal object, such as a cover plate for an electrical socket, above but near the surface of the earth, we can take advantage of the earth’s nature as a huge object made largely of conducting material. If we touch a charged rubber rod to the metal cover plate just mentioned, and then withdraw the rod, the charge that is transferred to the metal plate spreads out over the earth to the extent that the cover plate is neutral. We use the expression “the charge that was transferred to the cover plate has flowed into the earth.” A conductor that is connected to the earth in the manner that the cover plate just discussed is connected is called “ground.” The act of touching a charged object to ground is referred to as grounding the object. If the object itself is a conductor, grounding it (in the absence of other charged objects) causes it to become neutral. CHARGING BY INDUCTION
When a charged particle is taken near to neutral metallic object then the electrons move to one side and there is excess of electrons on that side making it negatively charged and deficiency on the other side making that side positively charged. Hence charges appear on two sides of the body (although total charge of the body is still zero). This phenomenon is called induction and the charge produced by it is called induced charge.
Step 1 : + + + + + + + + + + + + Charged body is brought near uncharged body Charged body V'= +ve q'=0 Step 2 : + + + + + + + Uncharged body is connected to the earth q' = –ve V' = 0 e Step 3 : Uncharged body is disconnected from the earth q' = –ve V' = 0 + + + + + + + Step 4 : Charging body is removed q' = –ve V' = –ve
A body can be charged by means of (a) friction, (b) conduction, (c) induction, (d) thermoionic ionisation, (e) photoelectric effect and (f) field emission.
BASIC PROPERTIES OF ELECTRIC CHARGE
(1) Charge is a scalar and can be of two types (i.e. + ve or – ve). It adds algebraically.
(2) Charge is conserved. During any process (chemical, nuclear, decay etc.) the net electric charge of an isolated system remains constant.
In the process one body gains some amount of – ve charge while the other gains an equal amount of + ve charge. Pair - production, Annihilation are processes understand on basis of charge conservation.
(3) Charge is Quantized (exists as discrete "Packets") : Robert Millikan discovered that electric charge always occurs as some integral multiple of fundamental unit of charge (e).
q = Ne [N is some integer] Charge on a body can never be 1
3 e, 2 3
e etc. as it is due to transfer of electron.
(4) Through large number of experiments it is also well established that similar charges repel each other while dissimilar attract. Here it is worth noting that true test of electrification is repulsion and not attraction as attraction may also take place between a charged bodies.
(5) Charge is always associated with mass i.e. charge can not exist without mass though mass can exist without charge. (6) Charge is transferable. Process of charge transfer is called conduction.
(7) Charge is invariant i.e. it is independent on frame of reference. (8) Accelerated charge radiates energy.
Charge at rest produces Electric & magnetic effect.
Accelerate charge particle Electric & magnetic effect + radiate energy (According to electromagnetic theory)
+
produces E v = 0+
v = const. with no radiation B and produces E+
v const. radiates energy B and produces E and(9) Charge resides on the outer surface of a conductor (10) How to express charge.
The SI unit for measuring the magnitude of an electric charge is the coulomb (C). Current drift of charge per unit time
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1 coulomb 1 ampere × 1 sec
If a charge of 1 coulomb drift per second through cross-section of conductor current flowing is called 1 ampere. Charge on electron = – 1.6 × 10–19 C, Charge on proton = + 1.6 × 10–19 C
The coulomb is related to CGS units of charge through the basic relation. 1 Coulomb = 3 × 109 esu of charge (static coulomb or frankline) = 1
10 emu of charge Practical units of charge are amp × hr (= 3600 coulomb), Faraday ( = 96500 coulomb) Example 1 :
How many electrons are there in one coulomb of negative charge. Sol. Number N of electrons is
N = q e = 19 1.00C 1.6 10 C = 6.25 × 10 18 Charge on 6.25 × 1018 electrons = – 1 C Example 2 :
A copper penny (Z = 29) has a mass of 3g. What is the total charge of all the electrons in the penny ?
Sol. The electrons have a total charge given by the number of electrons in the penny, Ne, times the charge of an electron, –e. The number of electrons is 29 times the number of copper atoms N. To find N, we use the fact that one mole of any substance has Avogadro's number (NA = 6.02 × 1023) of molecules, and the number of grams in a mole is the molecular mass M, which is 63.5 for copper. Since each molecule of copper is just one copper atom, we find the number of atoms per gram by dividing NA atoms/mole by M grams/mole.
1. The total charge is the number of electrons times the electronic charge : Q = Ne (– e) 2. The number of electrons is Z time the number of copper atoms Na : Ne = ZNa 3. Compute the number of copper atoms in 3g of copper : Na = (3g)
23
6.02 10 atoms / mol 63.5 g / mol
= 2.84 × 1022 atoms 4. Compute the number of electrons Ne ,Ne = ZNa = (29 electrons/atom) (2.84 × 1022 atoms) = 8.24 × 1023 electrons.
5. Use this value of Ne to find the total charge : Q = Ne (– e) = (8.24 × 1023 electrons) (– 1.6 × 10–19 C/electron) = – 1.32 × 105 C Example 3 :
A glass rod is rubbed with a silk cloth. The glass rod acquires a charge of + 19.2 × 10–19 C.
(i) Find the number of electrons lost by glass rod. (ii) Find the negative charge acquired by silk. (iii) Is there transfer of mass from glass to silk ?
Sol. (i) Number of electrons lost by glass rod is n = q e = 19 19 19.2 19 1.6 10 = 12
(ii) Charge on silk = – 19.2 × 10–19 C
(iii) Since an electron has a finite mass (me = 9 × 10–31 kg), there will be transfer of mass from glass rod to silk cloth.
Mass transferred = 12 × (9 × 10–31) = 1.08 × 10–29 kg.
The mass transferred is negligibly small. This is expected because the mass of an electron is extremely small. DETECTING CHARGE
Charge can be detected and measured with the help of gold-leaf electroscope, voltameter, ballistic galvanometer. Gold leaf electroscope consist of two gold leaves attached to a conducting post that has a conducting disc ball on top. The leaves are otherwise insulated from the container. Gold leaf electroscope can be used in 2 ways.
Uncharged electroscope when uncharged, the leaves hang together vertically. (a) If a charged body is brought near to it, charge on the ball of electroscope will be
opposite to that of body & on leaves similar to that of body and leaves will diverge.
+ + + + ++ + + + +
(b) If a charged body is touched : Ball & leaves both acquire similar charge and leaves will diverge.
From above method you will not be able to tell nature of charge (it may be +ve or – ve) in both case leaves will diverge.
+ + + + + + ++ + + + + + + +
GyaanSankalp Charged electroscope :
If a charged body is brought near a charged electroscope, the leaves will further diverge if the charge on the body is similar to that on the electroscope and will usually converge if opposite. Thus we will be able to determine nature of charge on a body.
+ + + + + ++ + + + + + + + + + + ++ + + + + + Example 4 :
What will happen if x-rays are incident on a charged electroscope.
Sol. Due to ionisation of air by x-rays the electroscope will get discharged and hence its leaves will collapse. Example 5 :
What will happen if x-rays are incident on a charged evaluated electroscope.
Sol. X-rays will cause photoelectric effect with gold and so the leaves will further diverge if it is positively charged (or uncharged) and will converge if it is negatively charged.
TRY IT YOURSELF
Q.1 Does the attraction between the comb and the piece of papers last for longer period of time ? Q.2 Is repulsion a true test of electrification ?
Q.3 What is the total charge, in coulombs, of all the electrons in three mole of hydrogen atom ? Q.4 The existence of a negative charge on a body implies that it has –
(A) Lost some of its electrons (B) Lost some of its protons
(C) Acquired some electrons from outside (D) Acquired some protons from outside Q.5 Lighting rods are made of –
(A) Porcelain (B) Bakelite (C) Plastic (D) Metal
Q.6 A positively charged body is brought near an uncharged gold leaf electroscope, then –
(A) No charge is induced in the leaves (B) Positive charge is induced in both the leaves
(C) Negative charge is induced in both the leaves (D) Positive charge is induced in one leaf and negative in the other. Q.7 Static electricity is produced by –
(A) Friction only (B) Induction only (C) Friction & induction both (D) Chemical reaction only
Q.8 Five balls respectively from 1 to 5 are suspended from different threads. If pair of balls (1,2), (2,4) and (4,1) represents attraction while pair (2,3) and (4,5) represents repulsion then on ball 1.
(A) Positive charge (B) Negative (C) Neutral (D) Made of metal Q.9 On charging two metallic spheres of same
mass-(A) Mass of positively charged sphere will be more (B) Mass of positively charged sphere will be less (C) Mass of negatively charged sphere will be more (D) Mass of negatively charged will be less
(A) 1, 2 (B) 2, 3 (C) 3, 4 (D) 1, 4
Q.10 The current produced in wire when 107 electron/sec. are flowing in
it-(A) 1.6 x 10–26 A (B) 1.6 x 1012 A (C) 1.6 x 1026 A (D) 1.6 x 10–12 A ANSWERS
(4) (C) (5) (D) (6) (B) (7) (C)
(8) (C) (9) (B) (10) (D)
COULOMB'S LAW
Force between two point charges (interaction force) is directly proportional to the product of magnitude of charges (q1 and q2) and is inversely proportional to the square of the distance between them i.e., (1/r2). This force is conservative in nature. This is also called inverse square law. The direction of force is always along the line joining the point charges.
F 1 22 q q r ; F = K 1 2 2 q q r where K is a constant K = 0 r 1 4 [K = 9 × 10 9 C2/N-m2 ] ;
0 = permittivity of free space = 8.85 × 10–12 N-m2/C2, r = relative permittivity (dielectric constant of medium) Coulomb’s Law in Vector Form
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1 2 12 3 1 2 0 1 2 q q 1 F r r 4 | r r | Similarly, electric force on q2 due to charge q1 is
1 2 21 3 2 1 0 2 1 q q 1 F r r 4 | r r | Here q1 and q2 are to be substituted with sign. Position vector of charges q1 and q2 are r1x i1ˆy j z k1ˆ 1ˆ and r2 x i2ˆy j z k2ˆ 2ˆ respectively. Where (x1, y1, z1) and (x2, y2, z2) are the co-ordinates of charges q1 and q2.
Superposition Theorem
The interaction between any two charges is independent of the presence of all other charges. Electrical force is a vector quantity therefore, the net force on any one charge is the vector sum of the all the forces exerted on it due to each of the other charges interacting with it independently i.e.
Net force on charge q,
F
= F1 + F2 + F3 + ... Example 6 :In a hydrogen atom, the electron is separated from the proton by an average distance of about 5.3 × 10–11 m. Calculate the magnitude of the electrostatic force of attraction exerted by the proton on the electron.
Sol. Substitute the given value into Coulomb's law : F = k | q q |1 22 r = 2 2 ke r = 9 2 2 19 2 11 2 (8.99 10 N.m / C ) (1.6 10 C) (5.3 10 m) = 8.19 × 10 –9 N Example 7 :
Compute the ratio of the electric force to the gravitational force exerted by a proton on an electron in a hydrogen atom. Sol. We use Coulomb's law with q1 = e and q2 = – e to find the electric force, and Newton's law of gravity with the mass of the proton,
mp = 1.67 × 10–27 kg, and the mass of the electron, me = 9.11 × 10–31 kg.
1. Express the magnitude of the electric force Fe and the gravitational force Fg in terms of the charges, masses, separation distance r, and electrical and gravitational constant s:
Fe = 2 2 ke r , Fg = p e 2 Gm m r
2. Take the ratio. Note that the separation distance r cancels :
e g F F = 2 p e ke Gm m
3. Substitute numerical values :
e g F F = 9 2 2 19 2 11 2 2 27 31 (8.99 10 N.m / C ) (1.6 10 C) (6.67 10 N.m / kg ) (1.67 10 kg) (9.11 10 kg) = 2.27 × 10 39 Example 8 :
What is the smallest electric force between two charges placed at a distance of 1.0 m. Sol. Fe = 0 1 4 . 1 2 2 q q r ... (i) For Fe to be minimum q1 q2 should be minimum.
We know that (q1)min = (q2)min = e = 1.6 × 10–19 C Substituting in Eq. (i), we have
(Fe)min = 9 19 19 2 (9.0 10 ) (1.6 10 ) (1.6 10 ) (1.0) = 2.304 × 10–28 N.
GyaanSankalp Example 9 :
Electric force between two point charges q and Q at rest is F. Now if a charge – q is placed next to q what will be the (a) force on Q due to q (b) total force on Q ?
Sol. (a) As electric force between two body interaction, i.e., force between two particles, is independent of presence or absence of other particles, the force between Q and q will remain unchanged, i.e., F.
(b) An electric force is proportional to the magnitude of charges, total force on Q will be given by :
F Qq q 0
0
F Qq q q
[as q' = q + (– q) = 0] i.e., The resultant force on Q will be zero.
Example 10 :
Four identical point charges each of magnitude q are placed at the corners of a square of side a. Find the net electrostatic force on any of the charge.
Sol. Let the concerned charge be at C then charge at C will experience the force due to charges at A, B and D. Let these forces respectively be FA, FB and FD thus forces are given as
2 2 0 A AC q 4 1 F along AC = 2 jˆ 2 iˆ a 2 4 q 2 0 2 2 2 0 B BC q 4 1 F along BC = ( jˆ) a πε 4 q 2 0 2 FD FA FB A B C D q q q q y x 2 2 0 D DC q 4 1 F along DC = (iˆ) a πε 4 q 2 0 2 D B A net F F F F 1 2 2 1 jˆ 1 2 2 1 iˆ a πε 4 q 2 0 2 2 0 2 net a 4 q 1 2 2 1 2 F 2 0 2 a 4 q 2 2 1 Example 11 :
Five point charges, each of value + q are placed on five vertices of a regular hexagon of side L m. What is the magnitude of the force on a point charge of value – q coulomb placed at the centre of the hexagon ?
Sol. If there had been a sixth charge + q at the remaining vertex of hexagon force due to all the six charges on – q at O will be zero (as the forces due to individual charges will balance each other), i.e., FR 0
Now if f is the force due to sixth charge and F due to remaining five charges,
F f 0 i.e., F f or F = f = 0 1 4 2 q q L = 0 1 4 2 q L Example 12 :
Two charged spheres of radius 'R' are kept at a distance 'd' (d > 2R). One has a charge +q and the other –q. The force between them will
be-+ be-+ + + + + + + + R R d (A) 2 2 0 1 q 4 d (B) 2 2 0 1 q 4 d (C) 2 2 0 1 q 4 d (D) None of these
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TRY IT YOURSELF
Q.1 Two identical balls each having a density are suspended from a common point by two insulating strings of equal length. Both the balls have equal mass and charge. In equilibrium each string makes an angle with vertical. Now, both the balls are immersed in a liquid. As a result the angle does not change. The density of the liquid is . Find the dielectric constant of the liquid. Q.2 The ratio of gravity force to the electric force between two
electrons-(A) 10–36 (B) 10–42 (C) 1042 (D) 10–47
Q.3 A charge Q1 exerts some force on a second charge Q2. If a 3rd charge Q3 is brought near, the force of Q1 exerted on Q2 – (A) Will increase (B) Will decrease (C) Will remain unchanged
(D) Will increase if Q3 is of the same sign as Q1 and will decrease if Q3 is of opposite sign.
Q.4 Two identical pendulums A and B, are suspended from the same point. The bobs are given positive charges, A having more charge than B. They diverge and reach equilibrium with A and B making angles and 1 with the vertical respectively. Then–2 (A) 1 > 2 (B) 1 < 2 (C) 1= 2
(D) The tension in A is greater than tension in B
Q.5 Force of attraction between two point charges placed at a distance d is F. What distance apart should they be kept in the same medium so that the force between them is F/3 ?
Q.6 A particle of mass m carrying charge +q1 is revolving around a fixed charge –q2 in a circular path of radius r. Calculate the period of revolution.
Q.7 Two pieces of copper, each weighing 0.01 kg, are placed at a distance of 0.1m from each other. One electron from per 1000 atoms of one piece is transferred to other piece of copper. What will be the Coulomb force between two pieces after the transfer of electrons ? The atomic weight of copper is 63.5 g/mole. Avogadro number = 6 × 1023.
Q.8 Two point charges of +2µC and +6µC repel each other with a force of 12 N. If each is given an additional charge –4µC, then force will
become-(A) 4 N (attractive) (B) 60 N (attractive) (C) 4 N (Repulsive) (D) 12 N (attractive) Q.9 Three equal charges (q) are placed at corners of a equilateral triangle. The force on any charge is
-(A) zero (B) 3 2 2 Kq a (C) 2 2 Kq 3a (D) 3 3 2 2 Kq a
Q.10 Five point charges, each of value – q coulomb, are placed on five vertices of a regular hexagon of side L meter. The magnitude of the force on a point charge of value –q coul. placed at the center of the hexagon is
-(A) 2 2 kq L (B) 5 2 2 kq L (C) 3 2 2 kq L (D) zero
ANSWERS
(1) K = / – d (2) (B) (3) (C) (4) (C) (5) 1.732 d (6) 0 1 2 mr T 4 r q q (7) F = 2.06 × 1014 N (8) (C) (9) (B) (10) (A) ELECTRIC FIELDThe physical field where a charged particle, irrespective of the fact whether it is in motion or at rest, experiences force is called an electric field. The concept of electric field was given by michael Faraday. Characteristics of electric field :
(1) Electric field intensity (shortly we will call electric field). (2) Electric potential. (3) Electric lines of forces. Electric field intensity E :
Electric field intensity at a point is equal to the electrostatic force experienced by a unit positive point charge both in magnitude and direction
If a test charge q0 is placed at a point in an electric field and experiences a force F due to some charges (called source charges), the electric field intensity at that point due to source charges is given by
0 F E q .
The presence of the charge q0 will generally change the original distribution of the other charges, particularly if the charges are on conductors. However, we may choose q0 to be small enough so that its effect on the original charge distribution is negligible.
q 0 0 0 E E lim q
GyaanSankalp Electric field due to a point charge
The electric field produced by a point charge q can be obtained in general terms from Coulomb's law. First, note that the magnitude of the force exerted by the charge q on a test charge q0 is F = kqq0/r2. Then, divide this value by q0 to obtained the magnitude of the field. Since q0 is eliminated algebraically from the result, the electric field does not depend on the test charge:
Point charge q : E = 2
kq r
If (x, y, z) are the co-ordinates of the observation point P, then
ˆ ˆ ˆ rxiyj zk Also, r = (x2 + y2 + z2)1/2 and r3 = (x2 + y2 + z2)3/2 Now, E(r) = 0 1 4 2 2 2 3/ 2 q (x y z ) ˆ ˆ ˆ (xiyj zk) Source Charge qO r P q 0
The three rectangular components of E (r) are as follows : Ex(r ) = 0 1 4 2 2 2 3/ 2 q x (x y z ) , Ey(r) = 0 1 4 2 2 2 3/ 2 q y (x y z ) and Ez( r ) = 0 1 4 2 2 2 3/ 2 q z (x y z ) Electric field due to Discrete distribution of charge :
Point charges placed at different position, use vector approach (Better term : super position rule) n 1 2 i i 1 E E E ... E
with i i 3 i 0 i q 1 E r 4 r Corona Discharge : Dielectric strength of medium mean minimum field required for ionisation of a medium. If value of E increases above dielectric strength of medium, medium gets ionised and charge leak out into the medium from body generally it happen at the corner where E is high. This leakage process is called corona discharge.
For air dielectric strength = 3 × 106 v/m
The electric field near a high-voltage power line can be large enough to strip the electrons from air molecules, thus ionizing them and making the air a conductor. The glow resulting from the recombination of free electrons with the ions is an example of corona discharge. Break-down in air is witnessed during atmospheric lighting.
Motion of a charged particle in a uniform electric field
If force of gravity does not exist or is balanced by some other force say reaction or neglected then F qE a constant m m [as F qE ]
Here equations of motion are valid.
(i) If the particle is initially at rest then from v = u + at, we get v = at = qEt m And from Eqn. s ut 1at2
2 we get s 1at2 1 qEt2 2 2 m + + + + + + + d PD=V E F +q + + + + + + + Y -q L D v0
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The motion is accelerated translatory with a t°; v t and s t² Here 2 2 1 1 qE W KE mv m t 2 2 m also W = qEd = qV
(ii) If the particle is projected perpendicular to the field with an initial velocity v0 From Eqn. v = u + at and s = ut + 1
2at² , ux = v0 and ax = 0, ux = v0 = constant and x = v0t
for motion along y-axis as uy = 0 and ay = qE m , y qE v t m and 2 1 qE y t 2 m
So eliminating t between equation for x and y, we have
2 2 2 0 0 qE x qE y x 2m v 2mv If particle is projected perpendicular to field the path is a parabola. Example 13 :
When a 5nC test charge is placed at a certain point, it experiences a force of 2 × 10–4 N in the x-direction. What is the electric field E at the point ?
Sol. E = F / q 0 = [(2 × 10–4 N)i] / (5 × 10–9 C) = (4 × 104 N/C)i)
Example 14 :
Four particles, each having a charge q, are placed on the four vertices of a regular pentagon. The distance of each corner from the centre is a. Find the electric field at the centre of the pentangon.
Sol. Let the charges be placed at the vertices A, B, C and D of the pentagon ABCDE. If we put a charge q at the corner E also, the field at O will be zero by symmetry. Thus, the field at the centre due to the charges at A, B, C and D is equal and opposite to the field due to the charge q at E alone.
The field at O due to the charge q at E is 2
0 q 4 a along EO. E A B C D O
Thus, the field at O due to the given system of charges is 2
0 q
4 a along OE. Example 15 :
Two positive charges Q1 and Q2 are placed on a line as shown in figure. Determine the position of point O, where the net electric field is zero.
Sol. Let position of P is at a distance x from Q1. Then the fields at P due to Q1 and Q2 are in opposite directions. They will add up to give zero, only if their (electric field's) magnitude are equal. That is
1 2 kQ x = 2 2 kQ (Rx) R x x = 2 1 Q Q or x = 2 1 R 1 Q / Q Q1 P 2 x R Q
The distance of point P from charge Q is d = R – x =
1 2
R 1 Q / Q
If two negative charges are placed on a line (instead of positive charges), then the position of point P where the net electric field is zero, is again
GyaanSankalp Example 16 :
Calculate the electric field strength required to just support a water drop of mass 10–7 kg and having charge 1.6 × 10–19 C. Sol. Here, m = 10–7 kg, q = 1.6 × 10–19 C
Step 1 : Let E be the electric field strength required to support the water drop. Force acting on the water drop due to electric field E is
F = qE = 1.6 × 10–19 E
Weight of drop acting downward, W = mg = 10–7 × 9.8 newton. Step 2 : Drop will be supported if F and W are equal and opposite. i.e., 1.6 × 10–19 E = 9.8 × 10–7 or E = 7 19 9.8 10 1.6 10 = 6.125 × 10 12 N C–1. Example 17 :
Two charges of + 10C and + 40C respectively are placed 12cm apart. Find the position of the point where electric field is zero. Sol. Let P be the point at a distance x from the charge + 10C where electric field due to two charges + 10C and + 40C is zero.
Electric field intensity due to q1 at P, E1 =
0 1 4 1 2 q x ; along PB
Electric field intensity due to q2 at P, E2 =
0 1 4 2 2 q (rx) ; along PA.A. Clearly, field at P will be zero if E1 = E2
i.e. 0 1 4 1 2 q x = 0 1 4 2 2 q (rx) or q x 1 2 = 2 2 q (rx) Here, q1 = 10C = 10 × 10–6 C ; q2 = 40C = 40 × 10–6 C 6 2 10 10 x = 6 2 40 10 (r x) (r – x)2 = 4x2 or (r – x) = 2x 3x = r or x = r 3 But r = 12cm (given) x = 12 3 = 4.0 cm.
Thus electric field will be zero at a distance of 4.0 cm from the charge + 10µC. Example 18 :
Can a metal sphere of radius 1cm hold a charge of 1 coulomb. Sol. Electric field at the surface of the sphere.
E = 2 KQ R = 9 2 2 9 10 1 (1 10 ) = 9 × 10 13 V m
This field is much greater than the dielectric strength of air (3 × 106 v/m), the air near the sphere will get ionised and charge will leak out. Thus a sphere of radius 1 cm cannot hold a charge of 1 coulomb in air.
TRY IT YOURSELF
Q.1 Two charges of opposite nature having magnitude 10 µC are 20 cm apart. The electric field at the centre of line joining these charges will
be-(A) 9 x 106 N/C in the direction of positive charge (B) 18 x 106 N/C in the direction of negative charge (C) 18 x 106 N/C in the direction of positive charge (D) 9 x 106 N/C in the direction of negative charge
Q.2 A point charge A of charge +4 µC and another point charge B of charge –1 µC are placed in air at a distance 1 meter apart. Then the distance of the point on the line joining the charges and from the charge B, where the resultant electric field is zero, is- (in metre)
(A) 2 (B) 1 (C) 0·5 (D) 1·5
Q.3 Four charges each +q, are placed at the four corners of a regular pentagon as shown in the fig. The distance of each corner from the centre O is r. Then the electric field at the center will
be-(A) 0 q 4 r² towards OA (B) 0 q 2 2 r² towards OA A O q q q q (C) Zero (D) 0 q r² towards OA
12 GyaanSankalp
Q.4 A charge Q is divided into two parts such that charge on each part is q and Q/q = 2. The Coulombic force between the two charges q and q when placed some distance apart –
(A) is maximum irrespective of the medium in which they are placed (B) is minimum
(C) is equal in magnitude for both opposite in direction (D) is depenent on the medium in which charges are placed
Q.5 Two free point charges +4Q and +Q are placed at a distance r. A third charge q is so placed such that all the three are in equilibrium– (A) q is placed at a distance r/3 from 4Q (B) q is placed at a distance r/3 from Q
(C) q = 4Q/9 (D) q = – 4Q/9
Q.6 A charge q = 1 µC is placed at point (1m, 2m, 4m). Find the electric field at point P (0m, – 4m, 3m).
Q.7 A copper ball of diameter d is immersed in an oil of density 0. There is a homogeneous electric field E directed vertically upwards such that the copper ball is suspended in the oil. Density of copper is C. The charge on the ball is –
(A) 3 c 0 d ( )g 3E (B) 3 c 0 d ( )g 2E (C) 3 c 0 d ( )g 6E (D) 3 c 0 d ( )g 12E
Q.8 A charge 10–9 coulomb is located at origin is free space and another charge Q at (2, 0, 0). If the x-component of the electric field at (3, 1, 1) is zero, calculate the value of Q. Is the y-component zero at (3, 1, 1) ?
Q.9 A charged particle of mass m = 2 kg and charge 1 µC is thrown from a horizontal ground at an angle = 45° with speed 10 m/s. In space a horizontal electric field E = 2 × 107 N/C exists. The range of the projectile is –
(A)20m (B) 60m (C) 200m (D) 180m
Q.10 In the fig. distance of the point from A where the electric field is zero is-(A) 20 cm (B) 10 cm (C) 33 cm (D) None of these 80cm 10 C 20 C A B
ANSWERS
(1) (B) (2) (B) (3) (A) (4) (A) (5) (BD) (6) E ( 38.42iˆ 250.52 j 38.42k)ˆ ˆ N C (7) (C) (8) 3/ 2 9 3 3 10 C 11 (9) (A) (10) (C)ELECTRIC FIELD DUE TO CONTINUOUS CHARGE DISTRIBUTION
The evaluate the electric field of a continuous charge distribution, the following procedure is used. First, we divide the charge distribution into small elements each of which contains a small charge q. Next, we use Coulomb's law to calculate the electric field due to one of these elements at a desired point. Finally, we evaluate the total field at a point due to the charge distribution by summing the contributions of all the charge elements (that is, by applying the superposition principle.)
Let us consider some cases
Case 1 : The electric field of a uniform ring of charge.
A ring of radius a has a uniform positive charge per unit length, with a total charge Q. To find the electric field along the axis of the ring at a point P lying a distance x from the center of the ring follow the procedure.
The magnitude of the electric field at P due to the segment of charge q is E = k 2
q r This field has an x component Ex = E cos along the axis of
the ring and a component E perpendicular to the axis. But as we see in figure, the resultant field at P must lie along the x axis since the perpendicular components sum up to zero. That is, the perpendicular component of any element is canceled by the per-pendicular component of an element on the opposite side of the ring. Since r = (x2 + a2)1/2 and cos = x/r, we find that
Ex = E cos = 2 2 2 3/2
q x kx k q r r (x a ) E1 E2 1 2 In this case, all segments of the ring give the same contribution
to the field at P since they are all equidistant from this point. Thus, we can easily sum over all segments to get the total field
GyaanSankalp at P : Ex = 2 kx2 3/2
(x a ) q = 2 2 3/2 kx (x a ) Q This result shows that the field is zero at x = 0.
At large distances from the ring (x > > a) the electric field along the axis approaches that of a point charge of magnitude Q. Case 2 : The electric field of a uniformly charged disk :
A disk of radius R has a uniform charge per unit area . To find the electric field along the axis of the disk, a distance x from its center follow the procedure.
Consider the disk as a set of concentric rings. We can then make use result of case 1,which gives the field of a given ring of radius r, and sum up contributions of all rings making up the disk. By symmetry, the field on an axial point must be parallel to this axis. The ring of radius r and width dr has an area equal to 2r dr. The charge dq on this ring is equal to the area of the ring multiplied by the charge per unit area, or dq = 2r dr. Using this result in equation (with a replaced by r) gives for the field due to the ring the expression. dE = 2 kx2 3/2 (x a ) (2r dr) r dq x
To get the total field at P, we integrate this expression over the limits r = 0 to r = R, noting that x is a constant, which gives
E = kx R 2 2 3/ 2 0 2r dr (x r )
= kx 0 Rz
(x2 + r2)–3/2 d(r2) = kx R 2 2 1/2 0 (x r ) 1 / 2 = kx 2 2 1/ 2 x x | x | (x R ) The result is valid for all values of x.
The field close to the disc along an axial point can also be obtained from equation by letting x 0 This gives E = 2k =
0 2
where 0 is the permittivity of free space, the same result is obtained for the field of a uniformly charged infinite sheet. Case 3 : E due to an infinite plane of charge
The field of an infinite plane of charge can be obtained from field by disc by either letting R go to infinity or letting x go to zero. Then Ex = 2k, x > 0
Thus, the field due to an infinite-plane charge distribution is uniform; that is, the field does not depend on x. On the other side of the infinite plane, for negative values of x, the field points in the negative x direction, so Ex = – 2k, x < 0
Case 4 : Spherical distribution of charge
(a) Conducting sphere (Hollow, solid) (b) Non-conducting sphere (Hollow, solid) Case (a) Charge on surface. Case (b) Volume distribution of charge. (a) : Hollow/solid conductor or hollow non-conductor
Imagine a sphere passing through desired point (point where E is to be calculated), calculate charge inside it and assume it to be concentrated at centre and use point charge formula. Inside sphere r < R Q r R + + + + + + + + + + + + + + + + +
E = 0 (No charge inside imagined sphere)
outside r > R Q r R + + + + + + + + + + + + + + ++ E = 2 KQ r surface r = R , E = KQ2 R
14 GyaanSankalp Graphically E KQ/R2 r = R r E 1/r2 (b) Inside r < R
Uniform volume distribution : charge inside volume 4 3r 3 3 Q 4 R 3 4 3 r 3 = Q' or Q' = 3 3 Qr R , E = 1 2 KQ r = 3 KQ R r + + + + + + + + + + + + + + + + + Q R r 0 ˆ E r 3 Outside, E = 3 KQ r r > R + ++ + + + + + + + + + + + + + + Q R r Surface E = KQ3 R , Graphically E r = R r E 1/r2 E r
Electric field intensities due to various charge distributions are given in table. Name/Type Point charge Infinitely long line charge Semi-infinite Finite change of charge + + + + + + + + + x P E E Formula 2 3 Kq Kq ˆ .r r | r | r 0 ˆ 2K r ˆr 2 r r 2k r , x y K k E , E r r x K E [sin sin ] r y K E [cos cos ] r If = || 0 E 0, E 2 r Particular q is source charge. r
is vector drawn from source charge to the test point.
Electric field is nonuniform, radially outwards due to + charges & inwards due to – charges.
is a linear charge density (assumed uniform)
r is perpendicular distance of point from line charge.
ˆris radial unit vector drawn from the charge to test point.
At a point above the end of wire at an angle 45°
Where is the linear charge density
r E
r E
GyaanSankalp Infinite non-conducting thin sheet Uniformly charged ring Infinitely large charged conducting sheet
Uniformly charged hollow conducting/ nonconducting/ solid conducting sphere Uniformly charged solid nonconducting sphere (insulating material)
Uniformly charged cylin-der with a charge density (R = radius of cylinder)
Uniformly charged cylin-drical shell with surface charge density is
is surface charge density (assumed uniform)
ˆn is unit normal vector.. Electric field intensity is independent of distance.
Q is total charge of the ring. x = distance of point on the axis from centre of the ring.
Electric field is always along the axis. Maximum at xR / 2
is the surface charge. ˆn is unit normal vector perpendicular is the surface. Electric field intensity is independent of distance.
R is radius of the sphere. r
is a vector drawn from centre of sphere to the point.
Sphere acts like a point charge, placed at centre for points outside the sphere.
E is always along radial direction. Q is total charge ( 4 R2) ( = surface charge density)
r
is a vector drawn from centre of sphere to the point.
Sphere acts like a point charge, placed at centre for points outside the sphere.
E is always along radial direction. Q is total charge ( . 4 R )4 3
3
( = volume charge density) Inside the sphere E r Outside the sphere E 1/r² 0 ˆn 2 2 2 3/ 2 KQx E (R x ) centre E 0 0 ˆn (i) for r R 2 kQ ˆ E r | r | (ii) for r < R E 0 (i) for r R 2 kQ ˆ E r | r | (ii) for r R 3 0 KQr r E 3 R for r < R, in 0 r E 2 for r > R, 2 0 R E 2 r for r < R, Ein = 0, for r > R, 0 r E r r E 0 2
E 1/r16 GyaanSankalp Example 19 :
Electric charge is uniformly distributed around a semicircle of radius a, with total charge Q. What is the electric field at the centre of curvature ?
Sol. Consider a small segment of angular width d, located at an angle from the x-axis. The length of the segment is ds = ad. The charge on the segment is dQ Qad Qd
a
The magnitude of the electric field at P is given by 2 2
0 0 1 dQ 1 Q dE d 4 a 4 a
This electric field has y-component y 2 2
0 Q dE dE sin sin d 4 a
The x-component of the field from the right hand half of the ring cancels with that of the left-hand half of the ring. The resultant electric field is thus in the y-direction, and is given by adding up the dEy from each segment in the ring. This is done by integrating from to rad : y y 2 2 0 0 Q E dE sin d 4 a
y 2 2 2 2 0 0 0 Q Q E sin d cos 4 a 4 a
; y 2 0 2 0 Q E [( 1) (1)] 4 a ; y 2 0 2 Q E 2 a Example 20 :A thin non-conducting ring of radius R has linear charge density 0cos, where is a constant, is the azimuthal angle.0
Find the magnitude of the electric field strength at the centre of the ring. Sol. The situation is shown in figure.
The half ring on the right hand side will be positive while on the half left side will be negative. The reason being that cos for first and fourth quadrants is positive while for 2nd and 3rd quadrants is negative.
Consider a small element dx of the ring. Here dx = R cos Charge on small element dq = = dx = 0 cos (R d)
0 2 0 R cos d 1 dE 4 R
Electric field along x-axis due to this element x dE dE cos = 0 2 0 R cos d 1 cos 4 R = 2 0 0 (cos d ) 4 R = 0 0 1 cos 2 d 4 R 2
Electric field due to positive part along x-axis,
/ 2 0 1 0 / 2 1 cos 2 E d 4 R 2
GyaanSankalp or 0 0 1 0 0 E 4 R 2 8 R
Similarly, the electric field due to negative charge along x-axis 0 2 0 E 8 R Enet = E1 + E2 = 0 0 0 0 0 0 8 R 8 R 4 R
TRY IT YOURSELF
Q.1 A spherical volume contains a uniformly distributed charge of density . The electric field inside the sphere at a distance r from center is – (A) 0 r 3 (B) 0 r 4 (C) 0 r (d) 0 1 r 4
Q.2 A point charge 50µC is located in the XY plane at the point of position vector r0 2iˆ3jˆ. What is the electric field at the point of position vector r8iˆ5jˆ –
(A) 1200 V/m (B) 0.04 V/m (C) 900 V/m (D) 4500 V/m
Q.3 A solid metallic sphere has a charge +3Q concentric with this sphere is a conducting spherical shell having charge –Q. The radius of the sphere is a and that of the spherical shell is b (>a). What is the electric field at a distance r (a < r < b) from the centre– (A) 0 1 Q 4 r (B) 0 1 3Q 4 r (C) 2 0 1 3Q 4 r (D) 2 0 1 Q 4 r Q.4 The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R will be –
(A) 2 0 1 q 4 3 3R (B) 2 0 1 2q 4 3R (C) 2 0 1 2q 4 3 3R (D) 2 0 1 3q 4 2 2R
Q.5 Two conducting plates X and Y, each having large surface area A (on one side) are placed parallel to each other. The plate X is given a charge Q where the other is neutral. The electric field at a point in between the plates is given by –
(A) Q
2A (B) 0
Q
2A towards left (C) 0 Q
2A towards right (D) 0 Q
2 towards right
Q.6 Two infinitely long parallel wires having linear charge densities and 1 respectively are placed at a distance R meter. The force2
per length on either wire will be
0 1 K 4 (A) 1 22 2 K R (B) K2 1 2 R (C) K 1 22 R (D) K 1 2 R Q.7 A wheel having mass m has charges + q and – q on diametrically opposite
points. It remains in equilibrium on a rough inclined plane in the presence of uniform vertical electric field E =
(A) mg q (B) mg 2q +q –q E (C) mg tan 2q (D) none
ANSWERS
(1) (A) (2) (D) (3) (C) (4) (C) (5) (C) (6) (B) (7) (B)18 GyaanSankalp ELECTRIC FIELD LINES
Electric lines of forces :
(i) The concept of electric field was introduced by Michael Faraday.
The magnitude of electric field strength at any point is measured by the number of electric line of force passing per unit small area around that point normally and the direction of field at any point is given by the tangent to the line of force at the point. (ii) An electric line of force is that imaginary smooth curve drawn in an electric field along which a free isolated unit positive (initially
at rest) charge moves.
E at any point on a line of force. Properties :
(1) The lines of force diverge out radially from a +ve charge and converge at a – ve charge. More correctly the lines
of force are always directed from higher to lower potential. + –
(2) The tangent drawn at any point on line of force gives the direction of force acting on a positive charge placed at that point. (3) Two lines of force never intersect. If they are assumed to intersect. There will be two directions of electric field at the point of
intersection : which is impossible.
(4) These lines have a tendency to contract in tension like a stretched elastic strong. This actually explains attraction between opposite
charges. –
Attraction
+
(5) These lines have a tendency to separate from each other in the direction perpendicular to their length. This explains repulsion between like charges.
Repulsion
+
+
(6) The no. of lines originating or terminating on a charge is proportional to the magnitude of charge. In rationalised MKS system
0
(1 / ) electric lines are associated with unit charge. So if a body encloses a charge q. Total line of force associated with it (called flux) will be
0 q .
(7) Total lines of force may be fractional as lines of force are imaginary.
q >qA B
+ –
A B
(8) Lines of force ends or strarts normally on the surface of a conductor. (9) If there is no electric field there will be no lines of force.
(10) Lines of force per unit area normal to the area at a point represents magnitude of intensity, crowded lines represent strong field while distant lines represent weak field.
(11) Electric lines of force differ from magnetic lines of force.
(a) Electric lines of force never form closed loop while magnetic lines are always closed or extended to infinity.
Electric line of force (A)
Magnetic line of force (B)
GyaanSankalp
(b) Electric lines of force always emerge or terminate normally on the surface of charged conductor, while magnetic lines emerge or terminate on the surface of a magnetic material at any angle.
(c) Electric lines of force do not exist inside a conductor but magnetic lines of force may exist inside magnetic material.
Lines of force do not exist inside a conductor (as field inside a conductor is zero) the plates is as shown. (Electrostatic shielding)
12. Neutral point : Where electric field intensity is zero (test charge does not experience any force)
13. The number of lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge. This means, for example that if 100 lines are drawn leaving a + 4µC charge then 75 lines would have to end on a –3µC charge. ELECTRIC FLUX ( ) :
If the lines of force pass through a surface then the surface is said to have flux linked with it. Mathematically it can be formulated as follows :
The flux linked with small area element on the surface of the body : d =
E
.
d
s
Where
d
s
is the area vector of the small area element. The area vector of a closed surface is always in the direction of outward drawn normal. The total flux linked with whole of the body,=
E.ds total flux linked with closed surface, where
is referred to closed integral done for a closed surface. (i) Electric flux is a scalar quantity(ii) Units (V - m) or (N - m2/Cb),Dimensions : [M1 L3 T–3 A–1]
(iii) The value of '' does not depend upon the distribution of charges and the distance between them inside the closed surface. (iv) The value of is zero in the following circumstances :
(a) If a dipole is enclosed by a closed surface.
(b) Magnitude of +ve and –ve charges are equal inside a closed surface. (c) If no charge is enclosed by a closed surface.
(d) In coming flux (–ve) = out going flux (+ ve). GAUSS'S LAW
The total flux linked with a closed surface is
0
1
times the charge enclosed by the closed surface (Gaussian surface).
i.e.
z
E ds . = q 0Law is valid for symmetrical charge distribution and for all vector fields obeying inverse square law. Gaussian surface :
(a) Imaginary surface
(b) Is spherical for a point charge, conducting and non-conducting spheres.
(c) Is cylindrical for infinite sheet of charge conducting charge surfaces, infinite line of charges, charged cylindrical conductors, etc. For finite charge distribution use Coulomb's law.
For infinite charge distribution use Gauss theorem
E ds.
z
=qnet 0
Application : (1) To Calculate flux (2) To calculate Electric field Intensity Study following cases to learn application properly
Observe flux through common geometrical figures
20 GyaanSankalp (iii)T = 0 (iv) T = q 0 , hemisphere = q 20
(dotted part shows imaginary part to enclosed the charge completely)
(v) T = q 0 , cyl. = q 2 .0 (vi) T = q 0 cube = q 20 Charge position T Cube centre = q/. Face centre = q/2. At corner = q/8. At centre of edge = q/40. T = 0 1 8 i i q
; i = 1, 2, ... 8.Electric field due to a line charge :
Consider an infinite line which has a linear charge density . Using Gauss’ss law, let us find the electric field at a distance ‘r’ from the line charge.
The cylindrical symmetry tells us that the field strength will be the same at all points at a fixed distance r from the line. Thus, the field lines are directed radially outwards, perpendicular to the line charge.
The appropriate choice of Gaussian surface is a cylinder of radius r and
S3 S2 S1 E S d E E E S d S d + + + + + + + + + + + + + + + + + + + + r
length L. On the flat end faces, S2and S3, E is perpendicular dS, which means flux is zero on them. On the curved surface S1, E
is parallel dS, so that E, dSEdS. The charge enclosed by the cylinder is Q = L.
Applying Gauss’s law to the curved surface, we have
0 ε L λ ) rL π 2 ( E dS E or r 2 E 0 Electric field due to an infinite plane thin sheet of charge :
To find electric field due to the plane sheet of charge at any point P distant r from it, choose a cylinder of area of cross-section A through the point P as the Gaussian surface. The flux due to the electric field of the plane sheet of charge passes only through the two circular caps of the cylinder. Let surface charge density =
According to gauss law E .dS qin/ 0
++ +++ + +++ + + + + ++ E P E Q r Plane sheet of charge Gaussian Surface 0 I circular II circular cylindricalsurface surface surface
A E ds cos E ds cos E ds cos
or EA + EA + 0 = 0 A 2 or E = 20 GyaanSankalp Electric field intensity due to uniformly charged spherical shell :
We consider a thin shell of radius R carrying a charge Q on its surface (i) at a point P0 outside the shell (r > R)
According to gauss law , 0
S1 E .ds
= 0 Q or E0 (4r 2) = 0 Q S Pin R O S2 Ps S1 P0 E ds = 0º E0 = 2 0 Q 4 r = 0 2 2 R rwhere the surface charge density = surface areatotal charge = 2
Q 4 R
The electric field at any point outside the shell is same as if the entire charge is concentrated at centre of shell. (ii) at a point Ps on surface of shell (r = R) ES = 2
0 Q
4 r = 0 (iii) at a point Pin inside the shell (r < R)
According to gauss law
S2 E .ds
= in 0 q As enclosed charge qin = 0 , So Ein = 0 E=0 E=0 E E r2 E r2 1 1 E Q/4 0R2 O r < R r = R r > R distance from centre (r)The electric field inside the spherical shell is always zero.
Electric field intensity due to a spherical uniformly charge distribution :
We consider a spherical uniformly charge distribution of radius R in which total charge Q is uniformly distributed throughout the volume. The charge density = total charge
total volume = 3 Q 4 R 3 = 3Q3 4 R (i) at a point P0 outside the sphere (r > R)
according to gauss law E .ds0
= 0 Q or E0 (4r2) = 0 Q or E0 = 2 0 Q 4 r = 3 0 3 2 R r + + + + + + + + + + + + + + + + + + + + + + P0 Pin Ps E ds r O R(ii) at a point Ps on surface of sphere (r = R) Es = 2 0 Q 4 R = 3 0 R
(iii) at a point Pin inside the sphere (r < R) According to gauss law
E .dsin
= in 0 q = 0 1 . 4 3 r 3 = 3 3 0 Qr R O r = R r > R 2 r 1 E 2 r 1 E E r E r E r < R Ein(4r2) = 3 3 0 Qr R or Ein = 0 3 Qr 4 R = 3 0 r (Ein r) ELECTROSTATIC PRESSURETo find force on a charged conductor (due to repulsion of like charge) imagine a small part PR to be cut and just separated from the rest of the conductor MLN. The field in the cavity due to the rest of the conductor is E2, while field due to small part is E1. Then
M N P
R
22 GyaanSankalp Inside the conductor :
E = E1 – E2 = 0 or E1 = E2 Outside the conductor :
1 2 0 EE E / Thus 1 2 0 E E 2
To find force, imagine charged part PR (having charge dA placed in the cavity MN having field E2. Thus force
2 dF ( dA) E or 2 0 dF dA 2
The force per unit area or electric pressure is
2 0 dF dA 2
The force is always outwards as ()2 is positive i.e. whether charged positively or negatively, this force will try to expand the charged body.
A soab bubble or rubber balloon expands on given charge to it. (charge of any kind + or –) Energy associated per unit volume of electric field of intensity E is defined as energy density.
u = dw dv = 0 2 2 E = 2 0 J/m 3 U =
z
u . dv = 0 2z
v E2 dv ; v is the volume of electric field.
Equilibrium of charged liquid surfaces : Soap Bubble :
Pressures (forces) acts on a charged soap bubble, due to (i) Surface tension of a soap bubble PT (inward)
(ii) Air out side the bubble p0 (inward)
(iii) Electric charges (electrostatic pressure) Pe (outward) (iv) Air inside the soap bubble Pi (outward)
Hence, in state of equilibrium
inward pressure = outward pressure PT + P0 = Pi + Pe Excess pressure (Pex.) = Pi – P0 = PT – Pe But PT = 4T r , Pe = 2 0 Pex. = 4T r – 2 0 If Pi = P0, then 4T r = 2 0 Example 21 :
There is a solid sphere of radius R having volume charge density 0
r 1
R
, where is any constant and r is the distance0 from the centre of sphere. Find electric intensity E inside and outside the sphere.
GyaanSankalp Sol. (i) Inside : When r < R, electric flux through small area,
d E.dSE dS
E dSE
dSE 4 r2 But according to Gauss's law enclosed0 q
Charge q contained between radius x to x + dx.
2 0 x dq 4 x 1 R dx
Charge enclosed inside the gaussian surface,
r r 3 3 4 2 enclosed 0 0 0 0 x x x q 4 x dx 4 R 3 R
= 3 4 0 r r 4 3 4R But 3 4 2 0 0 4 r r E 4 r 3 4R 0 0 r 1 r E 3 4R 0 0 r 3r E 1 3 4R , when r R (ii) Outside : When r R, thenR R 3 4 3 4 enclosed 0 0 0 0 x x r r q 4 4 3 4R 3 4R 3 0 2 0 R E 12 r R r Example 22 :
A charge of 4 × 10–8 C is distributed uniformly on the surface of a sphere of radius 1cm. It is covered by a concentric, hollow conducting sphere of radius 5 cm.
(a) Find the electric field at a point 2 cm. away from the center (b) A charge of 6 × 10–8 C is placed on the hollow sphere. Find the charge on the outer surface of the hollow sphere. Sol. (a) Let us consider figure (a). Suppose, we have to find
field at the point P. Draw a concentric spherical through P. All the points on this surface are equivalent and by symmetry, the field at all these points will be equal in magnitude and radial in direction.
The flux through this surface
P
(a) (b)
=
E.dS
E.dSE dS
4 x E2 where x = 2 cm = 2 × 10–2 mFrom Gauss’s law, this flux is equal to the charge q contained inside the surface divided by 0. Thus, 2 0 q 4 x E or 8 9 2 2 5 2 4 2 0 q 4 10 C E (9 10 Nm / C ) 9 10 N / C 4 x 4 10 m
