sometimes if you touch a metallic doorknob or any metallic object you might experience a mild shock this is because the metallic object was charged in other words it carried an electric charge objects that carry an electric charge are said to be electrically charged we know that when a comb was rubbed against dry hair it begins to attract small pieces of paper when two pieces of silk cloth I rubbed against two glass rods individually and the glass rods are kept beside each other but Lars rods repel each other now if one of the silk clots and a glass rod are kept beside each other they attract each other now bring one charged glass rod in contact with the pit ball and then separate them then bring the other charged glass rod in contact with other pit ball and then separate them to when these big balls are kept close to each other without contact we observe that they repel each other the physical quantity responsible for all these phenomena is known as electric charge electric charge can be defined as the property of matter that exhibits its electrostatic interaction with other matter you from previous examples bodies like combs Rod's silk cloth pieces bit balls etc are said to be electrified or are called charged bodies the study of electric charges at rest and the physical quantities associated with electric charges at rest is called electrostatics or static electricity these experiments prove that a body can be charged by rubbing against another body or by simply bringing a charged body in contact with another uncharged on neutral body when the glass rod and silk cloth were first dropped they get electrified but when they are brought in contact again and separated they lose the electrification or charge and unutilized in other words the charge on them is nullified ends we can conclude that one of them may be positively charged and the other negatively charged when they are rapped against each other you from these observations we see that there exists only two types of electric charges in the nature conventionally they are called positive and negative it is also observed that the light charged bodies repel each other for example two big balls both carrying like charges repel each other unlike charged bodies attract each other for example a bit ball carrying a positive charge will attract a negatively charged pit ball we know that all matter is made up of atoms or molecules also within an atom the total negative charge is balanced by the total positive charge in it when we rub one material against another material the material in which electrons are relatively loosely bound to its atoms readily loses a few of its electrons to the other material the material that lost electrons now has a deficit of electrons and hence becomes positively charged and the one that gained electrons becomes negatively charged in this case the glass rod lost electrons and became positively charged and sick gained those electrons and became negatively charged you to know whether a body has been charged we use an instrument called a goldleaf electroscope it consists of an insulated metal rod with two gold leaves or voids at one end and a metal disc at the other end this setup is enclosed in a glass jar with the help of a tight with the disc placed out of the jar to test a material like a glass rod for electrification the object should be brought in contact with the metal disc of the goldleaf electroscope the contact should be momentary if there is no change observed in the position of the gold leaves it indicates that the glass rod is uncharged or is neutral if the gold leaves repel each other then it indicates that the glass rod is charged the amount of divergence provides a rough measure of the charge on the glass rod assuming that the glass rod was positively charged the goal leaves now carry a positive charge now if we touch the metal disk with the silk cloth used to rub the glass rod and take it away we observed that the gold leaves fall back to their original position does we can see the silk clot was negatively charged on contact the opposite charges on the gold leaves and silk cloth nullified each other we know that electric charges are conventionally called positive and negative charges and that like charges repel each other and unlike charges attract each other however if the size of charged bodies is very small but the distance between them is comparatively larger than their sizes such charged bodies are treated as point charges Gulam see is the SI unit of electric charge however smaller units such as micro coulomb and milli Coulomb are also used in electrostatics electric charge is a scalar quantity which has only magnitude but no direction hence the total charge on the body is equal to the algebraic sum of the individual charges given to it let us illustrate this with an example [Music] consider a body carrying a charge cq1 we add another charge q2 to this body the total charge on the body is the algebraic sum of both charges that is total charge Q is equal to q1 plus q2 similarly if a body carries n number of charges q1 q2 and so on up to QN then the total charge on the body Q is equal to Q 1 plus Q 2 plus Q 3 and so on till QN why adding the charges proper science should be used of sign for a positive charge and a negative sign for a negative charge we can determine whether a body is charged positively or negatively by finding the antibiotic sum of all charges on the body for example if a body is given six charges all in Coulomb say plus one minus two plus three - for - five plus six the algebraic sum would be minus one Coulomb which means it is a negatively charged body another property of electric charges is that the total charge in an isolated system is always conserved when at last rod is rubbed with a sake clock electrons are transferred from the glass rod to the silk cloth now both the glass rod and sake plot are charged here there is only a transfer of charge taking place neither is a new charge created nor is any charge destroyed the number of electrons lost by the glass rod is equal to the number of electrons gained by the silk cloth if we consider the glass rod and sake plot as an isolated system then the total charge of this system before and after rubbing remains the same therefore we can conclude that the total charge of an isolated system is always conserved similarly charges are also conserved in natural processes such as the beta decay of a neutron during the beta DK a neutron breaks into a proton and an electron releasing a neutrino here the neutron and neutrino are charged less particles while the proton and electron are oppositely charged particles whose combined net charge is zero hence we observe that the total charge in this process is conserved the charge on an object is measured as an integral multiple of the basic unit of charge that the proton or electron carries and is denoted by E [Music] this property is called quantization of charge the magnitude of the basic unit of charge E is approximately equal to 1.6 into 10 raised to power minus 19 coulomb in an atom when we refer to the charge on a proton value is positive and is referred to as plus e [Music] whereas the value of e for charge on an electron is negative and is referred to as minus e does the charge on an object q is equal to the product of the number of electrons gained or lost by it in and the basic unit of charge II now let us calculate the number of electrons one Coulomb of negative charge carries consider a body possessing a charge Q equal to minus one Coulomb that the number of electrons it has gained to become negatively charged bien as the charge Q on a body is quantized we have Q equal to n e where the basic unit of charge E has a magnitude of 1 point 6 into 10 power minus 19 coulomb and Q equal to minus 1 Coulomb substituting the values of Q and E and simplifying we have n equal to minus six point two five into ten power 18 here n is a negative integer indicating that the body has gained a negative charge so we see that one Coulomb of negative charge contains six point two five into ten power 18 electrons which is a very large number of electrons in the practical examples we have seen for charging a body we deal with bodies possessing a charge of about a few micro Coulomb the charge of micro coulomb that is ten to the power of minus six Coulomb it's very large compared to the basic unit of charge E [Music] when such a microscopic level of charges involved the concept of quantization of charge can be ignored since it has no practical significance where we see the charges always added or removed in integral multiples of the basic unit of the charge quantization of charge is significant only in the microscopic sense when the charge possessed is due to a few hundred times the basic unit of charge II or even less than that let us now learn about conductors and insulators we know that atoms are the basic building blocks of all matter in an atom the protons and neutrons are the constituents of the nucleus while the electrons revolve in orbits around the nucleus for some materials generally metals the electrons in the outermost orbit are loosely bound to the nucleus and these electrons are called free electrons these electrons are relatively free to move around the material such materials with a large number of free electrons are called conductors and as such these materials readily allow the flow of electric current through them and cause very little resistance to the flow of electricity metals like copper nickel and the human body and earth etc are some conductors some materials have a very small number of free electrons and hence offer very high resistance to the flow of electricity through them such materials are called insulators nonmetals such as porcelain plastic glass would and nylon are some of the best insulators it is advisable to wear rubber Footwear or gloves or stand on a wooden platform why handling electrical equipment since they are insulators and protect us from the harmful effects of the electric current in the absence of the gloves any leakage in the current from the electrical equipment is transferred through the body to the earth and this is known as grounding or earthing in any household electric circuit in the mains supply we see three wires namely live neutral and Earth life and neutral wires carry the electric current whereas the earth wire is used as a safety mechanism all earth wires terminate in a thick metal plate buried deep in the earth when a fault occurs all the life wire touches the metallic body of an electric instrument the charge immediately and safely flows into the earth because of the earth connection through the earth wire this prevents damage to human life and property there are three methods to charge a party namely judging by friction conduction and induction let us first look at charging a body by friction when two uncharged or neutral objects need of different materials see a glass rod and a silk cloth erupted against each other both objects get electrified due to a transfer of charges this process is called charging by friction when drugged the glass rod loses a few electrons to the silk o'clock this deficit of electrons on the glass rod makes it positively charged and the excess of electrons mix the silk clot negatively charged in this method there is a transfer of electrons from one body to another the material that gains electrons is negatively charged and the one that loses electrons is positively charged in this case silk is negatively charged and glass rod is positively charged let us now look at how a body can be charged by conduction if we momentarily touch the positively charged glass rod to an uncharged width borne the big ball becomes positively charged to this process of charging by bringing a charged body in contact with the neutral body is called charging by conduction in this case some electrons are transferred from the pit ball to the glass and does the glass becomes less positively charged by gaining the electrons we observe that both bodies involved in this method of charging by conduction acquire the same type of charge once the process is complete let us illustrate this by a numerical example let us assume that initially glass has a deficit n/g of 200 electrons using the quantization of charge principle that is Q is equal to n e we see that the charge on the glass is Q G equal to plus 200 e now when glass rod was in momentary contact with the pit born let us assume that because of the electrostatic attraction a few electrons say npg equal to 50 are transferred from the pit ball to the glass rod now the pith ball has the deficit NP of 50 electrons which implies that the charge on the pit ball QP is equal to plus 50 e similarly the glass now has a charge of plus 200 e minus 50 E we're minus 50 E is the charge due to the 50 electrons gained by the glass therefore the effective charge on the glass is plus 150 e hence it has become less positively charged compared to its original state we can also charge a body without contact as in the first two methods the third method to charge a body is induction when a charged object II it's placed near an uncharged object be object a induces an unlike charge on the near side of B and a light charge on its far site in this case the charges inside be our only polarized and no transfer of charges takes place this process of polarization of the charge on an uncharged body when a charged body is held close to it is called induction of charge for example bringing a negatively charged rod near a neutral metal sphere will induce a positive charge on the near surface of the metal sphere and a negative charge on the far side when the negatively charged rod is brought near the metal sphere a few free electrons in the metal sphere move away to the far side of the metal sphere due to repulsion leaving a positive charge on the near site this charge is temporary that is if we move the glass rod away the sphere returns to its original uncharged state when the negatively charged rod is held close to the metal sphere the charges in it are polarized due to induction now if we earth the far side of the metal sphere using a conducting wire the electrons on the far side of the metal sphere flow into the earth however the positive charges on the near side of the metal sphere are held at rest by the negatively charged rod now if we take away the glass rod from the sphere the positive charge on the metal sphere distributes uniformly across the sphere in this way we can retain the charge on the metal sphere in the method of charging by induction we observe that the induced charge is always equal to the inducing charge for a conductor now we keep two metal spheres a-and be supported on insulated stands in contact with each other when we bring a positively-charged glass rod near sphere a negative charges accumulate on its near site and positive charges accumulate on the far side of sphere B according to the principles of attraction and repulsion respectively now separate the spheres by a very small distance keeping the charged glass rod near them there will be a force of attraction between a and B then if we move away the glass rod spear II will retain the negative charge and sphere B will retain the positive charge we observe that the excess charge of the metal spheres distributes uniformly later Coulomb's law stated by Charles Coulomb a French physicist to find the electrostatic force of attraction and repulsion between two charged particles this law was defined after an extensive study of Falls between the charges using a torsion balance torsion balance has an insulating rod with a metal coated ball at one end and a balancing sphere at the other this rod is suspended with the help of a suspension fiber the graded circular scale gives a measure of the displacement of the sphere due to repulsion another static sphere which is identical in size to the metal coated ball is placed at the zero mark a micrometer screw at the top helps adjust the distance between the three and rotating spheres the static sphere is first given a charge Q by bringing a charged object in contact with it then the rotating sphere is brought in contact with the static sphere using the micrometer screw due to conduction the charged static sphere now loses half of its charge that is Q by two to the other sphere this happens because the two metal spheres are identical in size now both the spheres are identically charged with a charge of Q by two each they start repelling each other since like charges repel the measure of repulsion is the amount of deflection or the angle through which the sphere attached to the rod is displaced statics fear is now discharged by grounding to recharge the static sphere the rotating sphere is brought in contact with the static sphere using the micrometer screw again due to conduction the spheres now have a charge Q by for each as both the spheres are identically charged they repel each other however the angle of deflection observed is lesser than the previous observation for charge q by 2 on each sphere therefore it can be concluded that as the charge on the spheres decreases the force of repulsion between them also decreases from these experiments with charged spheres at constant distances it was concluded that the electrostatic force between two charges is directly proportional to the product of the charges if the charges are Q 1 and Q 2 then electrostatic force F between them is proportional to Q 1 into Q 2 that this B equation one keeping the charge on each sphere constant if the distance between the charged spheres is varied using the micrometer screw the angle of deflection is found to be inversely proportional to the square of the distance R between the charges since the force of repulsion is proportional to the angle of deflection we can see that the electrostatic force F between the charges q1 and q2 is inversely proportional to the square of the distance R between the charges let this be equation 2 thus according to Coulomb's law the electrostatic force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges hence the electrostatic force F between charges q1 and q2 separated by a distance R is proportional to Q 1 Q 2 by R square that this be equation three a constant of proportionality K is used to get the equation s is equal to k q-1 q-2 by r square let this be equation 4 if the charges are separated in free space that is air or vacuum the value of the constant K is equal to 1 by 4 PI epsilon not where epsilon naught is called permittivity of free space thus the electrostatic force F is equal to 1 by 4 PI epsilon naught Q 1 Q 2 by r square in SI units the value of the proportionality constant K is approximately equal to 9 into 10 power 9 newton meter square per Coulomb square and the value of epsilon naught is equal to 8 point 8 5 into 10 power minus 12 coulombs square per newton meter square hence f is equal to 9 into 10 power 9 into q1 q2 by r square using Coulomb's law a unit charge can be defined as one that experiences an electrostatic force of 9 into 10 / 9 Newton when placed at a distance of 1 meter from another unit charge in free space let us now study Coulomb's law in vector form let the charges q1 and q2 be at positions specified by the position vectors R 1 and R 2 with respect to the origin Oh the vector r2 one is equal to R two minus R one joins q1 and q2 the force 1 Q 1 due to Q 2 is f1 2 force on Q to do to q1 is F 2 1 in vector form we can express Coulomb's law as shown if q1 and q2 are like charges they repel in this case the product of the charges q1 and q2 will be positive the unit vector r1 to cap is given by equation five here the force f1 - will be in the direction of the unit vector r1 - cap if q1 and q2 are unlike with respect to each other the force f1 2 will be in the direction of the unit vector R 2 1 cap as given in equation 6 since unlike charges attract each other in this case the product of the charges q1 and q2 will be negative the unit vector r2 one cap is given by equation seven according to Newton's third law you two experiences an equal and opposite electrostatic force to the one experienced by q1 we can write f 2 1 is equal to minus F 1 2 let us now consider a system of charges q1 q2 q3 and so on till QN as shown in the figure with position vectors r1 r2 r3 till r n with respect to the origin o we know that electrostatic force as given by Coulomb's law exists between every pair of charges of the combination to find the net force on any charge in such a system we first calculate the electrostatic force on that charge by each and every other charge of the system then the net electrostatic force on that charge is equal to the vector sum of the individual electrostatic forces on that charge due to every other charge of the system of charges the individual forces are unaffected by the presence of other charges in the configuration this principle is called the principle of superposition for example if the charge q1 is considered then according to principle of superposition the net electrostatic force on the charge q1 is f1 equal to f1 2 plus F 1 3 plus so on till f1 n that this be equation eight f12 is the force on the charge q1 due to Q 2 and is equal to 1 by 4 PI epsilon naught into q1 q2 by r1 2 square into R 1 2 cap r12 is the distance between q1 and q2 and r12 cap is the unit vector in the direction of the force f1 to f13 is the force on the charge q1 due to Q 3 and is equal to 1 by 4 pie epsilon-not into q1 q3 by art 1/3 square into R 1 3 cap r-13 is the distance between q1 and q3 and r-13 cap is the unit vector in the direction of the force f1 3 proceed in this manner we obtain f1n is the force on the charge q1 due to Q n and is equal to 1 by 4 PI epsilon not Q 1 Q n by r 1 n square into R 1 n cap r1n is the distance between q1 and Q n and r1 n cap is the unit vector in the direction of the force f1 in now using F 1 2 F 1 3 F 1 n in the principle of superposition we have F 1 as an equation 9 thus net electrostatic force on Q 1 is F 1 is equal to 1 by 4 pie epsilon-not into Q 1 into Sigma I equal to 2 to n Qi by R 1 I square into R 1 I cap let this be equation 10 to understand the superposition principle more clearly consider the charge configuration consisting of charges q1 q2 and q3 with position vectors r1 r2 and r3 respectively with respect to the origin o f12 is the force on the charge q1 due to Q 2 and is equal to 1 by 4 PI epsilon not Q 1 Q 2 by r1 2 square into r1 2 cap r12 is the distance between q1 and q2 and r12 cap is the unit vector in the direction of the force f1 to f13 is the force on the charge q1 due to Q 3 and is equal to 1 by 4 pie epsilon-not into q1 q3 by r1 3 square into R 1 3 cap r-13 is the distance between q1 and q3 and our 1/3 cap is the unit vector in the direction of the force f1 3 to find the net force f1 on q1 due to the other charges in the system we should apply the principle of superposition that is we should have f1 is equal to f1 2 plus F 1 3 vectorially this we can do by using parallelogram law of vectors as shown then the direction of f1 will be along the direction of the diagonals of that parallelogram the magnitude of f 1 will be equal to root over F 1/2 square plus F 1 3 square plus 2 F 1 2 F 1 3 cos theta where theta is the angle between F 1 2 and F 1 3 by knowing the values of the parameters involved the net force f1 on q1 can be calculated now let us consider the electrostatic force between two electrons having a charge equal to basic unit of charge E which is equal to minus 1 point 6 into 10 power minus 19 Coulomb and separated by 1 meter in vacuum the magnitude of the electrostatic force between them according to Coulomb's law is given as f e equal to 1 by 4 PI epsilon naught e square by R square where R is equal to 1 meter substituting the values of the various parameters we have the electrostatic force between the two electrons is equal to twenty three point zero four into 10 power minus 29 Newton let this be equation 11 let us now compare the electrostatic force between these electrons with the gravitational force between them and force F between the electrons is G into M 1 M 2 by R square where M 1 and M 2 are the masses of the electrons which are equal to 9 point 1 into 10 power minus 31 kilogram each substituting the values of the babies parameters we have the gravitational force between them is equal to five hundred and fifty two point three four into ten power minus seventy three Newton let this be equation 12 defining equation 11 by equation 12 we have the ratio between electrostatic and gravitational forces and we can see that the electrostatic force is enormous ly greater than the gravitational force consider a charged particle with the charge Q held at rest at the origin of an imaginary coordinate system if we place a charge q-not at a point P nil charge Q and release it then the charge q-not accelerates away from the origin this acceleration is due to a force on charge q-not now let us remove charge Q from the origin and reintroduce charge q-not at the same point B charge Q naught does not accelerate now and remains at rest this means that the force experienced by Q naught is due to charge Q we see that since charge q-not is in the electric field of charge Q it experienced a force an electric field is a region in which the influence of the charge is felt since charge Q produces this electric field this charge is called the source charge charge q-not is called a test charge as it is used to study the effect of the electrical field created by charge q we can use Coulomb's law to calculate the force on Q not due to Q the electrostatic force on Q not due to charge Q is F is equal to 1 by 4 PI epsilon not Q Q naught by r square into r cap where our cap is the unit vector along the direction of OB let this be equation one now if Q naught is a unit positive charge that is Q naught is equal to plus 1 Coulomb then the electrostatic force as denoted by equation 1 becomes 1 by 4 PI epsilon naught Q by r square into r cap let this be equation 2 equation 2 indicates the electric field strength or intensity of electric field e which is the force on a unit positive charge acting at a particular point in the electric field therefore the electric field strength due to charge Q at point B is e equal to 1 by 4 PI epsilon not Q by r square into r cap let this be equation three electric field strength II is a vector quantity pointing in the direction of the force on a unit positive charge at a point in the electric field in Si it is measured in Newton per Coulomb or volt per meter from equations 1 & 3 the folds on the charge q-not can now be expressed as f is equal to Q naught E let this be equation for hence we can write the electric field strength at P as E is equal to f by q-not let this be equation 5 the above equation is valid as long as the test charge q-not is very small and its own electric field is negligible hence at Point P we write the electric field strength e as f by q-not limit Q not tending to zero for a given distance R from the source charge Q the value of 1 by 4 pie epsilon-not is always constant and hence e as given by equation 3 will also be constant hence the electric field strength II at any distance R from the source charge Q is independent of the test charge q-not it varies with our cap and is inversely proportional to R square the electric field at any point in the electric field is finite and is zero only when distance R is equal to infinity in the electric field of a positively charged particle the electric field strength vector e at any point of the field always points away from the charged particle in a radially outward direction the point charge q-not will be accelerated in the direction of the electric field as it experiences an acceleration a due to the force due to the electric field and the acceleration is expressed as a is equal to Q naught e by M where m is the mass of the particle in the field of a negatively charged particle the electric field strength vector at any point in the field always points towards the charged particle innovatively inward direction here the point charge q-not in this electric field will be accelerated in the direction opposite to the electric field the magnitude of acceleration of Q naught is the same regardless of whether the electric field is due to SOSE large plus Q on minus q for a charge Q at all points on an imaginary sphere of radius R around it the electric field strength will be equal in magnitude and will be equal to 1 by 4 PI epsilon naught Q by R square there is a spherical symmetry with respect to the magnitude of the electric field strength E in the electric field of charge Q consider a region consisting of point charges Q 1 Q 2 Q 3 Q 4 and so on in air or a vacuum with position vectors R 1 R 2 R 3 and so on relative to the origin all let P be a point in this region the net electric field strength II at point B is the superposition of all the electric field strengths II won e 2 e 3 E 4 and so on due to the individual charges the net electric field strength II at Point P will be equal to the vector sum of all the electric field strengths II won e 2 e 3 E 4 and so on then e is equal to 1 by 4 pie epsilon-not into Q 1 by r 1 p square into R 1 P cap + Q 2 by r 2 p square into R 2 P cap + q 3 by R 3 P square into R 3 P cap and so on here R 1 P is the distance from Q 1 to P and R 1 P cap is the unit vector in the direction of Q 1 - P r2p is the distance from q2 to P and R - t cap is the unit vector in the direction of Q to to P and so on the net electric field strength II can now be written as 1 by 4 PI epsilon naught Sigma I 1 to n Qi by r IP square into r IP cap where our IP cap is the unit vector in the direction of the net electric field strength e consider a situation where charge Q is accelerating then it produces electromagnetic waves which propagate at the speed of light here Q naught is also accelerating in the electric field of Q and there will be a force on Q naught due to the electric field of Q this falls on cue not due to queue is actually caused by the electromagnetic waves generated by the accelerating charge Q and which travel from Q to Q not at the speed of light does an exhilarating charged particle is a source of electromagnetic waves for example electric current applied to a solenoid creates an electromagnetic field around it consider a region consisting point charges q1 q2 q3 and so on in air or in vacuum with position vectors R 1 R 2 R 3 and so on relative to the origin Oh let P be a point in this region the net electric field strength II at the point B is the superposition of all the electric field strengths II one e 2 e 3 and so on due to the individual charges Q 1 Q 2 Q 3 and so on respectively now one p is the distance from q1 to p and r1 p cap is the unit vector in the direction of Q 1 to P R 2 P is the distance from Q 2 to P and R 2 P cap is the unit vector in the direction of Q 2 to P R 3 P is the distance from Q 3 to P and R 3 P cap is the unit vector in the direction of Q 3 to P the net electric field strength II at the point P will be equal to the vectorial sum of all the electric field strengths II 1 e 2 e 3 and so on therefore e is equal to 1 by 4 pie epsilon-not into Q 1 by r 1 p square into R 1 P cap plus Q 2 by r 2 p square into R 2 P cap plus Q 3 by R 3 P square into R 3 P cap and so on the net electric field strength II can now be written as 1 by 4 pie epsilon-not into Sigma I equal to 1 to n Qi by r IP square into our IP cap in many practical situations electric fields are produced by continuous charge distribution on the surfaces of the bodies of finite size and not due to the charge distribution discussed earlier a few examples of some practical situations are a line charge distribution a sheet of charge or a volume charge distribution since it is difficult and impractical to work with discrete charges in terms of the microscopic locations of the charge distributions to evaluate the electric field at a point we adopt a different approach in such situations the entire charge distribution is assumed to be made of infinitesimally small elements and the electric fields trend at any point in the electric field will be equal to the vectorial sum of the electric field strength due to each such small element let us look at a line charge distribution of length L carrying a charge Q as shown if we consider a small element of such a charge distribution having a charge Delta Q and of length Delta L the linear charge density lambda off such an element is equal to Delta Q by Delta L since it is the charge per unit length the SI unit of linear charge density lambda is Coulomb per meter it is a scalar quantity if line charge distribution is carrying a non uniformly distributed charge then it's linear charge density lambda for an individual element is equal to Delta Q by Delta L a conductor of length L is carrying a uniformly distributed charge Q then it's linear charge density lambda is equal to Q by M now let us consider a sheet of charge carrying a total charge Q on a surface area a in such situations the electric field strength at an imaginary point P can be calculated by assuming the charge distribution to be divided into infinitesimally small areas each having an area Delta e and charge distribution Delta Q surface charge density Sigma is the charge per unit surface area therefore surface charge density for an individual's surface element is expressed as Sigma is equal to Delta Q by Delta a it is a scalar quantity measured in coulombs per meter square in SI unit if charges are non uniformly distributed surface charge density for an individual surface element is expressed as Sigma is equal to Delta Q by Delta a if the charges are uniformly distributed over the surface then the surface charge density Sigma is equal to Q by a now let us consider a volume charge distribution consider a volume V carrying a charge Q that is assumed that the total volume is divided into a number of small elements each of volume Delta V having a charge Delta Q then volume charge density or charge density draw of such a volume charge distribution is defined as the charge per unit volume does rho is equal to delta q by delta v and is measured in coulombs per meter cube in SI unit it is a scalar quantity if charges are not uniformly distributed the charge density row for an individual volume element is equal to Delta Q by Delta V if the charges are uniformly distributed the volume charge density of the conductor is expressed as ro is equal to Q by V now consider the small volume element Delta V of charge distribution carrying a charge Delta Q we know that the charge density Rho is equal to Delta Q by Delta V does charge on the volume element is then Delta Q equal to Rho Delta V the position vector of the volume element Delta V with respect to the origin o is R consider a point P in the electric field of the volume charge distribution which is at a distance of R - from the volume element Delta V the electric field strength at Point P due to the volume element is Delta e is equal to 1 by 4 PI epsilon naught Delta Q by R - square into R - cap where R - cap is the unit vector in the direction of the straight line from the volume element towards the point P this can be written as delta e equal to 1 by 4 pie epsilon-not into Rho Delta v by r - square into R - cap the electric field strength e at be due to the total volume charge distribution is equal to the vectorial sum of the electric field strength due to all individual volume elements however this is impractical as there would be many infinitesimal volume elements at varying distances from the point P hence to obtain the electric field strength e due to the total volume charge distribution we integrate Delta e with in suitable limits a to be under the assumption that the volume elements have very small volume tending to zero consider a point charge plus Q and it's electric field let us introduce a unit positive charge at a point p1 in the electric field of Q as both the source charge and the test charge are positive this unit positive charge placed at point B one follows the path P 1a in the electric field due to the repulsive force on it exerted by the electric field of the source charge does but p1e represents an electric line of force or an electric field line an electric field line represents the path followed by a unit positive charge in an electric field repeating this procedure for various points in the electric field we obtain the electric field lines around the charge electric field lines are used to represent the electric field of a charge or a system of charges pictorially in fact the electric field lines follow the same direction as the electric field vector at any point in the electric field as any electric field line is the path followed by a unit positive charge in an electric field if we draw all the electric field lines for the electric field of the given source charge Q the electric field lines appear as shown one of the most important characteristics of electric field lines is that they help us estimate the electric field strength the relative density or closeness of the field lines at different points gives an indication of the relative strength of the electric field at those points the field strength is stronger at places having more crowded field lines than at places where they are less crowded to illustrate this let us now consider two regions a and B of same area placed normal to the field lines in the electric field the field lines in region a are relatively closer than those in region B the number of field lines passing normal to an area is proportional to the magnitude of the electric field strength add these points this indicates that the strength of the electric field at a is greater than the field at B as the electric field strength is stronger in region a and weaker in region B the electric field shown is a non-uniform electric field as the field strength varies from region to region to understand the dependence of field lines on area let us understand the relation between the area of the element and the solid angle created for this first let us understand the term solid angle solid angle is a measure of a cone when the intersection of a corn with the sphere of radius R is considered angle Delta Omega subtended by the area element is defined as Delta s by R square where Delta s is the area on the sphere which is the area element cut out by the cone let this be equation 1 therefore delta s is equal to R square into Delta Omega let this be equation 2 for a given solid angle Omega the number of radial field lines is the same at two different points say X and Y in the electric field which are at distances r1 and r2 from the charge we observed that the area of the elements subtending the same solid angle Delta Omega will be Delta s1 equal to Delta Omega into r1 square at X and Delta s2 equal to Delta Omega into r2 square at Y respectively the number of field lines see n passing normal to Delta s 1 and Delta s 2 is also the same the number of failed lines passing normal to a unit area is equal to the electric field strength II therefore the electric field strength II won at X will be equal to number of field lines per unit area at X which is equal to n by Delta s 1 or n by r 1 squared into Delta Omega using the value of Delta s 1 let this be equation three similarly electric field strength e to at Y is equal to n by Delta s to equal to n by r 2 square into Delta Omega let this be equation for if we divide equations three and four we see that the ratio of the electric field strength at x and y is equal to r2 square by r1 square hence we can conclude that the electric field strength due to a point charge varies inversely to the square of the distance from the charge from equations three and four we see that even though the number of field lines in and the solid angle delta omega are the same at X&Y the electric field strength II we'll be equal to number of field lines per unit area is proportional to 1 by R square where R is the distance of the point from the charge Q earlier you have learnt that the electric field strength vector of a positive charge say plus Q always points in a radially outward direction and the electric field strength vector for a negative charge say minus Q points in a radially inward direction thus we see that the electric field lines for a positive charge diverge normally from the charge and those for a negative charge converge normally towards the church let us now visualize the electric field lines for a system of two positive charges c plus Q each from the field lines we can see a very clear picture of the repulsion existing between the system of charges shown the electric field lines also help us in visualizing the electric field of a dipole consisting of charges plus Q and minus Q we observe that the electric field lines for an electric dipole are continuous lines extending from the positive charge to the negative charge from the field lines we see very clearly the force of attraction existing between the system of charges in the dipole thus see that the field lines can be a very clear picture of the nature of the electrostatic forces existing in any system of charges from these examples of various charge configurations we observe the following properties of electric field lines the tangent drawn to an electric field line at a point in the electric field gives the direction of the electric field strength II at that point the direction of the tangent at a point indicates the direction of the force on a unit positive charge at that point and hence it gives the direction of the electric field strength at that point for the electric field created by a system of positive and negative charges the field lines begin at the positive charge and end at the negative charge since the field lines diverge normally from a positive charge and converge normally at the negative charge for a single charge they may start or end at infinity electric field lines do not intersect each other if the electric field lines intersect each other then at the point of intersection there will be more than one direction for the electric field which is not possible the number of electric field lines is proportional to the magnitude of the charge the number of electric field lines is proportional to the magnitude of the charge for a charge configuration of magnitude Q if the field lines per unit area normal to it are say n then for a charge of magnitude 2q the number of field lines passing normal to a unit area is equal to the electric field strength electrostatic field lines never form closed loops due to the conservative nature of the electric field since the lines of force diverged from a positive charge and cannot converge at the same positive charge for a uniform electric field the electric field lines are parallel equidistant and unidirectional for a non-uniform electric field the electric field lines are not parallel in the charge-free region the electric field lines are continuous curves without any brakes [Music] [Applause] [Music] [Applause] [Music] [Applause] [Music]