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ELECTRIC CHARGE AND FIELDS CHAPTER-1 1. ELECTROSTATICS- It deals with the study of forces , fields and potentials arising from static charges. 2. ELECTRIC CHARGE – Electric...

ELECTRIC CHARGE AND FIELDS CHAPTER-1 1. ELECTROSTATICS- It deals with the study of forces , fields and potentials arising from static charges. 2. ELECTRIC CHARGE – Electric charge is the property associated with matter due to which it produce and experience electric and magnetic effect. There is two types of charges – 1. Positive charge – Lack of electron in a matter is called positive charge. 2. Negative charge – Excess of electron in a matter is called negative charge. # like charges repels and unlike charges attracts each other. (Fundamental law of electrostatics ) # SI unit of charge =coulomb (C) 3. TYPES OF MATTER - According to flow of charges , there is three types of matters. 1. Conductor – The substances through which electric charges can flow easily are called conductor. like iron , aluminium , copper , silver etc. 2. Insulator – The substance through which electric charges can not flows , called insulators. like – plastic , wood etc. 3. Semiconductor – The substances which are behave as insulator at low temperature and behave as a conductor at high temperature. called semiconductor. like germanium , silicon etc. #when some charges is given to surface of a conductor then due to property of conduction that charges distributed uniformly at the surface of that conductor. But if we give some charges to insulators then due to its property the charges does not distributed to its surface. it remain at the same point where we give that charges. 4. TYPES OF CHARGING - 1. BY RUBBING METHOD – When we rub two substances then due to friction heat is produced. then one of substance absorb that heat and emit electron , and second one absorb that electron. The one which emit electron due to lack of electron it become positive charged substance and second one due to excess of electron become negative charged substance. Following table represents negative and positive charge body when it is rubbed - Positive charge Negative charge Glass rod Silk cloth Flannel or cat skin Ebonite rod Woollen cloth Amber rod Woollen coat Plastic seat 2. BY CONTACT METHOD – When two identical shape and size conductor touches each other. where one of conductor is charged and second one is neutral. then by touching charge will transfer form charged conductor to neutral conductor. and by this both will have same charge. 3. BY INDUCTION METHOD – When charged body brings nearer to a conductor. then due to attraction of unlike charges , unlike charges accumulated at closer surface of conductor and due to repulsion like charges accumulated at rear surface. this temporary electrification of a conductor is called electrostatic induction. 4. BASIC PROPERTIES OF ELECTRIC CHARGES – 1. ADDITIVITY OF ELECTRIC CHARGES –  According to this property if a system contain n charges Q1, Q2 ,Q3 ,Q4 ,Q5 ……..then total charges of system is Q= Q1 +Q2 +Q3 +Q4 +Q5 …… # Charge has magnitude but no direction. # Proper sign have to be used while adding the charges in a system 2. CHARGES ARE CONSERVED – According to this property no new charges either created or destroyed. Total magnitude of charges in initial and final condition is equal. 3.QUANTISATION OF CHARGES – It is established that all free charges are integral multiple of a basic unit of charge denoted by e , thus charge Q on a body always given by Q=ne n=integer number e=1.6 x 10-19 C This fact is called quantisation of charges. 5. COULOMB’S LAW – According to coulomb , if two charges q1 and q2 are kept at r distance. then forces between these two charges is – 1. Directly proportional to product of two charges. it means F ∝ q1 x q2 2. reciprocal to square of distance between these charges. it means 1 F ∝ 𝑟2 By combining above – q1 x q2 F ∝ 𝑟2 where k= Coulomb constant. k x q1 x q2 𝑁𝑚2 F= k= 9 x 109 𝑟2 𝐶2 1 Case (1) when both charges situated in vacuum , then k = 4𝜋Ԑₒ , Ԑₒ = permittivity of free space 𝐶2 = 8.85 x 10-12 𝑁𝑚2 1 Case (2) when both charges situated in medium ,then k = 4𝜋Ԑ , Ԑ = permittivity of medium. Relation between Ԑₒ and Ԑ Ԑ = Ԑₒ x Ԑr , Ԑr = relative permittivity Some values of Ԑr 1. vacuum =1 2. air = 1.00054 3. water = 80 4. metal = infinite 6. VECTOR NOTION OF COULOMB LAW – Let two charges q1 and q2 are located at r1 and r2 position vector with reference to point O. then joining vector along direction from r1 to r2 is r12 = r1 – r2 = -r21 𝑘𝑞1 𝑞2 F12 = 𝑟 2 𝒓̂𝟏𝟐 , 𝒓̂ 𝟏𝟐 = unit vector along r12 vector 𝑘𝑞1 𝑞2 F21 = 𝒓̂ 𝟐𝟏 , 𝒓̂ 𝟐𝟏 = unit vector along r21 vector 𝑟2  Now according to vector , 𝒓̂ 𝟏𝟐 = -𝒓̂ 𝟐𝟏 It means F12 = - F21 Above equation represents that coulomb rule follow newton third law of motion. # when sign of force is positive , then it represents repulsion. when sign of force is negative , then it represents attraction. 7. FORCES BETWEEN MULTIPLE CHARGES – Consider a system of n charges , where q1 ,q2 ,q3 ,q4 ,q5 ………qn charges located at r1, r2 ,r3 ,r4 ,r5……rn position vector. let us consider that an another charge qo is located at ro. then total force on qo charge is equal to vector addition of all forces , exerted by all individual charges.This is termed as principle of superposition. 𝑘𝑞0 𝑞1 1)force due to q1 charge on qo charge = F01 = 2 𝒓̂ 𝟎𝟏 𝑟01 𝑘𝑞0 𝑞2 2)force due to q2 charge on qo charge = F02 = 2 𝒓̂𝟎𝟐 𝑟02 𝑘𝑞0 𝑞3 3) force due to q3 charge on qo charge = F03 = 𝒓̂𝟎𝟑 𝑟2 03.. 𝑘𝑞 𝑞 n) force due to qn charge on qo charge = F0n = 𝑟02 𝑛 𝒓̂ 𝟎𝒏 𝑜𝑛 then total force exerted on the charge q0 is F0 = F01+ F02+ F03+ F04+ F05……………… F0n 𝑘𝑞 𝑞 Fo = 𝑟12 2 𝒓̂ 𝟎𝟏 + 01 𝑘𝑞1 𝑞2 𝑘𝑞1 𝑞2 𝑘𝑞1 𝑞2 2 𝑟02 𝒓̂ 𝟎𝟐 + 2 𝑟03 𝟎𝟑 …………. 𝒓̂ 2 𝑟𝑜𝑛 𝒓̂ 𝟎𝒏 𝑘𝑞𝟎 𝑞𝑖 Fo = ∑𝒏𝒊=𝟏 2 𝒓̂𝒐𝒊 𝑟𝑜𝑖 The vector sum is obtained as usual by the parallelogram law of addition of vector. # according to above derivation , it is clear that force between two charges does not affected by the presence of any other charges. 8. ELECTRIC FIELD – A region around the charged particle where any other charged particle experience electric force is called electric field.It is a vector quantity and denoted by E. 𝑭 E=𝑞 , unit = N / C Dimension = [ M1 L1 T-3 A-1 ] According to formula , if a test charge q0 is located at r distance from q1 charge then ratio of force applied on that test charge and magnitude of rest charge is called electric field intensity. 𝑘𝑞0 𝑞1 𝑟2 𝑘𝑞1 E= = 𝑞0 𝑟2 It means electric field depends on two factors (1) magnitude of charge (2) distance of point from the charge. For the positive charge , direction of electric field is radially outward and for negative charge , direction of electric field is radially inward.  𝑭 𝑘𝑞 Vector notification of electric field E = 𝑞 = 𝒓̂ 𝑟2 # A charge having unit magnitude and no electric field around it , is called test charge. 9.ELECTRIC FIELD DUE TO SYSTEM OF CHARGES – Consider a system of n charges , where q1 ,q2 ,q3 ,q4 ,q5 ………qn charges located at r1, r2 ,r3 ,r4 ,r5……rn position vector. let us consider that a point located at ro position vector. then total electric field on p point is equal to vector addition of all electric field , produced due to all individual charges.This is termed as principle of superposition of electr 𝑘𝑞 Electric field due to q1 charge is E1 = 21 𝒓̂𝟏 𝑟1 𝑘𝑞2 Electric field due to q2 charge is E2 = 𝒓̂𝟐 𝑟22 𝑘𝑞3 Electric field due to q3 charge is E3 = 𝒓̂𝟑 𝑟32.. 𝑘𝑞 Electric field due to qn charge is En = 𝑟 2𝑛 𝒓̂𝒏 𝑛 Then total electric field is equal to vector sum of all electric field = E = E1 + E2 + E3 + …….En 𝑘𝑞 𝑘𝑞 𝑘𝑞 𝑘𝑞 E = 𝑟 21 𝒓̂𝟏 + 𝑟 22 𝒓̂𝟐 + 𝑟 23 𝒓̂𝟑 … … …. 𝑟 2𝑛 𝒓̂𝒏 1 2 3 𝑛 𝑘𝑞𝒊 E= ∑𝒏𝒊=𝟏 𝒓̂𝒊 𝒓𝟐𝒊 E is a vector quantity that varies from one point to another point in space and is determined from the positions of the source charges. 10. ELECTRIC FIELD LINES – electric field is represented by some imaginary and continues lines. These lines is called electric field lines. Properties of electric field lines is as below – 1. These lines are imaginary and continues lines without any break. 2. Electric field lines are start from the positive charge and end at negative charge.  3. If we draw a tangent at any point of electric field line it represent the direction of electric field at that point. 4. The point where density of electric field lines is higher than at that point magnitude of electric field is more and vice versa. 5. For a point charge electric field lines is radially move outside or inside acc. to nature of charge.. 6. If we represent the field lines between the unlike charge then it have a nature to being compressed. It represent attraction. 7. If we represent the field lines between the like charges then It have a nature to being expansion. It represent repulsion. 8. Two electric field lines cannot intersect each other. Because at intersecting point there is two direction of electric field that can’t be possible. Means at a point acc. to law of superposition only when direction of electric field can be possible. 9. Electric field lines does not form close loop. 11. ELECTRIC DIPOLE – An electric dipole is a pair of equal and opposite charges +q and –q separated by a distance 2a. The midpoint of location of –q and q is called the centre of the dipole. The total charge of electric dipole is zero. (A) ELECTRIC DIPOLE MOMENT – Product of electric charge and distance between two charges is called electric dipole moment. it is denoted by P. it is a vector quantity. and the direction of electric dipole is from -q to +q. P = q x 2a (B) ELECTRIC FIELD AT AXIS DUE TO ELECTRIC DIPOLE –  Consider , two charges +q and -q is located at 2a distance. and a point P is located at r distance from centre of dipole and axis of dipole. then 𝑘𝑞 Electric field on P point due to +q charge is E1 = (𝑟+𝑎)2 (towards O to P) 𝑘𝑞 Electric field on P point due to -q charge is E2 = (𝑟−𝑎)2 (towards P to O) Then total electric field is = E = E2 -E1 𝑘𝑞 𝑘𝑞 E = (𝑟−𝑎)2 - (𝑟+𝑎)2 1 1 E= kq[(𝑟−𝑎)2 − (𝑟+𝑎)2 ] (𝑟+𝑎)2 −(𝑟−𝑎)2 E= kq[ (𝑟−𝑎)2 (𝑟+𝑎)2 ] 𝑟 2 +𝑎2 +2𝑎𝑟−𝑟 2 −𝑎2 +2𝑎𝑟 E= kq[ (𝑟 2 −𝑎2 )2 ] 4𝑎𝑟 E= kq[(𝑟 2 −𝑎2 )2 ] If r >>>>> a then 4𝑎𝑟 E = kq [ 𝑟 4 ] 𝑘𝑞. 4𝑎 E= 𝑟3 2𝑘𝑃 E = 𝑟3 [ where P = 2aq ] In vector significance – 2𝑘𝑷 E = 𝑟3 (C) ELECTRIC FIELD AT EQUATORIAL DUE TO ELECTRIC DIPOLE – Consider two charges +q and -q is situated at 2a distance , and a point C is located at r distance from centre point of dipole on distance equatorial axis 𝑘𝑞 Electric field on Point C due to +q charge E1= 𝑎2 +𝑟 2 (in direction of BC) 𝑘𝑞 Electric field on Point C due to -q charge E2= 𝑎2 +𝑟 2 (in direction of CA) Now these electric fields are not linear. so for solving this ,we have to use vector method. if we find component then we have four component E1sinθ , E2sinθ ,E1cosθ ,E2cosθ. Now E1 =E2. so according to diagram component of sinθ cancel out each other and component of cosθ add each other. So resultant electric field component= E = 2E1cosθ 𝑘𝑞 E = 2 x 𝑎2 +𝑟 2 x cosθ 𝑘𝑞 𝑎 E = 2 x 𝑎2 +𝑟 2 x √𝑎2 +𝑟 2 𝑘𝑞𝑎 E= 2x 3 (𝑎2 +𝑟 2 )2  𝑘𝑝 E = 3 [2qa = P] (𝑎2 +𝑟 2 )2 If a