Title Class 12 Physics Derivations Shobhit Nirwan.pdf ✅

CHAPTER #L : Electric Charges and Fields # Coulomb's law of Electrostatics:(In vector form) consider two charges + q and - q separated by distance ' o ! where , F)2 = force exerted on q, by 92 Fa, = force exerted on qz by 9s F. ←⊕. 9-Ñ ma 1<-181-I i. FI = kq.iq#r?-rI) ¥ñ =%÷f ⇐ ñ=¥, ) brackets#ET =¥÷㱺:÷÷, i:*. ¥::::: Ñ2Ñt : E. = ,k¥÷㱺'ñ-ñ > Heme PED similarly , Force applied by FI , FI =kq;,9÷(É% =%÷ I = :÷÷, ÷ FI = ,!%;→cñ-ñ> HencePed Here , we can clearly observe that → FI = - FI , i.e- 3rd law of newton is valid in electrostatics also. # we Electric have two Field opposite due to dipole on axis : → charges separated by P 4- - - - - - - - ; a distance of ' 21 ; which makes it a dipole . 7- l→ ←I→ A is a random point on axis at a distance g- ate +721¥ ' ayaan' ' sit ÑN→ r ' from centre of dipole . teacher impress IT 1 Now, Field due to ' -9' at A → E-5- -1T¥, , similarly , Field due to ' +9' at A - 1-+ g. = ,k÷e, ,
So, Net Field 㱺 f- f-qtftq =÷÷ii÷. = -Ñq(8+172 ((rtl) (8-1)/2 = 21%21%-2 㱺 2kPr WI Now, if 8>> l , we can ignore 12 in the denominator, E-=2¥㱺 E- 21¥ μenie%¥ " A # Electric field due to dipole on equitorial line : " ' , , , , , E-q=¥→, 1-+9=4%2 . . ' " i -q it % C-net = J+E+iE e-e-e-e→ =¥÷i¥÷;iÑ "+nisa, d Hypo __ TH ( ° : pytha) so , cos 2 = ¥+2, - 2 = ÷÷+÷÷:i = JY¥㱺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 .
Hence, f-= -kg¥ -here - resign denotes the direction which is anti-parallel to direction of dipole / - veto +ve) Hence PIM # 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 acting on dipole is equal in magnitude and opposite in direction at the two ends . Therefore it will behave like a couple . As couple Ps acting on dipole , so it produces torque . I = ( magnitude of we know, either force ) ✗ ( to distance from line of action of F) = F ✗ (BC) = qE ✗ (21 sin ⊖) I = PE Sino ( % P=qke)) E=FxÉ Hence Pd Caste : when 2 = 0° 㱺 % sin 0=0 ; which means K=. this condition is called stable equilibrium because when the dipole is displaced from this orientation,Ptamback to same configuration. fassett: when ⊖-480° 㱺 . : sin 1800--0 ; which means 12=07 this condition is called unstable equilibriumcause once displaced the dipole never comes backlothientaninsted it aligns itself parallel to the field . _É¥teaur CEIL: when -0=9-8 㱺 i. sin90=1 ; which means T.is maximum. unstable af ☒ e✗Ptd" 1$47 I2&¥ 㱺 G- F- PEsÉd 㱻* paper Ñ㱻 II=Pcm→ d- An teacher Miguel FEI 34TH / ot㱻¥%
# Gauss law verification using Coulomb's law: late know, the net electric field through a closed surface B.D) Ps ¥ times the net charge enclosed by the surface . Hosed = %÷= § F. dA→ Verification : According to electric flux , ☒c. = §, Ed? = §, Edscoso we know, intensity of electric field IÉI at same distance from charge q ;gt will remain constant , also for spherical surface 0=00 To electric flux :- ∅e = C- Ads cos0° ∅c- = c- fgds ( As § ds means area = 4*82) To ∅e = f 4-11-82 -2 Now, according to Coulomb's law 9 C- = {◦ % (putting in 2 , we get:- ∅e=¥¥× c- * " A- = ¥9 ◦ og Ole = i. ✗ ( enclosed charge) HavePwd