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# Coulomb's Law, Electric Potential

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### Coulomb's Law, Electric Potential

1. 1. Name : Roll No. : Topic : Electric Field & Mohammed Potential Asif Ph : 9391326657, 64606657 Coulomb’s Law For IIT-JEECoulomb’s Law If two stationary and point charge Q1 and Q2 are kept at a distance r, then it is found that force at attraction or repulsion between them is Q1Q2 kQ Q F∝ 2 i.e. F = 12 2 r r (k = Proportionality constant) Q1 Q2 In C.G.S. (for air) k = 1, F = Dyne r2 In S.I. (for air) 1 Q1 Q2 ⇒F = . 2 Newton (1 Newton = 105 Dyne) 4π ε0 r ε0 = Absolute permittivity of air of free space C2  Farad  = 8.85 ×10 −12 =  . It’s i N − m2  m  dimensional formula is [M-1 L-3T4A2]Vector form of coulomb’s law : vector from of Coulomb’s law is Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 1 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
2. 2. → Q1Q2 → Q1 Q2 ^ ^ F1 2= K 3 r1 2= K. 2 r1 ,2 w h r1e 2 ris ethe unit vector from first charge to second charge r r along the line joining the two charges.Effect of Medium: When a dielectric medium is completely filled in between charges rearrangement of the charges inside the dielectric medium takes place and the force between the same two charges decreases by a factor of K(dielectric constant) F 1 Q Q i.e. Fmedium = air = . 12 2 K 4π ε K r 0 (Here ε 0 K = ε 0ε r = ε = permittivity of medium) If a dielectric medium (dielectric constant K, thickness) is partially filled between the charges then effective air separation between the charges becomes (r −t +t K ) 1 Q1 Q2 Hence force F = 4π ε K . 0 ( r −t +t K ) 2Principle of superposition: According to the principle of super position, total force acting on a given chargedue to number of charges is the vector sum of the individual forces acting on that charge due toall the charges Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 2 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
3. 3. Net force on Q will be → → → → → Fnet = F1 + F2 +.......... + Fn−1 + Fn The magnitude of the resultant of two electric forces is given by → Fnet = F12 + F22 + 2 F1 F2 cos θ F2 sin θ And tan α = F1 + F2 cos θ For problem solving remember following standard results.Q.1. Two identical charged spheres are suspended by string of equal length. The string make an angle of 300 with each other. When suspended in a liquid of density 800 kg/m3. The angle remains the same. What is the dielectric constant of the liquid? The density of the material of the sphere is 1600kg /m3.Solution: Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 3 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
4. 4. Let T, T’ be the tensions in the string when spheres are in air and in liquid respectively. T sin θ = F T sin θ = F T cos θ = mg B +T cos θ = mg ⇒ F = mg tan θ ⇒ F = ( mg − B ) tan θ F = electrostatic repulsion in ari q2 F= 4π ε a 2 0 In liquid this force is F’ and is given as q2 F= where k is a dielectric constant 4π ε ka 2 0 F ⇒F = form force diagram K F mg = F m − B where B is Buoyant Force g V dg d 1600 ⇒K = = = =2 V dg −Vfg d−f 1600 − 800Q. 2. A ring has charge Q and Radius R. If a charge q is placed at its centre then calculate the increase in tension in the ring?Solution: Consider a small element A. B θ is very small then AB = R (2 θ ). Q Qθ Charge on AB is dQ = 2π R ( 2 Rθ ) = π dQ.q Qq θ 2T sin θ = = 4π ε R 0 2 4π ε R 2 0 Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 4 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
5. 5. Qq θ Qq 2T θ = 8T = 4π 2 ε 0 R 2 8π 2 ε 0 R 2Electric Field The interaction between two charges is visualized in terms of the electric field concept. A charge produces and electric field around itself. This field then exerts force on the other charge. Thus, interaction between two charge is a two step process.Electric field intensity (E): → → → F → F E= E = Lim q0 q0 →0 q 0 Unit and Dimensional formula Newton volt Joule It’s S.I. unit - coulomb = meter = coulomb × meter And C.G.S. unit-Dyne / stat coulomb. Dimension: [E] = [MLT-3A-1]Direction of electric field: → Electric field (intensity) E is a vector quantity. Electric field due to a positive charge is always away from the charge and that due to a negative charge is always towards the charge.Relation between electric force and electric field: → → → In an electric field E a charge (Q) experiences a force F = Q E . If charge is positive then force is directed in the direction of field while if charge is negative force acts on it in the opposite direction of fieldSuper position of electric field: Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 5 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
6. 6. → → → → E = E1 + E 2 + E3 + .......... ... → Electric field due to continuous distribution of charge: E = ∫dEQ. 3. A cube of edge a meters carries a point charge q at each other. Calculate the resultant force on any one of the charges.Solution: Let us take one corner of cube as origin O(0, 0, 0) and the opposite corner as P(a, a, a) we will Calculate the electric field at P due to the other seven charges at corners. Let us take one corner of cube as origin O(0, 0, 0) and the opposite corner asP(a, a, a). We will calculate the electric field at P due to the other seven charges at corners. Expressing the field of a point charge in vector form. → q ^ E= .r 4π 0 rε3 (i) Field at P due to A, B, C → q → → → E1 = AP + BP + CP  4π ε a 3  0   q  ^ ^ ^ = 3  ia + ja + ka  4π 0 a   ε (ii) Field at P due to D, E, F DP = EP = FP = aV2 → q  → → →  E2 = DP + EP + FP  4π ε a 2  0 ( 3  )  Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 6 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
7. 7. q       ^^ ^^ ^^= 3ai+ k aj++ + ai k 4π 0a      q  ^ ^ ^= 2  i+ j+ k  4π 0 2 a   ε(iii) Field at P due to OOP = aV3 → q → E3 = .OP 4π ε a 3 a 0 3 ( ) q  ^ ^ ^= 3  ia + a j+ ak () 4π 0 3 3 a   ε Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 7 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
8. 8. → q  ^ ^ ^ E3 = 2  i + j + k  ( ) 4π 0 3 3 a   Resultant field at P ε → → → → E = E1 + E 2 + E 3   ^^^  iq j++ k →  1 1 =E  1+ + out ward along OP 2  4π 0 a  2 3 3 εElectric Potential Definition : Potential at a point in a field is defined as the amount of work done in bringing a unit positive test charge, from infinity to that along any arbitrary path (infinity is point of zero potential). Electric potential is scalar quantity, it is denoted by W V;V = q0 Unit and dimensional formula Joule S.I. unit : = volt Coulomb 1 C. G. S. unit: Stat volt (e.s.u) 1volt = Stat volt 300 Dimension : [V] = [ML2T-3A-1]Types of electric potential: According to te nature of charge potential is of two types (i) Positive potential : Due to positive charge (ii) Negative potential : Due to negative charge Potential of a system of point charges: Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 8 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
9. 9. V =k Q1 Q Q +k 2 +k 3 +k ( − Q4 ) + .......... .. r1 r2 r3 r4 X kQ1 In general V = ∑ i =1 r1 Electric potential due to a continuous charge distribution dQ V = ∫d V , = ∫ 4πε0 r Graphical representation of potential: As we move on the line joining two charges then variation of potential with distance is shown below Potential difference: In an electric field potential difference between two points A and B is defined as equal to the amount of work done (by external agent) in moving a unit positive charge form point A to point B W i.e., V B −V A = q 0Q. 4. A charge q =10 µC is distributed uniformly over the circumference of a ring of radius 3m placed on X-Y plane with its centre at origin. Find the electric potential at a point P(0, 0, 4m)Solution: The electric potential at P Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 9 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
10. 10. 1 q V = r0 = 3 2 + 4 2 = 5m 4π ε r0 0 9 ×10 9 ×10 −5 V = = 1.8 ×10 4 volt . 5Electric Field and Potential Due to Various charge DistributionPoint charge: Electric field and potential at point P due to a point Q is Q → Q ^ 1  Q E = k 2 o r E = k 2 r k =  , V = k r r  4π 0  ε r Graph Line charge: Electric field and potential due to a charged straight conducing wire of length l and charge density λ kλ Ex = ( sin α + sin β ) r kλ And E y = ( cos β − cos α ) r Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 10 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
11. 11. λ  r2 +l2 −l V = log e   2π ε 0  r2 +l2 +l   (i) If point P lies at perpendicular bisector of wire i.e. α = β ; 2kλEx = sin α and E y = 0 r π(ii) If wire is infinitely long i.e. l → ∞ so α = β = ; 2 2kλ λEx = and E y = 0 ⇒ E net = r 2π ε r 0 −λAnd V = 2π ε log e r + c 0(iii) If point P lies near one end of infinitely πLong wire i.e. α = 0, and β = 2 kλEx = E y = r 2 kλ⇒E net = E x + E y = 2 2 rCharged circular ring:At point P kQx kQE= ,V= (x 2 +R ) 2 3/ 2 x2 + R2 kQAt centre x = 0 so Ecentre = 0 and Vcentre = R Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 11 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
12. 12. kQ kQAt point on the axis such that x >>R E = 2 , V = x x RIf x = ± 2 Q QE max = and Vmax = 6 3π ε a 0 2 2 6π ε0GraphSome more results of line charge: If a thin plastic rod having charge density λ isbent in the following shapes then electric field at P in different situations shownin the following tatble. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 12 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
13. 13. Charged cylinder(i) Non-conducting uniformly charged cylinder(ii) Conducting charged cylinderIf point of observation (P) lies outside the cylinder then for both type ofcylindrical charge distribution λ −λEout = , and Vout = log e R + c 2πε0 r 2πε0If point of observation lies at surface i.e. r =R so for both cylinder λ −λ E Surface = and Vsurface = 2πε log e R + c 2πε0 R 0If point of observation lies inside the cylinder then for conducting cylinder Ein = 0 λrand for non-conducting Ein = 2π ε R 2 0Graph Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 13 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
14. 14. (A) For non-conducting cylinder(B) For conducting cylinderCharged conducting sphere (or shell of charge) : If charge on a conducting sphereof radius R is Q (and σ =surface charge density) as shown in figure then electricfield and potential in different situation are(i) Out side the sphere: If point P lies outside the sphere 1 Q σ R2 Euot = . = 4π ε r 2 ε 0 r 2 0 1 Q σ R2And Vuot = . = 4πε0 r ε0 r (Q = σ × A = σ × 4πR 2 )(ii) At the surface of sphere: At surface r = R 1 Q σSo, Es = . 2 = 4π ε R 0 ε0 1 Q σRAnd Vs = . = 4πε0 R ε0(iii) Inside the sphere: Inside the conducting charge sphere electric field is zeroand potential remains constant every where and equals to the potential at thesurface.Ein = 0 and Vin = constant = VsGraph Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 14 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
15. 15. Uniformly charged non-conducting sphere: Suppose charge Q is uniformlydistributed in the volume of a non-conducting sphere of radius R as shown below(i) Out side the sphere: If point P lies outside the sphere 1 Q 1 Q Eout = . and Vout = . 4πε0 r 2 4πε0 rIf the sphere has uniform volume charge density Q ρR 3 ρR 3 ρ= then Eout = and Vout = 4 3 3ε 0 r 2 3ε 0 r πR 3(ii) At the surface of sphere: At surface r = R 1 Q ρR 1 Q ρR Es = . 2 = and Vs = . = 4π ε R 0 3ε 0 4πε0 R 3ε0(iii) Inside the sphere: At a distance r from the centre 1 Qr ρr Ein = . 3 = { Ein ∝r} 4π ε R0 3ε 0 1 Q[3R 2 − r 2 ] ρ (3R 2 − r 2 )And Vin = . = 4π ε0 2R 3 6ε 0 3 1 Q 3At centre r = 0 so, Vcentre = 2 × 4πε . R = 2 Vs 0Graph Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 15 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
16. 16. i.e. Vcentre >Vsurface >VoutInfinite thin plane of charge: Consider a thin infinite non-conducting plane havinguniform surface charge density is σ . Electric field and potential near the sheetare as follows σ σr E= 2ε 0 (E ∝ r0 ) and V = 2ε 0 +CElectric field due to two thin infinite plane parallel sheet of charge: Consider twolarge, uniformly charged parallel. Plasts A and B, having surface charge densitiesare σ A and σB respectively. Suppose net electric field at points P, Q and R is tobe calculated 1At P, E P = −( E A + E B ) = − 2ε (σ A + σ B ) 0 1At Q, EQ = ( E A − E B ) = 2ε (σ A − σ B ) 0 1At R, E R = ( E A + E B ) = 2ε (σ A + σ B ) 0Special case(i) If σ A + σ B = σ then E P = E R = σ / ε 0 and Eq = 0(ii) If σ A = σ and σ B = −σ then E P = E R = 0 and EQ = σ / ε 0Hemispherical charged body σ σRAt centre O, E = 4ε , V = 2ε 0 0 Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 16 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
17. 17. Uniformly charged disc: At a distance x from centre O on it’s axis E= σ  1 − 2ε 0  x x +R  2 2   V = σ 2ε 0 [ x2 − R2 − x ] σIf x → 0, E ≈ 2ε i.e. for points situated near the disc, it behaves as an infinite 0sheet of chargePotential Due to Concentric Spheres(1) If two concentric conducting shells of radii r1 and r2 (r2 > r1) carryinguniformly distributed charges Q1 and Q2 respectively. Potential at the surface ofeach shell 1 Q1 1 Q2 V1 = . + . 4π ε r1 0 4π ε r2 0 1 Q1 1 Q2 V2 = . + . 4π ε r2 0 4π ε r2 0(2) The figure shows three conducting concentric shell of radii a, b and c (a < b <c) having charges Qa , Qb and Qc respectivelyPotential at A; 1  Qa Qb Qc  VA =  a + b + c  4π ε 0   Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 17 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
18. 18. Potential at B; 1  Qa Qb Qc  VB = + +  4π ε  b 0  b b Potential at C;  Qa Qb Qc  1 VC =  c + c + c  4π ε  0 (3) The figure shows two concentric spheres having radii r1 and r2 respectively (r2> r1). If charge on inner sphere is +Q and outer sphere is earthed then(i) Potential at the surface of outer sphere 1 Q 1 Q V2 = . + . =0 4πε0 r2 4πε0 r2 ⇒ =− Q Q(ii) Potential of the inner sphere V1 = 1 Q +. 1 . ( −Q ) = Q  1 + 1    4πε0 r1 4πε0 r2 4πε0 r1 r2 (4) In the above case if outer sphere is given a charge +Q and inner sphere isearthed then(i) In this case potential at the surface of inner is zero, so if Q’ is the chargeinduced on inner sphere 1 Q Q Then V1 =  +  =0 4πε0  r1 r2  r1i.e. Q = Q r2(Charge on inner sphere is less than that of the outer sphere).(ii) Potential a at the surface of outer sphere 1 Q 1 Q V2 = . + . 4πε0 r2 4πε0 r2 1  r1  Q  r1  V1 = − Q + Q  = 1 −  4π ε0  r2  4π ε r2 0  r2  Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 18 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
19. 19. Relation Between Electric Field and Potential (1) In an electric field rate of charge of potential with distance is known as potential gradient. (2) Potential gradient is a vector quantity and it’s direction is opposite to that of electric field. (3) Potential gradient relates with electric field according to the following relation dV volt E =− d r ; This relation gives another unit of electric field is meter . (4) In the above relation negative sign indicates that in the direction of electric field potential decreases. (5) Negative of the slope of the V-r graph denotes intensity of electric field i.e. V tan θ = = −E . r → ^ ^ ^ (6) In space around a charge distribution we can also write E = xiE + Ey j+ z kE . ∂V ∂V ∂V where E x = − ∂ x , E y = − ∂ y and E Z = − ∂ z dV (7) With the help of formula E = − d r , potential difference between any two points in an electric field can be determined by knowing the boundary conditions 2→ → 2 d V = −∫ E . dr = −∫ E.dr cos θ 1 1Q. 5. A uniform electric field of 100 v/m is directed at 300 with +ve x-axis as shown in figure. Find the potential difference VBA. If OA = 2m and OB = 4m.Solution: Here VA > VB & VB – VA will be –ve d AB = OA cos 30 0 + OB sin 30 0 =2× 2 3 1 +4× = 2 ( 3 +2 ) Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 19 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
20. 20. ∴VB −V A = −E d AB=− ( 100 2 + 3 ).Motion of charge particle in electric field(1) When charged particle initially at rest is placed in the uniform fieldSuppose a charge particle having charge Q and mass m is initially at rest in anelectric field of strength E. The particle will experience an electric force whichcauses it’s motion.(i) Force and acceleration: The force experienced by the charged particle is F =QE. F QEAcceleration produced by this force is a = = . m m(ii) Velocity: Suppose at point A particle is at rest and in time t, it reaches thepoint B where it’s velocity becomes v. Also if ∆V =Potential difference betweenA and B, S = Separation between A and B QEt 2 Q ∆V ⇒ = v = m m QEt(iii) Momentum: Momentum p = mv, p = m × =Q E t m 2 Q ∆VOr p =m × = 2 m Q ∆V m(iv) Kinetic energy: Kinetic energy gained by the particle in time t is 2 1 1 Q Et  Q 2 E 2t 2K= mv 2 = m  = 2 2  m  2m 1 2QVOr K = m × = Q ∆V 2 m(v) Work done: According to work energy theorem we can say that gain in kineticenergy = work due in displacement of charge i.e. W = ∆ Q V Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 20 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
21. 21. Where ∆ = Potential difference between the two position of charge Q. ( V → →∆V = E .∆ r = E∆r cos θ where θ is angle between direction of electric field anddirection of motion of charge).   → ^^^If charge Q is given a displacement r =  r1 + 2 + 3krjri  in an electric field    → ^ ^ ^E =  1iE + E2 j+ 3kE  . The work done is  → → W =QE . r  =Q( E1 r1 + E 2 r2 + E3 r3 ) .    Work done in displacing a charge in an electric field is path independent.(2) When a charged particle enters with an initial velocity at right angle to theuniform fieldWhen charged particle enters perpendicularly in an electric field, it describe aparabolic path as shown(i) Equation of trajectory: Throughout the motion particle has uniform velocityalong x-axis and horizontal displacement (x) is given by the equation x = utSince the motion of the particle is accelerated along y-axis 2 1  QE  x So y =    ; this is the equation of parabola which shows y ∝x 2 2  m  u  Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 21 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
22. 22. QEt (ii) Velocity at any instant: At any instant t, vx = u and v y = m → Q 2 E 2t 2 So v = v = v x + v y = u 2 + 2 2 m2 If β is the angle made by v with x-axis than vy QEt tan β = = vx muEquilibrium of charges Definition: A charge is said to be in equilibrium, if net force acting on it is zero. A system of charges is said to be in equilibrium if each charge is separately in equilibrium. Type of equilibrium: Equilibrium can be divided in following type. (i) Stable equilibrium: After displacing a charged particle from it’s equilibrium position, if it returns back then it is said to be in stable equilibrium. If U is the d2U potential energy then in case of stable equilibrium is positive i.e. U is d x2 minimum. (ii) Unsatble equilibrium: After displacing a charged particle form it’s equilibrium position, if it never returns back then it is said to be in unstable equilibrium and d2U in unstable equilibrium is negative i.e., U is maximum. d x2 (iii) Neutral equilibrium: After displacing a charged particle form it’s equilibrium position if it neither come back, nor moves away but remains in the position in which it was kept it is said to be in neutral equilibrium and in neutral equilibrium d2U is zero i.e. U is constant. d x2 Table: Different cases of equilibrium of charge Suspended charges system of three collinear charge mg Freely suspended charge In equilibrium QE = mg ⇒ E = Q Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 22 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
23. 23. In the following fig. three charge Q1, Q and Q2 are kept along a straight line, charge Q will be in equilibrium if and only if | Force applied by charge Q1 | = | Force applied by charge Q2| Suspension of charge form string Q1 Q2 = Q2 Q 2 x1 = 2 x2 2 Q x  ⇒ 1 = 1  Q2  x2    This is necessary condition for Q to be in equilibrium. If all the three charges (Q1, Q and Q2) are similar, Q will be in stable are similar while charge Q is of different nature so Q will be in unstable equilibrium.Time period of Oscillation of a charged Body (1) Simple pendulum based: If a simple pendulum having length 1 and mass of bob m oscillates about it’s mean position than it’s time period of oscillation l T = 2π g Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 23 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
24. 24. Case-1: If some charge say +Q is given to bob and an electric field E is appliedin the direction as shown in figure then equilibrium position of charged bob (pointcharge) changes from O to O’.On displacing the bob from it’s equilibrium position 0’. It will oscillate under theeffective acceleration g’. wheremg = ( mg ) 2 + ( QE ) 2 ⇒g= g 2 + ( QE / m ) 2Hence the new time period is l lT1 = 2π = 2π g ( g + (QE / m 2 ) )1/ 2 2Since g’ > g, so T1 < T i.e. time period of pendulum will decrease.Case-2: If electric field is applied in the downward direction then.Effective accelerationg’ = g + QE / mSo new time period lT2 = 2π g + ( QE / m )T2 < TCase-3: In case 2 if electric field is applied in upward direction then, effectiveacceleration.g’ = g + QE / m Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 24 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
25. 25. So new time period l T3 = 2π T3 > T g + ( QE / m ) (2) Charged circular ring: Having time period 4π ε0 mR 3 T = 2π QqNeutral Point and Zero Potential: A neutral point is a point where resultant electrical field is zero. (1) Neutral point Due to a system of two liked point charge: For this case neutral point is obtained at an internal point along the line joining two like charge. If N is the neutral point at a distance x1 from Q1 and at a distance x2(=x-x1) from Q2 then At N | E.F. due to Q1 | = | E.F. due to Q2| 2 1 Q1 1 Q2 Q x  i.e. . 2 = . 2 ⇒ 1 = 1  4π ε x1 0 4π ε x 2 0 Q2  x2    x x Short Trick: x1 = Q / Q + 1 and x2 = Q1 / Q2 + 1 2 1 Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 25 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
26. 26. (2) Neutral point due to a system of two unlike point charge: For this condition neutral point lies at an external point along the line joining two unlike charges. Suppose two unlike charge Q1 and Q2 separated by a distance x from each other. Here neutral point lies outside the line joining two unlike charges and also it lies nearer to charge which is smaller in magnitude. If |Q1| < |Q2| then neutral point will be obtained on the side of Q1, suppose it is at a distance l form Q1 so x l= Q2 / Q1 −1 (3) Zero potential due to a system of two point charge (i) If bothe charges are like then resultant potential is not zero at any finite point. (ii) If the charges are unequal and unlike then all such points where resultant potential is zero lies on a closed curve. (iii) A long the line joining the two charge, two such points exits, one lies inside and one lies outside the charges on the line joining the charges. Both the above points lie nearer the smaller charge. For internal point (it is assumed that |Q1| < |Q2| ) Q1 Q2 At P, = x1 ( x − x1 ) x ⇒ x1 = ( Q2 / Q1 + 1) For external point Q1 Q2 At P, = x1 ( x + x1 ) x ⇒ x1 = ( Q2 / Q1 −1)Electric potential Energy (1) Work done in brining the given charge from infinity to a point in the electric field is known as potential energy of the charge. Potential can also be written as W U potential energy per unit charge i.e. V = Q = Q (2) Potential energy of a system of two charge Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 26 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
27. 27. Potential energy of Q1 = Potential energy of Q2 = potential energy of system QQ U =k 1 2 r QQ In C.G.S U = 1 2 r (3) Potential energy of a system of n charge k n QQ  1  It is given by U= ∑ 1 2  k =  2 i, j r  4π 0  ε i≠ j The factor of ½ is applied only with the summation sign because on expanding the summation each pair is counted twice: For a system of 3 charges Q Q QQ QQ  U = k 1 2 + 2 3 + 1 3   r  12 r23 r13   (4) Work energy relation: If a charge moves from one position to another position in an electric field so it’s potential energy change and work done by external force for this change is W = Uf - Ui (5) Electron volt (eV): It is the smaller practical unit of energy used in atomic and nuclear physics. As electron volt is defined as “the energy acquired by a particle having one quantum of charge (le), when accelerated 1 volt” 1j i.e. leV =1.6 ×10 −19 C × =1.6 ×10 −19 J =1.6 ×10 −12 C (6) Electric potential energy of a uniformly charged sphere: Consider a uniformly charged sphere of radius R having a total charge Q. The electric potential energy of this sphere is equal to the work done in bringing the charges from infinity to 3Q 2 assemble the sphere. U = 20 πε0 R (7) Electric potential energy of a uniformly charged thin spherical shell: It is given Q2 by the following formula U = 8πε0 R (8) Energy density: The energy stored per unit volume around a point in an electric U 1 field is given by U e = = ε 0 E 2 . If in place of vacuum some medium is Volume 2 1 present then U e = ε0εr E 2 2Force Charged Conductor: Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 27 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
28. 28. To find force on a charged conductor (due to repulsion of like charges) imagine a small part XY to be cut and just separated from the rest of the conductor MLN. The field in the cavity due to the rest of the conductor is E2, while field due to small part is E1. Then Inside the conductor E = E1 – E2 = 0 or E1 = E2 σ σ Outside the conductor E = E1 + E2 = ε Thus E1 = E2 = 2ε 0 0 (1) To find force, imagine charged part XY (having charge σd A placed in the cavity MN having field E2). σ2 Thus force dF = (σ d A) E 2 or dF = dA . The force per unit area or electrostatic 2ε 0 dF σ 2 pressure p = . dA 2ε 0 (2) The force is always outwards as ( ± σ ) is positive i.e. whether charged 2 positively or negatively, this force will try to expand the charged body. [A soap bubble or rubber balloon expands on charging to it (charge of any kind + or –)Equilibrium of Charged Soap Bubble (1) For a charged soap bubble of radius R and surface tension T and charge T density σ . The pressure due to surface tension 4 and atmospheric pressure R Pout act radially inwards and the electrical pressure (Pel) acts radially outward. (A) Uncharged bubble (B) Charged bubble Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 28 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
29. 29. 4T σ 2 (2) The total pressure inside the soap bubble Pin = Pout + − R 2ε 0 4T σ 2 (3) Excess pressure inside the charged soap bubble Pin − Pout = Pexcess = − R 2ε 0 (4) If air pressure in inside and outside are assumed equal then Pin = Pout i.e. Pexcess = 0, 4T σ 2 So, + − R 2ε0 4T σ 2 8ε 0T 2T (i) Charge density: Since − ⇒σ = = R 2ε 0 R πkR 8ε 0T (ii) Radius of bubble R = σ2 σ 2R (iii) Surface tension T = 8ε 0 (iv) Total charge on the bubble Q =8π 2ε0T R R 8T 32 π k T (v) Electric field intensity at the surface of the bubble E = ε R = 0 R 8 RT (vi) Electric potential at the surface V = 32 πRTk = ε0Electric Dipole: System of two equal and opposite charges separated by a small fixed distance is called a dipole. (1) Dipole moment: It is a vector quantity and is directed from negative charged → to positive charge along the axis. It is denoted as p and is defined as the product of the magnitude of either of the charge and the dipole length i.e. →  → p = q2 l . Its S.I. unit is coulomb-meter of Debye (1 Debye = 3.3 x 10-30C   x m) and its dimension are M0L1T1A1. (2) When a dielectric is placed in an electric field, its atoms or molecules are considered as tiny dipoles. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 29 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
30. 30. Water (H2O), Chloroform (CHCl3), Ammonia (NH3), HCl, CO molecules are sameexample of permanent electric dipole.(3) Electric field and potential due to an electrical dipole: If a, e and g are threepoints on axial, equatorial and general position at a distance r form the centre ofdipole. 1 2p(i) At axial point: Electric field and potential are given as E 0 = 4π ε . r 3 (directed 0form –q to +q) 1 2p → →Va = . . Angle between E a and p is 00. 4π ε r 2 0 1 2p(ii) At equatorial point: E 0 = 4π ε . r 3 (directed form +q to –q) and Ve = 0. Angle 0 → →between E e and p is 1800. p cos θ 1(iii) At general point: E g = 4π ε . r 3 2p (3 cos θ + 1) and V g = 2 1 4π ε . . Angle 0 0 r2 → → 1between Ee and p is (θ + α ) (where tan α = 2 tan θ )(4) Dipole in an external electric field: When a dipole is kept in an uniformelectric field. The net force experienced by dipole is zero as shown in figure. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 30 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
31. 31. The net torque experienced by the dipole is τ = pE sin θ , .Hence due to torque so produced, dipole align itself in the direction of electricfield. This is the position of stable equilibrium of dipole.(i) Work done in rotation: Suppose initially, dipole is kept in a uniform electricfield at an angle θ . Now to turn it through an angle θ2 (with the field). Work 1done W = pE ( cosθ 1 − cosθ 2 ) .If θ 1 = 0 and θ 2 = θ i.e. initially dipole is kept along the field then it turn through 0θ so work done p W (1 − cos θ)(ii) Potential energy of dipole: It is defined as work done in rotating a dipole froma direction perpendicular to the field to the given direction, i.e. form aboveformula of work. If θ 1 = 90 and θ 2 = θ ⇒ W = U = − pE cosθ 0 θ = 00 θ = 90 0 θ = 1800 Stable equilibrium Not in equilibrium un stable equilibrium τ =0 τ max = pE τ =0 W = 0 W = pE Wmax = 2pE Umin=-pE U = 0 Umax= pE(iii) Equilibrium of dipole: When θ = 0 0 i.e. dipole is placed along the electric fieldit is said to be in stable equilibrium, because after turning it through a smallangle, dipole tries to align itself again in the direction of electric field.When θ = 180 0 i.e. dipole is placed opposite to electric field, it is said to be inunstable equilibrium. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 31 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
32. 32. (iv) Oscillation of dipole: In a uniform electric field if a dipole is slightly displacedfrom it’s stable equilibrium position it executes angular SHM having period ofoscillation. 1T = 2π pE where 1 = moment of inertial of dipole about the axis passing throughit’s centre and perpendicular to it’s length.(5) Electric dipole in non-uniform electric field: In non-uniform electric fieldFnet ≠ 0, τ net ≠ 0Motion of the dipole is combination of translatory and rotatory motion. Table Dipole-dipole interaction 1 6 p1 p 2 1 2 p1 p 2 . .4π ε0 r4 4π ε0 r3 (attractive) Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 32 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
33. 33. 1 3 p1 p 2 . 4π ε0 r4 1 3 p1 p 2 . 4π ε0 r3 1 3 p1 p 2 . 4π ε0 r3 (perpendicular)Important Points∗ With rise in temperature dielectric constant of liquid decreses.∗ Two point charges separated by a distance r in vaccum and a force F acting between them. After filling a dielectric medium having dielectric constant K completely between the charges, force between them decreases. To maintain the force before separation between them has to be changed to r k . This distance known as effective air separaton.∗ For ta short dipole, electric field intensity at a point on the axial line is double the electric field intensity at a point on the equatorial line of electric dipole i.e. E axial = 2 E equatorial .∗ Coulombs law is valid at a distance greater than 10-15m.∗ Ratio of gravitational force and electrostatic force between (i) Two electrons is 10-43/1. (ii) Two protons 10-39/1∗ At the centre of the line joining two equal and opposite charge V = 0 but E ≠ 0 .∗ Electric field intensity and electric potential due to a point charge q, at a distance t1 + t2 where t1 is thickness of medium of dielectric constant K1 and t2 is thickness of medium of dielectric constant K2 are: 1 Q 1 Q E= V = 0 1( 4π ε t K + t K 2 2 ) 2 ; ( 4π ε t K + t K 0 1 2 2 ) 2 Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 33 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
34. 34. ∗ If an electron (charge and mass m) is moving on a circular path of radius r about a positively charge infinitely long linear charge, (charge density λ ) then the eλ velocity of electron in dynamic equilibrium will be v = 2 πε0 m∗ A metal plate is charged uniformly with a surface charge density σ . An electron of energy W is fired towards the charged metal plate from a distance d, then for Wε 0 no collision of electron with plate d = . eσ∗ Newton’s third law is inapplicable in certain cases with reference to electrostatic or electrodynamics. For instance, if a charge q1 is placed inside a conducting shell while charge q2 is outside the shell as shown, then the force of q1 on q2 ≠ 0 while force of q2 and q1 is zero.Questions:Q. 1. (a) Two similar point charges q1 and q2 are placed at a distance r apart in air. If a dielectric slab of thickness t and dielectric constant K is put between the charges. Calculate the coulomb force of Repulsion. (b) If the thickness of the slab covers half the distance between the charges the coulumb Repulsive force is reduced in the ratio 4 : 9. Calculate the dielectric constant of slab? 1 q1 q 2Ans: a) 4π ε b) 4 ( 0 r −t +t K ) 2Q. 2. A thin non-conducting ring of Radius R has a linear charge density λ = λ0 . cos φ where λ is a constant φ is the azimuthal angle. Find the magnitude of the 0 electric field strength on the axis of the ring as a function of the distance x form its centre. Investigate the obtained function at x >>R. λ0 R 2Ans: E = . For x >>R the strength ( 4ε 0 x 2 + R 2 ) 3/ 2 1 E≈ where P = πR 2 λ0 4π ε0Q. 3. Four point particles, each of mass m and charge q are initially held in a plane at the four corners of the square of side l0. If the particle are simultaneously released they fly apart. Determine the velocity and acc. of each particle as it Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 34 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
35. 35. moves away. What will be the final velocity of each particle as they separated infinitely apart? Q2 1 1 Ans: a= 2  + 2 4π ε m 2 r  2 0  V2 = ( ) Q 1+ 2 2  1 1  2  −  and V2 = ( Q2 1+ 2 2 ) 8π ε m  r0 r  0   8π ε mr0 0Q. 4. Figure shows a charged rod, bent in the form of an arc of a circle. The charge distribution on the rod is shown in the figure the assembly is kept in a uniform electric-field. Show that for small angular displacement the system will perform S.H.M. Determine i + s –period. mAns: T = λ sin 2 θ0 EPassage -1: The imaging drum of a photocopier is +vely charged to attract –vely charged particles of toner. Near the surface of the drum. Its electric field has magnitude 1.4 x 105N /C. A toner particles is to be attached to the drum with a force that is 10 times the weight of the particle. Assume toner particles are made of carbon 12C.1. Find charge to mass ratio of the charged toner particle? a) 7 x 10-4c/kg b) 7 x 10-3c/kgc) 7 x 10-5c/kg d) 7 x 10-6c/kg2. Find the number of carbon atoms that for each excess electron on a toner particles? a) 1.15 x 108 b) 1.15 x 107 c) 1.15 x 109 d) 1.15 x 1010Passage-2 A region in space contains a total +ve charge Q that is distributed spherically R such that the volume charge density δ ( r ) = α for r ≤ 2 Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 35 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
36. 36.  r R δ ( r ) = 2α1 −  for ≤ r ≤R  R 2 Where α is a +ve constant f(r) = 0 for r ≥ R R1. What fraction of total charge is constained in the region r ≥ ? 2 a) 4/ 15 b) 8 /15 c) 7 / 15 d) None2. If an electron is placed at the centre and slightly displaced it will execute S.H.M. R find the time period of oscillation assuming x < 2 ε0π mR 3 15 ε0π mR 3 7ε 0π mR 3 a) 2π b) 2π c) 2π d) None 8Q e 8Q e 8Q e R3. The electric-field in a region < r < R is 2 αR 3 2α r  r αR 3 2α r  r a) + 1 −  b) + 1 −  24 r ε 0 2 3ε 0  R 8 r ε0 2 6ε0  R αR 3 2α r  r c) + 1 −  d) None of these 16 r ε 0 2 3ε 0  RMultiple Matching1. Match Column-I with Column-II Column-I Column-I I A) Force on an electron in an atom P) Gravitational force B) Force between a proton and a neutron inside Q) Strong force C) Force between a proton and proton inside R) Coulomb force D) Conservative force S) Electric forceAns: (A-P, R, S) (B – P) (C – P, Q, R, S) (D - P, Q, R, S)2. Match Column-I with Column-II Column-I Column-I I Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 36 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com
37. 37. A) Electric field due to a dipole at any point P) Zero B) Electric field between the plates of a capacitor Q) αx − 3 C) Electric field due to a long charged plate R) αd , the distance between the plates D) Electric field due to a ring at its centre S) Independent of distanceAns: (A-Q) (B – S) (C –S) (D - P) Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 37 Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com