Electrostatics Target 70 Marks PDF

Electrostatic 1 TETRAHEDRON GROUP OF INSTITUTIONS Tangi, Cuttack-754022 TARGET : 70 MARKS ELECTROSTATIC...

Electrostatic 1 TETRAHEDRON GROUP OF INSTITUTIONS Tangi, Cuttack-754022 TARGET : 70 MARKS ELECTROSTATIC ELECTROSTATICS 2ND Year Properties of Charge : ELECTRIC CHARGES AND FIELDS i. Quantisation of charge : Charge: The property of charge by virtue of which the The intrinsic property of matter which total charge of body is integral multiple of produces the electrostatic force is called basic unit of charge. “charge”. Q =  ne Cause of Charging: n = 1, 2, 3, 4 ---- The cause of Charging is due to transfer of Q = Total charge electrons. e = basic unit of charge or charge of Q. Is the mass of the body is affected by charg- electron ing ? ii. Conservation of charge : Ans: Yes, the mass of a body is affected by charg- The total charge in an isolated system is al- ing as e(-) have certain mass. ways constant. Some important facts associated with charge : Charges neither created nor destroyed but they get transfered from one body to another. A body said to be neutral as it contains same Ex : 1) Decay of 92 U 236 no. of proton & electron. 92 U 236 90 Th 232  2 He 4 A body possesses +ve charge if it loses Before Decay no. of charge particle = 92 electrons. So the mass of the positively After decay no. of charge particle = 92 charged body is less than the nutral body. So, charge is conserved A body posseses -ve charge if it gains Ex : 2) fission of 92 U 235 electrons. So the mass of the negatively 92 U 235  0 n1 56 Ba142  36 kr 92  30 n1 charged body is more than that of nutral Before fission no. of charge particle = 92 body. After fission no. of charge particle = 92 The basic unit of charge is the charge of an So, charge in conserved. electron or proton i.e., iii. Aditive in nature : 1.6 × 10-19C. The total charge of a system is the algebric sum of individual charges present inside the Mass of electron: me = 9.1 x 10-31 kg system. Mass of proton: mp = 1.67 x 10-27 kg Ex : Consider a system contain the charges q1, q2, Some important facts associated with induction- q3, q4 Inducing body neither gains nor Total charge = Q = q1 + q2 + q3 + q4 loses charge charge is a scalar quantity The nature of induced charge is iv. Invariance of charge : always opposite to that of inducing charge The charge of a body doesn’t depend upon Induction takes place only in bodies the frame of reference, so it is invariant. (either conducting or non conducting) and not in particles. TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK Problems based on quantisation: Exercise : Example-1 1. A glass rod is rubbed with a silk cloth. How many electrons are remove from a T he gl a s s r od a cqui res a ch arge of body so that it acquires a charge of 8 +19.2×10 –19 C. micro coulomb. Solution : i) Find the number of electrons lost by glass Q = ne rod. Q 8  10 6 8  10 6 ii) Find the negative charge acquired by silk. n   19  5 10 13 e e 1.6  10 iii) Is there transfer of mass from glass to silk? Example-2 Ans. i) 12, ii)-19.2×10 –19C, iii) Yes Which is bigger, charge of an electron or 1 2. If a body gives out 10 electrons every 9 coulomb of charge? second, how much time it requires to get a Solution : toal charge of 1C from it ? 1 coulomb of charge is bigger than charge of 3. How many electrons must be removed an electron. from a piece of metal to give it a positive Example-3 charge of 1  C ? How many electron form 1C of charge ? Coulombs law: Solution : The force of attraction or repulsion Q = ne between two point charges at rest is directly Q n  1 6.25 1018 electrons proportional to the product of magnitude of e charges and inversly proportional to square Example-4 of distance between them. A polythin is robbed with wool so that it contains – ve charge of 3 x 10-7 C Estimate the no. of e(-) transfer from which substance to which substance. Ans: Q = ne Q 3  10  7 Consider two charges q1 & q2 are separeted n  by distance r. e e According to coulmbs, law the force be-  3  10 7  6.25  1018 tween two charges  18.75  1011 1 F  q1q2 , F  n  1.875  10 12 r2 1.875 × 1012 electrons are transfered from q1q2 F wool to polythine r2 Example-5 q1q2 Is a charge of 4.5×10–19C possible ? F k r2 Ans : Here , q = 4.5×10 –19 C ; Where k is the proportionality constant called e = 1.6×10 –19 C electrostatic force constant or coulmb con- stant. q 4.5 10 19 Now q= ne n    2.8 The value of k depends on the system e 1.6  10 19 of unit and the nature of medium. Since n is not an integer, this value of charge is not possible.  Electrostatic 3 Value of K If air or vacuum is replaced by any other (i) In SI medium then the coulomb force between In free space or in air or vacuum two charges decreaes. The value of 1 permitivity is minimum in air. K  9  10 9 Nm 2 C 2 4 0 Coulomb force between two charges is an Where 0 is the permittivity in free space action reaction pair. or vacuum. Coulombs force is a central force because it 0  8.85  10 12 C 2 N 1 m 2 act along the line joining the centre of two For other medium charges. Coulomb force between the two charge is in- 1 K valid if the distance between two charges be- 4  Where  is the permittivity in that medium. comes less than1 fermi (10-15m) (ii) IN C. G. S: It is also called inverse square law. Permittivity: In CGS unit of K free space is ‘1’. Permitivity is the property of medium that Unit of permitivity: affects the force between the two charges in 1 qq that medium. F. 1 22 Relative permittivity or dielectric constant: 4 0 r Absolute permittivity of the medium qq r  0  1 2 2 Absolute permittivity of free space 4Fr  C.C r or K  0  0 Nm 2 Relative permittivity of a medium is defined  C 2 N 1m 2 as the ratio of absolute permitivity of the Dimension of permitivity: medium to the absolute permitivity of free 1 qq F. 1 22 space. 4 0 r Another Defination of r : qq 0  1 2 2 Force between two charge separated by a 4 F r distance in free space or volume 0  AT AT  1 q1 q 2 M 1L1T 2  L2 Fair  4 0 r 2  0  M 1 L3T 4 A2  Force in medium Some important facts on Coulomb’s Law : 1 q1 q 2 Coulombs law is an experimental law Fm  4  r 2 This law is applicable for point charges at rest. 1 q1 q 2    Fm   r   The force between two charges is medium 4 0 r r 2  0  dependent. Fair F Fm  or air The force between the two charge is inde- r K pendent of the presence of other charges in Fair their neighbourhood. r or K  Fmed TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Relative permittivity or dielectric constant x2 of a medium is defined as the ratio of force  4 dyne. 10 between two charges separated by a certain x2 distance in free space or vacuume to the force  4  10 5 Newton  (2) 10 between same two charges separated by same Equating equestion (1) & (2) distance in that medium. x2 Note:For a conductor the value of dielectrc con- 9  109   10 5 10 4 stant is   x 2  9  1018 The value of dielectric constant in the different  x  3 10 9 medium. 1C  3  109 stat coulomb Med iu m K Coulomb’s Law in vector form: Va ccu m / a ir 1 Wa t er 80 Mica 6 Meta l 8 Consider two similar charges q1 & q2 sepa- Unit of charge rated by distance r. In S.I  Let F12  Force on q1 due to q2 The S.I Unit of charge is coulomb (C).  1 q1 q 2 In C.G.S F12   rˆ21 4 0 r 2 a) Electrostatic unit of charge or statcoulomb: 1 coulomb=3×109 stat coulomb r̂21 is the unit vector pointing from q2 to q1  F21  Force on q2 due to q1 b) Electromagnetic (emu) unit of charge or  1 q1 q 2 abcoulomb: F21   rˆ12 1 4 0 r 2 1C  abcoulomb 10 r̂12 is the unit vector pointing from q1 to q2 Relation between coulomb and statcoulomb: Both r̂21 and r̂12 are oppositely directed. Let 1C= x stat coulomb r̂12 =- r̂21 Consider two charges each of magnetude 1C  1 q1q 2 and separated by distance 1m in free space. F12    rˆ12  4 0 r 2 Force between two charge in SI   F12   F21 1 q1q2 F Coulombs law agrees with Newton’s Third 4π 0 r 2 Law. 1 1 Importance of coulomb’s law F  9  109    From equestion (1) & (2) r12   r21 so. 12    9  109 N  (1) If F12 is action F21 is its reaction which is Force between the charge in C.G.S (esu) accordance with netwons 3rd law of motion. 1 q1q2 F 4π 0 r 2 x.x  1  in cgs k  1 1002  Electrostatic 5 Coulomb’s law in term of position vector     charge q1 is F1  F12  F13  F14 ..... Force due to multiple charges : Consider a group of charges, q1, q2, q3 ------    qn are placed with position vector r1 , r2 ---- rn respectively. According to super position prin- ciple the net force of on q1 is    F  F12  F13 ......  F1n --- (1)  F12 = force on q1 due to q2 Consider 2 charges q1 & q2 are placed at A   1 q1q2 & B with position vector r1 and r2 respec- .rˆ21 4 0 r212 tively.   F12 = Force on q1 due to q2 1 q1q2 r21   4 0 r212 | r21 |  1 q1q 2 ˆ F12 4  r 2 r21 = 1 q1q2   0 21    3 (r1  r2 )  4 0 | r1  r2 | r21 since r̂21 = | r | similarly 21  1 q1q2    F13    2 (r1  r3 )  1 q1q2 r21 1 q1q2 4 0 | r1  r3 |     F12 = r21 4 0 r21 | r21 | 4 o r 3 2 21  1 q1q2   F1n    3 (r1  rn )  1 q1q2   4 0 | r1  rn | F12 = 4  | r  r |3 (r1  r2 ) 0 1 2 eqn (1) becomes  r21 | r1  r2 |  1 q1q 2   F   3 ( r1  r2 ) Similarly, 4 0 | r1  r2 |  1 q1q2     1 q1q3   F21 =   3 (r2  r1 )[ r  r2  r1 ]    3 (r1  r3 ) ..... 4 0 | r2  r1 | 4 0 | r1  r3 | Super position principle : 1 q1qn      3 (r1  rn ) F13 4 0 | r1  rn |  F  q1 n q   q3 F   i  3 ( r1  ri ) q1 4 0 | r1  ri | i  2  F12 Problems based on coulomb’s law r3  r1 Example : q2 The force between 2 charges is F when sepa-  r2 rated by distance ‘d’ what will be the dis- tance if the force becomes 9F. If a system contains no. of charges, the net 1 q1q2 Ans. F  force on a given charge is the vector sum of 4 0 r 2 all the indivisual forces due to other charges of that system. 1 q1q2 Consider a system contain the charges q1, F  (1) 4 0 d 2 q2, q3 ------- qn and so on. The net force on TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 6 Example : 1 q1q2 9F   ( 2) Two identical spheres having charges +8 c 4 0 x 2 & – 4 c are kept certain distance apart. Now Dividing equation 1 and equation 2 they are touched and after that again kept at same distance. Compare the force in both 1 1  2 2  d 2  9x2 cases? d 9x Ans. q1  8C 2 2 d d d q2  4C  x2  x x 9 9 3 K q1q2 Concept F1  Two charged bodies q 1 & q 2 are in r2 contact with each other then the charge of K 8  4  1012 two bodies after sharing F1 .....(1) r2 q1  q2 q q After contact q '1  , q2'  1 2 2 2 84 Example : q1  q 2   2 C 2 Two identical conducting sphere A & B hav- ing charge -q & +3q they are got in contact K q1q2 F2  with each other then separated by a certain r2 distance ‘d’ apart. Find the n a t u r e K 2  2  10 12 of the coulomb force between them. F2 .....(2) r2 Ans. q1  q, q2  3q F1 82 After contact   8 :1 F2 4  q  3q 2q q1    q (Repulsive) Example : 2 2 Two identical spheres P & S with charge Q Example : on each other, exert a force F. A third iden- Two point charges are placed at distance tical uncharge conducting sphere R is suc- ‘r’ in air, exerts a force ‘F’ on each other. At cessively brought woth contact with two spheres. What is the new force of repulsion what distance will these charges experience between P & S ? same force in a medium of dielectric con- Ans. When R is brought in contact with P, charge stant k. ? Q is equally distributed charge on P & 1 q1q 2 Q0 Q R  Ans. Fa  4  r 2  (1) 2 2 0 Now R is kept contact with S, charge on 1 q1q 2 F1   ( 2) Q 4 0 K x 2 Q = 2  3Q 1 q1q 2 1 q1q 2 2 4   4 0 r 2 4 0 K x 2 Initial Force 1 1 Q.Q   F  K.............. (1) r 2 Kx 2 r2 Final force  r 2  Kx 2 Q 3Q r2 r  x  F  K. 2 4............ (2) K K r2 TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 7 Q 3Q  135  10 2  F 2 2 3  13.5N  2  F Q 8 Example : A certain charge Q divided into two parts. 3 How the charges are related in a way that F  F 8 they experience maximum force ? Example : Ans. Let the charges are q and Q-q and distance The force between two charged objects between them is x is get unchanged even though the charge q.Q  q  FK on one of the object is half keeping other x2 same. if d is the original separation then Here force F is the function of charge q what is the new separation. Force will be maximum when Kq1q2 Kq1q2 dF Ans. F    (1) 0 r2 d2 dq Kq1q2 F .....( 2) d Kq Q  q  2x2 0 dq x2 Kq1q2 Kq1q2  d d2 2x 2 q Q  q   0 dq  d 2  2x 2 Q  2q  0 d2 x  2 Q 2 q 2 d Example : x 2 A charge q is placed at the centre of the line Example : joining two equal charge Q. Show that the Charge of 10 c and 50 c are placed 4m system of three charges will be in apart. Calculate the force exerted on the Q equillibrium, if q   charge of 15 c placed midway between 4 them. Ans. The system is equillibrium means net force on each charge must be zero K q1q2 Net force on charge ‘Q’ is Ans : F1  r2 KQq KQ.Q  2 0 K 10 15 1012   r 2 2 r  22 4.KQ.q KQ 2 135  2  10 2 r2 r 4 Q 9 109 15  50 10 2 q F2  4 22 Example : Two charges, each of charge Q, are placed 135  5   10  2 at two opposite corners of a square. Two 4 charges each of q are placed at each of the F  F2  F1 ot h er t w o cor n ers. I f th e r esu l tant electric forc on Q is zero, what will be the 135   10  2 5  1 relation between Q & q ? 4 TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 8 force due to 20c Ans : k 40  1 f2  (0.12  x) 2  F1  F2 K10 k 404   x2 (0.12  x) 2 1 4   x 2 (0.12  x) 2 kQq FB  a2 1 2   kQq x 0.12  x FD  a2  0.12  x  2 x kQq  0.12  3x fc  0.12 2a 2 x  4c.m 3 Fnet  FB 2  FD 2  FC  0 The force will be zero at point x = 4c.m kQq kQ 2 from the charge 10c.  2 2 2 0 Example : a 2a Two charges +20 C & –50 C are placed kQq kQ 2 at a distance 80 cm apart. Calculate the dis-  2   a2 2a 2 tance of the point where a third charge if Q Q placed, the force will be zero. q 2 q (Ans) 2 2 2 Null Point method or equillibrium point : The point where a charge placed so that the net force on it will be zero due to other charges Let a unit charge place at a distance x from the charge 20 c where the net force will be is called ‘null point.’ zero i. For similar or like charges null point lies k q  20  10 6 inside line joining the 2 charges and near Force due to f1  x2 smaller magnitude of charges. k q  50 10 6 ii. For dismilar charges null point lies outside Force due to f 2  (0.8  x 2 ) the line joining the 2 charges & near smaller  f1  f 2 magnitude of charges. Example : k q 2 106 k q 25 106   Two spheres of charge 10C & 40C are placed x2 ( x  0.6) 2 12cm apart. Find the position of point be- tween them, when force on charge 1C is zero. 1 25   10C 40C x 2 ( x  0.8) 2 1 5   x x  0.8 Ans : Let the position is at x m. from charge 10c. Force due to 10c.  x  0.8  5 x k10  0  4 x  0.8 f1   x  0.2m  2cm x2 TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 9 Example : K q1q2 An infinite no. of charges each equal to Ans. F  r2 4 C are placed along x-axis, at x=1m, x=2m, x=4m, x=8m and so on. Find the to- F r2  q1q2  tal force on the charge placed at the origin. K 0.075  9 Solution:  9  109  1 1 1 1  F  K  4  10 6 1  2  2  2  2 .... 25  1 2 4 8   12  75  10 12 10   q1  q2  20c  20  10 6 6  1  F  K  4 10  q1q2  75  1012 1 1    4 q1  q2 2  q1  q2 2  4q1q2     2  20  10 6  4  75  10 12  1   400  10 12  300  10 12  9 109  4 10 6  4 1     100 10 12  4  q1  q2  100  10 12  48 10 N3 Example :  10  10 6  10 C Two charges each of Q unit are placed along q1  q2  20C a line. A third charge q, is placed between them. What will be the relation between q q1  q2  10C and Q that the system is in equilibrium. After solving q1  15C , q2  5C Q Q Q. The charges separated by a distance ‘r’ in a medium of dielectric constan k. What will The 3rd charge placed at the mid point on line be the distance in free space that if the same joining the two charges. two charges are placed so that force in both Since the system is in equiullibrium the net cases are same. force of each charge must be zero  Net force on Q K 0 kQq kQ 2 F  2 0  K 0 (r/ 2 ) 2 r 1 q1q2 kQq  kQ 2 Fair   2  4 0 ra2 r /4 r2 1 q1q2  4q  Q Fmed  4  ra2 Q q 4 Q. Two point charges q1& q2 are 3 m apart & 1 q1q2  their combined charge is 20 C. If one 4 0 K ra2 repeles the other with a force of 0.075 N, 1 q1q2 what are the values of two charges. Fair 4 0 ra2  Fmed 1 q1q2 4 0 K ra2 TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 10 2) For -ve point charge: r2 Direction of electric field for -ve charge is 1 2 ra K towards the charge. Field intensity due to single point charge: 1 1  2  2 r ra K  ra  r K ELECTRIC FIELD Consider a +ve point charge +q placed at point The region around any charge in which its ‘0’ we have to determine electricfield intencity influence can be realised is called electric field. at point ‘p’ at distance ‘r’ from Q. Source charge : Now placed a test charge q0 at p. so electrifield The charge whose electric field is to be deter- intensity at p is mined is called source charge. F Test charge : E The charge by which electric field can be de- q0 termined is called test charge. kqq0 Note:Test charge (q0) is always taken as small & F +ve. r2 k q q0 ELECTRIC FIELD INTENSITY E q0 r 2 Electric field intensity at a point in an elec- kq tric field is defined as the force experience by E r2 a unit +ve test charge at that point 1 q Mathematically it is given by E 4 0 r 2 F E q0 Unit : T h e S. I u n i t o f e l e c t r i c f i e l d i n - t e n s ity is N/C or NC-1 or J/cm Other SI Unit of Electic Field : N V J or or C m Cm Dimension:  Vector form of E : F E  q0 1 Q E rˆ 4 0 r 2 M 1 L1T 2   AT   1 Q r  4 0 r 2 | r |   M 1 L1T 3 A1   1 Q E r Direction of E : 4 0 r 3 1) For +ve point charge: Example : Direction of electric field for +ve charge is Find electric field intensity at a point 2 cm away from the charge. from a charge of 10 C TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 11 kQ k  10  10  10 6 Ans: E  2 Ans: E1  r x2 k  40 10 6 E2  9 109 10 106  12  2    x 2 10  2 2  100  9  10 4  10 4 12  x 4  100 2 x  2.25  108 N C 12 Example :   x  2x  100 Find E at 10-9 m from ‘He’ nucleus. 12 Ans: r  10 9 m   3x 100 kQ E 1 r2 x  0.4m. 25 9 109  2 16 1019 Example :  (109 ) 2 Two charges +16 C & –9 C are separated  18  16  10 19  108  1019 by 8 cm apart. Find the p o s i t i o n w h e r e t h e n e t f i e l d i n t e n s i ty will be zero.  228  10 2  2.28 k  9  10 6 Ans: E1  Example : x2 A charge of 5C experience 1mN at a point k  16  10 6 in an electric field.Find electric field at that E2  point. (8  x) 2 3 3 4 Ans: E  F  F  1 10  0.2  103  200 N / C   x 8 x q0 5  10 6  24  3x  4 x Example :  x  24 cm A charge of  3C is placed at the origin. Find the magnitude & direction of the elec- Example : tric field at the 2cm from the charge on x- Infinite no. of charges each of q are placed axis ? along x-axis at x= 1 x=2, x=4, x=8 & so on. Ans: q  3C a) Find the electric field at x=0 due to this set up this charge ? r  2 c.m b) What will be the electric field in the above kQ setup if the conjucative charges have oppo- F r2 site sign ? 9  109  3  10 6  10 4 kQ  Ans: (a) E  4 r2  6.75  10 7 N / C 1 1 1   9 109 q 2  2  2     Direction along – ve x-axis. 1 2 4  Example : Two charges 10 C & 40 C are placed    1   a  12cm apart. Find the point where electric  9 10 q 9   s  1  r  field intensity is zero? 1 1     4 TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 12 q  12  10 9 q  3 0 kQ (b) F r2 1 1 1 1  kq  2  2  2....  1 2 4 8  q  1 1 1   1    ..... 4 0  4 16 64  iii) Electric lines of force are discontinous & open   iv) Lines of force starts from +ve charge and end q  1     on -ve charge. 4 0  1  1  v) If lines of force are curved then the tangent at    4 any point on it gives the direction of electric q field intensity.  vi) Two lines of force never intersect. 5 0 If the two lines of force intersect each other Electric line of force : then at the point of intersection their will be Electric line of force is an imaginary line or two tangent that means two values of electric curved path along which a unit +ve charge field intensity at a single point which is im- would move if free to do so in an electric field. possible. Note: * Electric line of force gives the direction of the electric field of intensity. vii) Properties of line of force i) The direction of line of force for an isolated +ve point charge is radilly outwards. Lines of force have tendency to contract length wise. This property shows attraction between two dissimilar charges. viii) ii) The lines of force for isolated -ve point charge is radially in wards. Lines of force of force have tendency to expand laterally. This properties shows the repulsion between two similar charges. TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 13 ix) Lines of force are always perpendicular to the 91   10 11 surface of conductor. 16 x) The region where the line of force is crowded  5.6  10 11 Example : the field is strong & where the field lines are A charge particle of mass 2mg is balanced spread out, the field is weaker. in a region of electric field of 4N/C. Find xi) Lines of force can’t pass through a conduc- the magnitude of charge. tor, So inside a conductor the electric field Ans: Weight of charged body  = Electrical force exerted on it intensity ( E ) is zero. Representation of a unifom electric field: mg  qE The field is said to be uniform if its magni- mg 2 10 5  9.8 tude and direction are same at every point. q  E 4  4.9  10 5 C Electric dipole : A system consist of two equal and opposite charges separated by small distance is called “electric dipole” Equillibrium of charge particle in electric field.: Let a charge particle has charge ‘q’ mass ‘m’  and strength ' E '. So in equillibrium weight of the charge body is balanced by the electri- cal force i.e. qE = mg Electric dipole moment : Electric dipole moment is defined as the prod- uct of magnitude of either charge & the sepa- ration distance between them.   Mathematically , P  q  2a Note: Electric dipole moment is a vector quantity Note : For +ve charge & it is directed from -ve charge to +ve charge. For +ve charge direction of field is upward to The S.I unit of electric dipole moment balance it weight. For -ve charge: is coulomb × metre (Cm) For -ve charge direction of field is downwards Ideal electric dipole : to balanced it weight. Dipole is said to be ideal if the separation Example : Find the value of electric field in which distacne tends to zero & each charge tends the weight of an electron is balanced. to infinite. Ans: qE = mg mg 9.11031 10 E  q 1.6 1019 91 TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 14 Electric field intensity due to 1 2q  2a r  dipole : a) At a point on axial line or end on posi-  4 0 r 2  a 2 2  tion. 1 2 Pr E  4 0 r  a 2 2  2 Where, P=q×2a= dipolemoment For short/ideal dipole: r >>a, so the term a2 can be ne- g l e cted 1 2 pr Consider an electric dipole having two E charges +q at ‘A’ & -q at ‘B’ separated by dis-   4 0 r 2 2 tance ‘2a’. 1 2P E Let ‘p’ be the point at distance ‘r’ from the 4 0 r 3 centre of dipole where electric field intensity In vector form is to be determind. 1 2P From figure, E 4 0 r 3 AP = r - a Note : The angle between electric field intensity at BP = r + a axial point and dipole moment is zero. The electric field intensity due to charge +q is B) At a point on equitorial line/ Broad side 1 q on position: E1  along AP 4 0 AP 2 1 q  along AP 4 0 r  a 2 The electric field intensity due to charge -q is 1 q E2  along PB 4 0 BP 2 1 q  along PB 4 0 r  a 2 The resultant intensity at ‘P’ is  E  Consider an electric dipole consisting of two E  E1  E2 1  E2 charges +q at A and -q at B separated by distance 2a. 1 q 1 q   Let P be the point on equitorial line at distance 4 0 r  a  4 0 r  a 2 2 ‘r’ from center of dipole. We have to determine the electric field q  r  a   r  a   2 2    intensity at ‘P’ due to the dipole. 4 0  r  a 2 r  a 2  OP = r q 4ra  AP  BP  r 2  a 2  4 0 r 2  a 2  2 The electric field intensity at P due to +q is, q 2 2 r  a  1 q  E1 .  4 0 r 2  a 2 2  4 0 AP 2 TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 15  1 q 1 P E1 . 2 along PK  E. 3 along Pl 4 0 ( r  a 2 ) 4 0 r The electric field intenscity at P due to -q is In vector form,  1 q E2 . 2 1 P 4 0 BP E . 4 0 r 3  1 q The angle between resultant intensity at E2 . 2 along PM 4 0 ( r  a 2 ) equitorial point and dipole moment is 180o The angle between E1 and E2 is 2 Electric field intensity at any point due to dipole:- So, According to parallelogram law of vector Consider a short dipole consisting of two addition, resultant intensity at P is, charges +q at A and -q at B having dipole mo- ment P E  E12  E22  2 E1 E2 cos 2   E12  E12  2 E1 E2 cos 2 E E2 E1  2 E12 1  cos 2   E1  E 2  2 E12  2 cos 2   2 E1 cos P cos θ 1 q θ E  2. 2. cos  4  0 r  a 2   O P From  OAP P sin θ OA cos   AP Let P be the point at distance ‘r’ from cnter a ‘O’ of dipole.  r 2  a2 The line OP markes an angle  with dipole moment P. 1 q a E  2. 2.  4 0 r  a 2 r  a2 2  The dipole moment P ressolved into its two component i.e. P cos along OP and 1 P  E. along PL P sin  perpendicular to OP. 4 0 r  3 2 a 2 2 The electric field intensity at P due to the di- pole of dipole moment P cos is where, P = q × 2a For short dipole, 1 2 P cos  E1 . 4  0 r3 r  a So the term a2 can neglected The electric field intensity at P due to the di- pole of dipole moment P sin  is, 1 P E. 1 P sin  r  4 0 2 32 E2  4 0. r3 Since E1 and E2 are perendicular, then their resultant is TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 16 E  E12  E22 2 1.5 107  4 E  9 109   1 2 P cos   1 2 P sin   2 4 2  32  2  E  .   .   4 0 r 3   4 0 r 3  9  109  2  6 10 7  P 16  92 E 4 cos 2   sin 2  4 0 r 3 108 10 2 E NC 1 P 49  E. 1  3 cos 2  Torque experienced by electric dipole in uniform 4 0 r 3 Electric field: Direction of E E2 K P sin  / r 3 tan ß   E1 K 2 P cos  / r 3 tan   tan ß  2 Special case i. At a point on axial line,  0 2P Consider an electric dipole AB consisting of E 4 0.r 3 two charges -q and +q separated by distance ii. At a point on equitorial line 2a   90 Let the dipole is placed in an uniform electric P field E in such a way that the dipole mo- E 4 0.r 3 ment makes an angle  ,with the direction of Example: electric field. Two charges  0.2 C and  0.2 C aree Force on charge +q = +qE along the direction placed 10-6cm apart. Calculate the E at an of field. axial point at a distance of 10cm from their Force on charge -q = -qE opposite the direc- mid point. tion of field. Ans:- P  q  2a  0.2  10 6  10 6 10 2 Net force on dipole = +qE - qE = 0  2  10 15 These two forces are equal and opposite and act in different line of action. So, these two 2  2 10 15 E  9 10  9 forces are called couple. 10 10  2 3 This couple exert a torque on the dipole and 36  10 6 rotates the dipole in clockwise direction.  36 10 3 Torque   = magnitude of either 103 force x perpendicular distance.  3.6  10  2 N / C Example:    qE  BC Two charges  25nC are placed 6m apart.    qE  2a sin  Find the E at a point 4m from the dipole on axial line    q  2a E sin  From center of dipole    PE sin  , P  5  10 9  6  1.5  10 7 TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 17 where, P = q × 2a = dipole moment Example: Charge distributed in cube or solid sphere In vector form volume charge density    P E q dq Special case   or v dv i) If   0 ELECTRIC FLUX   PE sin  Area vector   0 Area is a scalar quantity, but in some cases it When the dipole is placed along the direction is considered as a vector. of electric field, Torque acts on dipole is zero. Its direction is along the outward drawn nor- This is the condition of stable equillibrium. mal to the area Electric flux ii) At   90 The electtric flux linked with any surface in    PE an electric field is the measure of total num- When the dipole placed perpendicular to the ber of lines of force passing normally through electric field, torque acts on dipole is maxi- the surface. mum. iii) At   180  O When thedipole placed opposite direction of electric field, it experience zero torque. This is the condition of unstable equillibrium. Continuous charge distribution In this case the electric field is perpendicular 1. Linear charge distribution to the serface. So electric flux through that Charge is distributed uniformly over a line or surface is. wire   ES Ex: Linear charge density: q dq   or l dl q q For ring   l  2 r 2. Surface charge distribution: When charge is distributed over a surface then it is called surface charged distribution. Example: Charge on plane sheet the surface charge In this case electricfield is not perpendicular density, to the surface. q dq But the component E cos  is perpendicular   or s ds to the surface so, electric flux is linked with that surface dq For elementary area. d  E cos .ds ds 3. Volume charge distribution: d  Eds cos  When charge is distributed uniformly over a  volume, then it is called volume charged dis- d  E.ds  tribution.  d   E.ds     E.ds TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 18 for closed surface Example:  Consider a unif or m electr ic f ield.    E.ds  E  3 103 iˆN / C Electric flux is also defined as the surface a) What is the flux of this field through a integral of electric field square of 0.1m on a side whose plane is Case-1 (Maximum flux) parallel to the Y - Z plane. When electric field is perpendicular to the  b) What is the flux through the same square surface then, angle between E and ds is zero. if the normal to its plane makes a 60o angle with the X-axis.    Eds cos 0 o Solution: = ES (max)  a) E  3 103 iˆN / C , Electric field is directed towards X-axis Area S = (0.1)2 = 0.01m2 = 10-2m2 Area vector S = 10-2 S  10 2 iˆm 2  3  10 3  10  2  30 Nm 2C 1 b) Now the area vector makes an angle 60o with X-axis E  3 103 iˆN / C Case-2 Condition of zero flux S = 100cm2 = 10-2 m2 If electric field is parallel to the surface, angle   = 60o between E and ds is 90o so   E.S  3  10 3  10  2  cos 60 o  15 Nm 2C 1 Gauss’s Law  The total electricflux linked with a closed    E.ds 1  Eds cos 90o surface is equal to  times the total charge 0 0 enclosed by the surface. q Note: In free space :   0 For a closed body outward flux is taken as Some important points about Gauss’s law: positive and inward flux taken to be nega- tive Gauss’s theorem is applicable for all types of Electric flux is a scalar quantity closed surface, wheather regular or irregular, Unit of electric flux but surface must be fully closed. However to calculate electric field using The SI unit of electric flux is Nm2C-1   ES Gauss’s theorem only symmetrical surfaces should be considered. N  m 2 To apply Gauss’s theorem a closed surface C should be drawn around the charge body 2 1  Nm C called Gaussian surface. Dimenstion of electric flux If a closed body encloses no net charge, then   ES total flux linked with that closed body will be F zero.  S q The electric flux doesnot depend upon the shape or size of Gaussian surface  M 1 L1T 2 2 AT   L  M 1L3T 3 A1  Gaussian surface may be spherical cylindri- cal in most cases TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 19 Some important concept to determine flux 1. If charge is kept at the centre of cube Total flux over the entire surface of cylider is a) Then total flux through the cube    E.d s   E.d s   E.d s q 1 2 3  0     Eds cos 0   Eds cos 0   Eds cos 90 b) Flux linked through any face of cube 1 2 3 q    2 ES  6 0 According to Gauss’s Law q c) Flux through corner   q 80  0 q d) Flux through edge   q q  12 0  2 ES  E E 0 2 0 s 2 0 Example: A point charge +q is placed L/2 above the E is independent of ‘r’’ centre of a square having side L. Find the Electric field intensity due to infinitely long flux through the surface. straight uninformly charged wire (line charge): Solution: Imagine the charge is placed at the centre of a q cube. So total flux for cube    0 flux per face q  fau  6 0 Square is the one face of cube q  6 0 Electric field due to thin infinite plane sheet of Consider an infinite thin straight wire over charge - which charge is distribut uniformaly. Consider a thin infinite plane sheet of charge Let  bethe linear charge density of the with surface charge density . wire. We have to determine the electric field intensity at point ‘p’ situated at distabnce ‘r’ from the wire. Now a Gaussian surface is drown in the form of acylinder taking ‘r’ as radius & ‘l’ as length. The Gaussioan surface is devided into i) Topsurface ii) Bottom surface iii) Curved surface We have to find out field at the point ‘p’ at a    Ed s   Ed s   Ed s perpendicular distance ‘r’ from the sheet. Now | || ||| Imagine a cylinrical Gaussian surface of cross - sectional area A around ‘p’ and length r, pass-   Ed s cos   Ed s cos   Ed s cos | || ||| ing through the sheet.   Ed s cos 90o   Ed s cos 90o   Ed s cos 0o | || ||| TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 20  q 1C   Eds q  ne or n   electrons ||| e 1.6  10 19 C  ES 4. What is the SI unit of charge ?  E  2 r l ----- (1) Ans:- The SI unit of charge is coulomb (C). According to Gauss law, 5. What do you mean by quantization of q charge ?  0 -----(2) Ans:- The charge on a particle or body can be only From equation (1) and (2) integral multiple of least possible value of q E  2 r l  charge i.e., e 0 Charge on a body, q   ne q When n  1,2,3 or e  1.6  10 19 C E  2   0 rl 6. Can a body have a charge of 0.8  10 19 C ?  q 0.8  10 19 C E  2  0 r Ans:- No. q  ne or n    0.5. e 1.6  10 19 C 2 Since n is not a whole number the body can- E  4  0 r not have acharge of 0.8  10 19 C.  q  7. Can a body have a charge less than     linear charge density   l  1.6 10 19 C ? Q. An infinite line charge produces a field of 9 x Ans:- No. It is because the least possible value of 104 N/C at a distance of 2 cm. Find the linear charge on a body can be  1.6  10 19 C. charge density. 8. What is the cause of quantization of Ans: E      2 0 rE charge ? 2 0 r Ans:- During charging, only integral number of  1 2  9  10 9      2  10  2  9  10 4  10 7 cm 1 electrons are transferred from one body to another, Therefore, charge on a body can be only integral multiple of  1. 6  10   19 C. SHORT ANSWER QUESTIONS 9. A glass rod rubbed with silk acquires a 1. What is electrostatices ? charge of 1.6  10 19 C. What is the Ans:- The branch of physics which deals with the charge on the silk ? phenomena associated with charges at rest is Ans:- The silk will acquire an equal and opposite called electrostatics. charge i.e. 1.6  10 12 C. It is because silk 2. What is the least possible value of charge? gains as many electrons as lost by the glass Ans:- The least possible value of charge is the charge rod.   on an electron i.e.  16  1019 C or proton 10. Given two properties of electric charge. i.e.  16 10 19 C.  Ans:- (i) Electric charge is conserved. 3. How many electrons are present in 1 cou- (ii) Quantization of charge. lomb of charge ? 11. Give two difference between electric Ans:- Let n be the number of electrons in 1 cou- charge and mass. lomb and e be the charge on one electron. TETRAHEDRON GROUP OF INSTITUTIONS : l TANGI l CUTTACK  Electrostatic 21 Ans:- (i) Electric charge may be positive, nega- 21. When are charged bodies approximately tive or zero, Howeever, mass is a positive point charges ? Ans:- Charged bodies approximate to point charge quantity. (ii) Electric charge is always con- if charges if they are small compared to the served. Mass is not conserved because it can distance separating them. be converted into energy ( E  mc 2 where 22. Does Coulomb’s law of of electric force m is mass and c is velocity of light ) and obey Newton’s third law of motion ? vice-versa. Ans:- Yes. It is because the forces exerted by two charges on eachother are equal and oppo- 12. Does the electric charge vary with the site. speed of the charged body ? 23. Is elecric force between two charges a cen- Ans:- No, electric charge does not vary with the tral force ? speed of the charged body. Ans:- The electric force between two charges is a 13. What is the cause of charge on a body ? central i.e. it acts along the line joining the Ans:- The cause of charging is due to transfer of centers of the two charges. 24. If q1 q2  0 , what is the nature of force electrons. between the two charges? 14. How is the mass of a body affected on Ans:- If q1 q2  0 , it means that product of two charging? charges is positive. This is only possible if  Ans:- An electron has a mass of  9  1013 kg.  both charges are positive or negative. There- When electrons are removed from a body, fore, the nature of force between the charges it acquires a positive charge so that the mass will be repulsive. of a body decreases on getting positively 25. If q1 q2  0 , what is the nature of force charged. On the other hand, the mass of a between the two charges? bo