Electrostatics Mind Map PDF

ELECTROSTATIC POTENTIAL...

ELECTROSTATIC POTENTIAL EQUIPOTENTIAL SURFACE ENERGY ELECTROSTATIC POTENTIAL 2 point charges 2kqQ a) Q dmin= VA r r P Vp = 4pe 1 r v mv2 > Q ra 0 V=0 Q VB > E VA=VB>VC > VC Q1 Q2 [ m,q q Q Large distance fixed [ 1 1 Q A B VAB=VB-VA= 4pe rb - ra > like charges - positive (repulsive energy) rb 0 1 Q1 Q2 dmin Equipotential surfaces U= b) 4 0 r Unlike charges - negative (attractive energy) WORK DONE POTENTIAL OF CHARGED CONDUCTING SPHERE System of charges c) Field lines and equipotential VA q VB n(n-1) w = q[VB-VA] 1 2 surface are perpendicular to Usystem = Upair No. of pairs = A B V=KQ 2 q R each other at every location. 500 -500 Q1 r Q2 4 x 3 A B w = 1 x (-500-500)= -1000J E=0 3 No. of pairs = = 6 r 2 2 r=R d) work done in moving a charge r r KQ1 Q2 KQ1 Q4 Total P.E = + + KQ2 Q3 Superposition of potential - Algebraic sum case 1: Vinside =Const case 2: Vsurface = kQ Voutside = kQ on equipotential surface is 0 r 2 case 3: r r2 r2 R r KQ2 Q4 KQ3 Q4 KQ1 Q3 of all potentials = Vsurface= KQ R Q3 r Q4 + + + ZERO POTENTIAL Redistribution of Charge when two Conducting sphere are connected r r r Charge flow from higher WORK DONE IN REARRANGEMENT OF THE SYSTEM Q1 potential to lower potential ELECTRIC FIELD & POTENTIAL a) Like charge-no zero potential point Q2 Final potentials of spheres are equal Q r1 r2 1 Q‛1 1 Q‛2 b) Unlike charge- 2 points of zero potential on E=-dV - V - V - V Q = 4pe0 r1 4pe0 r2 Ex= Ey= Ez= r r 2r 2r 3kQ 2 3kQ 2 line joining (Q1+Q2) r2 dr x y z a b (Q1+Q2) r1 W = Uf- Ui = 2r - r Q1l= Q2l= r1+r2 r1+r2 Q Q = - 3kQ 2 Q1 r Q2 V=- E.dr Q₂ r r Q Q 2r 2r Q 1r Q 1‛ r E1 r r2 Inside point = a= Outside point = b= = 1, = 2, 1 = 3kQ2 3kQ2 Q1+Q2 Q1-Q2 Q2‛ r2 E2 r1 2 r1 Ui = Uf = r 2r ELECTROSTATICS PHYSICS WALLAH DIPOLE ELECTRIC POTENTIAL 2) Umaximum (Unstable equilibrium) Dipole moment P max = pE sinq = pE, (q = 90o) -Q kp Cos q VP= q = 180 P +Q P = q 2l 2l r2 = pE sinq = 0,(q = 0o) min Direction is from -Q to +Q q +q Vaxial= Kp = 0 Vequitorial cosq=-1 -q r2 P=ql 60o P=ql WORK DONE U = +pE 2q TORQUE l l P= 2qlcos30 = 2 3ql( ) o Uniform field W= pE (cosq1-cosq2) Force in Non- Uniform field Q. Fine net dipole moment - Rotational motion only qE -q -q p q - No translatory motion POTENTIAL ENERGY p qE2 ELECTRIC FIELD qE U= -pE cosq q E p qE1 =PxE r Kp = PE Sin q 1) Uminimum(stable equilibrium) Ep= 3cos2q+1 q r 3 q=0o Fx = p dE cos q +Q -Q cosq=1 dx Kp qE1 - Both rotatory and Eaxial= r3 3 +1 = 2Kp r3 Oo with dipole moment p translatory motion U = -pE Enet from -Q to +Q qE2 where, dE= small change in field Eequitorial = Kp 0 +1 = Kp 180o with dipole moment at the two locations r3 r3 Enet from +Q to -Q Non - Uniform field of the charges