Class 12 Physics Derivations PDF

 CHAPTER # L : Electric Charges and Fields # Coulomb 's law of Electrostatics :( In vector form) consider two distance o ! q...

 CHAPTER # L : Electric Charges and Fields # Coulomb 's law of Electrostatics :( In vector form) consider two distance o ! q and q separated by ' where , charges + - F)2 force exerted on q by 92 ⑨-Ñ = , Fa force exerted on qz by 9s ma = F. ⊕ ←. , I FI 1> l , we can ignore 12 in the E- =2¥ E- 21¥ µenie%¥ " " A field due to dipole equitorial line : ' # Electric on , " , E- q=¥→ 1-+9=4%2 ' , i. ,. , , % C-net = J+E+iE -q e-e- it e- e→ =¥÷i¥÷;iÑ " +nisa , Hypo TH ( __ d ° : pytha) ¥+2 ② so cos ② ÷÷+÷÷:i = - , = , = JY¥ f+ = 11,2%+97,12121050=(71+10520--2105-0) (2%+1-4152) = cos ⊖ putting lose here from ② , Enet -1%4%312 = T.fr > > d , 12 can be neglected in the denominator.  - here - denotes the direction which resign is anti-parallel -kg¥ Hence, f- direction of dipole / veto +ve) = PIM to - Hence # Torque on dipole in external field : _ _*÷. Figure shows an electric dipole with charges + q& q - at a separation of 2L placed in a uniform electric field CET. Dipole makes an angle ⊖ with electric field. FT = - qf → force on charge - q F) = qf → force on charge q FT = - Fi which means the force direction at the two ends acting on dipole is equal will behave in like magnitude and opposite in Therefore. it a couple. As couple Ps on acting dipole it produces torque , so. either force ) we know , I = ( magnitude of ( ✗ to line distance of action from of F) = F ✗ ( BC) = qE ✗ (21 sin ⊖ ) I = PE Sino ( % P=qke )) E=FxÉ Pᵈ Hence Caste : when ② = 0° % sin 0=0 ; which means K=. this condition is called stable equilibrium because when the dipole is displaced from this orientation ,Ptam back to same configuration. fassett : when ⊖ -480°. : sin 1800--0 ; which 12=07 means condition is called unstable cause once displaced this equilibrium the dipole never comes backlothientaninsted it aligns itself parallel to the field _É¥teaur. when -0=9-8 af e✗Pᵗᵈ ☒ " CEIL : i. sin 90=1 ; which means T.is maximum. unstable G- I2&¥ Ñ F- PEsÉᵈ 1$47 paper Miguel II=Pcm→ * teacher d- An FEI 34TH / ot ¥%  Gauss verification Coulomb's law : # law using late know, the net electric field closed surface B. D) Ps enclosed by the through a ¥ times the net charge surface. Hosed = %÷= § F. dA→ Verification : According to electric flux , ☒ § Ed? § Edsco so c. = , = , intensity of electric field IÉI distance from ;gᵗ we know at same , charge q will remain constant , also for spherical surface 0=00 To electric flux : - ∅e = C- Ads cos 0° ∅ c- = c- fgds ( As § ds means area = 4*82) To ∅e = f 4-11-82 - ② Coulomb's law 9 Now , according to C- = {◦ % ( putting in ②, we get :- c- "* ∅e=¥¥× A- = ¥9 ◦ og Ole = i. ✗ ( enclosed charge) Have Pwᵈ # Electric field due to a straight long charged 08 conductor Electric field due to a line of density'd! straight uniformly charged infinite* ' ' Consider part of length l uniform ↑ a on this conductor. c- 1- So Gaussian surface will be cylindrical in 1 A , this case. d I - # Let As direction da be the small areas conductor is positively of E- field will on charged be the this " , 't"" | ; ! radially , outwards. 1 I × da Now , ∅ / c- DA , = cos ⊖ G- 90T = , 1 i 1 02 =/ EDA [⊖ 90° ) - o cos ⊖ [ = Here A- linear charge den ∅ =/. ( 0=0] , C- DA cos ⊖ } 12=9/1 Hence Netflix , ÷ One + = ∅ -102+0} , = ffda cos 90 + ffdA cos 90° -1 /EDA cost = 0 + 0 + SEDA ∅ = C- A i. [ Total curved area of surface =2ñrl ] Now Ace to Gauss. Law : , ∅=9÷ from⑦ , C- (211-8/1) = 1¥ [: gin Al] __ E- 2¥74 P¥ʳᵈ = Hence Here , we can clearly see , C- ✗ ± ^ So Graphically Eats ÷ , c- > # Electric Field due to infinite phage she !et of density ' o ' : - i ¥ + Ida T→da - ¥ - - ± - d←§ - - - - - - + ' + +1 infinitely long charged with uniform plane sheet surface charge I 1 density → ( o). I _ Gaussian cylinder of of radius ' ' Draw a area 8. Take 3 sample small ' surfaces 'd A at ① ② , & ③. Total flux , ∅net = ∅ -10/2+03 , = ffd A cos 0° + / C- DA cos 0° / C- DA + cos 90° = / C- DA + / C- DA 0 + = C- A + C- A l∅A→ ② Acc to gauss law, ∅ =9g÷ 0¥ from ② d④ oq.t /C---oy-- Pw¥ᵈ ZEA = independent of , _ ↳  CHAPTER # 2 : Electric Potential and capacitance # Potential at a point due to point charge : +a ᵗ→p* -- charge ' Let there be P at distance ' from +9 a point a r. Electric potential means work done to bring a unit + recharge from infinite to the point P. N % Wip → a) = / Fext dr " = § kQ¥ D dr cos ⊖ [ I 0=180 ] ' dr §¥ = - KQ cos 180=-1 * 1¥15 - = KQ f-± t 's ) ] - = - KQ 1- ± I] = - + Wtp → a) ¥= 4¥ ¥ ,µueP¥ᵈ ✓ = ooo. # Potential due to dipole : (a) At a point on axial line : -9 + q ④-10 p - - - - - - - - - - a → Consider dipole with separated by distance of '2A ! a charges +94 q a - point the axial line at distance from centre of dipole ' A P dies on a ' r. Potential at P due +1 , ✓+ So to ¥ᵈa = , due to q , V -_a - = - % Net potential at Pg Vaxial = µ) + ( V ). =4¥aHH¥a )  = k9G+aj¥aY* = 12,91% Hence Vaxial ¥a2 = , for short dipole ( r >> a) 2s Vania, = KI 82 proved # p Her iii. Arta ar y ' (b) At a point on Equatorial line : , ' i. Let there be ' P' at distance 's ' equatorial line ' point ' a a on. ' " " ' ' ' ' - q ↑ + q so , as in diagram : a-a 1- a- a- Potential at P due to +9 4- !÷g = , , due to -9 , V. = ,ka¥ᵈ→ So , Net potential at P, Veg = (4) ( K) + =÷→+t÷ lVeg=07_µenoePnᵈ Hence , electric potential due to dipole at any point on eq dine will be. 0. (c) At any arbitrary point : " " ^ let A be any distance 's ' from ! arbitrary point at a o , centre of dipole making angle ⊖ with dipole axis Pcos ⊖ ' an. Observe the resolve figure carefully : If we ④ ☒ dipole moment(F) P into two rectangular components - q + q as shown. e- a → Éga → Then axial line of dipole with ' point A lies dipole moment pcos -0 " on , potential k(pcg at due to this so , a component = ' ' and point A lies on equatorialline of dipole with dipole moment sin-0 p but as discussed component above as A is , on eq line. i. potential due to this will be zero.  Hence , Vnet =kpc + O kp µeneePIeᵈ V = # Relation between Electric field and Potential : Consider two equipotential surface A and B separated by _y a distance of ' da let the potential ' , of surface B be VB=V and of A be Va = Vtdv Now , Work done to displace unit positive charge from Bto A : dw = Fdr cos 180° dw = - Fdn ( =Ee→j ) As c- i. C- = F so , dw = - Édse - ② Also , we know dW= q( Va - VB ) dw-tt.tl/Vtdv-HdW--dv - ④ from ② & - Édx = dv E- date ¥×µuP¥ᵈ - = c- = - # Potential Energy of system of two point charge ( in absence of C- f.) A- Initially there were no charge. at A and B. 9, 92 Firstly we'll , bring 9 from , to A. So work , done to place charge 9, to A g WA = % VA = 0 [% Va = 0 , ie - of static ] potential energy charge Now , we'll bring 92 from to B (and in this case q, is already at A) So , potential at B due to % at A , VB = k¥ - ② % Work done to place 92 at B, WB Ez VB ( from②) = = 94k¥ ) WB = k9jI  And as we know , sum of work done is equal to the potential energy of system Potential Energy A) k9i9÷ % = 0 1- U k9÷µwePYᵈ = # Potential Energy of a system of two charges in an external electric field : - Let A and B % potential at be Va and Vis respectively. , Now , , work done to place q , at A , Wa = 9, Va - ② ( % inHally 92 was not there ) % Work done to 92 at B. place ◦. WB = 92 VB t 1 ⊖ of dB→sPn - - - - - - - - t.rs#si-no----- " I sdB→ -ÉÉÉ I distort Iq - _ _ , Consider a circular loop of radius ' a' the axis of the circular loop at which we have to calculate the magnetic field due to the circular ' loop and ✗ is the distance between the loop and the point P'. According to Biot savants law , dB=¥¥Iᵈ¥irE  So the , magnetic field at P due to current element Idf : -1dL ] dB=%ˢ Idl{In9I [: ° r dB ME ,É+→ = ) Idt Magnetic field at P due to current element dB ' 14¥ Idlgi29I = dB ' ME {aᵈ÷×z ) = dB =D B ' we can see → Here , Resolving dB in two components we find that cos ⊖ component for , two diametrically opposite elements cancel each other. So that , magnetic field intensity at P will be only due to sin 0 component therefore , total magnetic field due to the whole coil. B→ = 5dB sin ⊖ ☒ = 11¥ IdlaÉ+n- B- µ%a?→fᵈl = B- 47%4×5*+7 fall = ⑤ #a ¥Y%¥×→a≠× = 's B- = %a?pk Hence %¥ it n> >> a , then a Ps neg legible ⑤ Mz¥ñ) = 31k if Mz÷s = # Ampere 's circuital Law : It states that the line integral of magnetic field intensity over a closed loop is µo times the total current threading the loop. ↳ i. e. § B-.de?--1UoI of Proof : Consider a straight conductor carrying as shown in the "ÉqÑ↑B - " figure Consider a circular American loop of radius r ! -. around the conductor ' - - - - - - -. As B- and DÑ are in same direction so angle between the miso. ^ % / B- diBdl = / cos 0° = fBdl = Bfdl =M¥j# ( : fdl means circumference =2#r) /Bdl = Not Pw¥ᵈ peen MF due to long straight current [ solenoid. conductor carrying Application of Ampere's circuital law → [ Toroid # Magnetic field due to an infinitely long straight current carrying conductor : along of a cross-sectional radius a' carrying steady ' we've given straight wire current I. This current is uniformly distributed across this cross-section. have to calculate field at a distance from Now But , we here we 'll have 3 cases : magnetic - o centre. ④ r > a ; i. e. point lies outside wire E) r a ; ie point lies on the wire =. Pl r< a ; ie point lies inside the wire. CA: - 8 >a at point Pi. Now , to find the magnetic field at point P,. - - - outside the wire make a circular loop - made of radius o as shown in figure ' '.  Using Ampere's law , ftp.di = µ I. § Bdl cos 0° = Not B Jodl = No I B (2*8) = MOI ( : Jodl ° means circumference -211-8) ② B=Y¥-g ( where is the distance of point from] - r centre Bttg ( for o> a) Cased : - r=a at point Pz Now to find , the magnetic field intensity at point B. on the surface of the wire. Make a circular loop of radius a) G-. % similarly like ②st we'll get → B- MI - 21T A cased : - Ra , at point Pz § To find the magnetic field intensity at point B inside the surface of the - - loop - - - circular cylindrical wire radius → make a made of 8 (rea) Now in this , case the enclosed current Ie is not I but less than the value. Since the current distribution is uniform the current enclosed , is , 2 Ie = I✗ at using Ampere's law , § B. DI Mo Ie = § Bdl =M°Ia¥ B § de = Mo Ia¥ B (2*4) = MoIa¥ 1321T Nigg B= µz¥% = Bar ↳  Field due to solenoid : # Magnetic @£%①①☆ Mmm MM " ⊕⊕É⊕⊕⊕⊕ק ← " 2 g-I → < No. Of turns = N < < D C ⑥ ③ ③ ① ③ ⑨ ③ ①③ ① : -13 A I ④ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ?⃝ let a solenoid consists of of turns per unit length and carry current ' n' no. I. Magnetic field inside the solenoid is uniform and strong. M F. outside the solenoid is weak Ialmost zero). Consider a close loop ABCD. % § Bode fB→ de→ JEDI fB?dT + + + fB?dT = AB BC CD DA Here , B. all = 0 [B outside -0] B- dt-afB.de?0fiB.tdeT Hence § # DI § B- di +0+0+0 , = - AB § B.DI / Edt = AB cos 0° § B- di = B- Jdt AB § Bill = BTL ) - ② According to Ampere's law : § B. all = No 7- Here , N number of turns , are present To § B. dl MONI = BCL ) MON I = - from ② 1401¥ B=  ⑦ = Mon I where ,n= no. of turns per unit length i.e. , n= NI I 4¥ " B=M B= Mon i i. > corner centre law , obtained # Using Ampere 's circuital the magnetic field inside toroid a Outside / Between) : - Case) Inside = - - _ , , ??; , from Ampere 's law : - '. §BdÑ= Motion [at Pi ) , 1 , t here Iin 0 : = ' ' , ' § B- di ' = 0 13--0 , " ' -4 if _ * B Casey ) - Between the turns : - _ - from Ampere 's law : - § Edt = No 1in fat B) § Bdl cos ② = Mo Lin B fall = MONI B (211-8) = MONI B- - M◦¥÷ or B- Mon I - [ ñn=N_ = Egg ] Case Outside :-( at B) § Bdl = Nitin 13--0 # Force acting on a current carrying conductor placed in Mf : - a B- Consider carrying a conductor of current I placed lengthinland a of section A area magnetic field at an ( I ¥↑ l - ⊖ shown It of electrons in the density angle number ← as. conductor Ps n then , total no. of electrons in the conductor Ps : Aln.  As the force acting one electron is f-=eVdBsPn⊖ where Vd is the drift velocity of electrons. So the total force acting on the