EAMCET

1 ELECTROSTATICS PREVIOUS EAMCET BITS 1. An infinitely long thin straight wire has uniform linear charge density of 1/3 coul.m–2. Then the magnitude of the electric intensity at a point 18 cm away is : (given 12 2 2 0 8.0 10 /C N m − ∈ = × − ) (2009 E) 1) 11 1 0.33 10 NC− × 2) 11 1 3 10 NC− × 3) 11 1 0.66 10 NC− × 4) 11 1 1.32 10 NC− × Ans : 1 Sol: Magnitude of electric intensity at a point due to an infinitely long thin straight wire of uniform linear charge density λ is 0 2 E r λ π = ∈ 9 0 1 9 10 4 π   = ×   ∈   9 21 18 10 3 18 10− × × = × = 11 0.33 10 × NC-1 2. Two point charges –q and +q are located at points (0,0-1) and (0,0,a) respectively. The electric potential at a point (0,0,z), where z >a is : ( 2008 E ) 1) 2 0 4 qa z π∈ 2) 0 4 q a π∈ 3) ( ) 2 2 02 4 qaz a π∈ − 4) ( ) 2 2 02 4 qaz a π∈ + Ans :3 Sol: Electric potential at the point P because of the charges –q and +q is 1 2 p V V V = + ( ) ( ) 2 2 0 0 0 1 1 . . 4 4 4 p q q q z a z a V z a z a z a π π π   − + + − +   ∴ = + =     ∈ + ∈ − ∈ −     ( ) 2 2 0 2 4 p qa V z a π ⇒ = ∈ − 3. Two charges q and -q are kept apart. Then at any point on the perpendicular bisector of line joining the two charges. (2008E) 1) the electric field strength is zero 2) the electric potential is zero 3) both electric potential and electric field strength are zero 4) both electric potential and electric field strength are non-zero Ans : 1 Sol: Potential at any point on the perpendicular bisector of the line joining the two charges V = 1 2 0 0 1 1 0 4 4 q q V V r r π π − + = + = ∈ ∈ [ r = distance between the charges and any point on perpendicular bisector which is same for both the charges] Electrostatics 2 4. A charge of1 C µ is divided into two parts such that their charges are in the ratio of 2 : 3. These two charges are kept at a distance 1m apart in vaccum. Then, the electric force between them (in newton) is (2008E) 1) 0.216 2) 0.00216 3) 0.0216 4) 2.16 Ans: 2 Sol: 1 25 q Cµ = 2 35 q Cµ = 1 2 2 0 1 4 q q F d π = ∈ ( ) 9 6 6 2 2 3 9 10 10 10 5 51 − − × × × × × = = 0.00216 N 5. Along the x-axis, three charges , 2q q − and 2q are placed at x = 0, x =a and x =2a respectively. The resultant electric potential at a point ‘P’ located at a distance r from the charge -q ( ) a r << is ( 0 ∈ is the permittivity of free space) (2007-E) 1) 2 0 4 qa r π∈ 2) 2 3 0 4 qa r π∈ 3) 2 3 0 4 4 a q r π      ∈ 4) 0 4 q r π∈ Ans :2 Sol: The potential at P is V which is given by 1 2 3 V V V V = + + ( ) ( ) 0 0 0 1 1 1 4 2 4 4 2 q q q V r a r r a π π π − = + + ∈ + ∈ ∈ − 0 1 1 2 1 4 2q V r a r r a π   = − +   ∈ + −   = ( ) ( )( ) ( ) ( )( ) 0 2 1 4 2 r r a r r a r a r a r q r r a r a π   − − + − + −   ∈ + −   = 2 2 2 2 2 2 0 1 2 2 4 2q r ra r a r ar r r a π   − − + + +   ∈   −       = 2 2 2 01 2 4 2q a r r a π     ∈   −       Electrostatics 3 = 2 3 0 1 4 qa r π∈ as a< (2009 M) Sheet 1 Sheet 2 1) 10 2 σ∈ 2) 20 2σ∈ 3) 1 2 0 2 σ σ +∈ 4) 1 2 0 2 σ σ −∈ Ans : 3 Sol: Applying Gauss law to the region I the Electric field intensity is Electrostatics 10 ( ) 1 2 0 1 2 E σ σ = + ∈ Where 1 σ and 2 σ are the surface charge densities. 25. A parallel plate capacitor with air as dielectric is charged to a potential ‘V’ using a battery. Removing the battery, the charged capacitor is then connected across an identical uncharged parallel plate capacitor filled with wax of dielectric constant ‘k’. The common potential of both the capacitor is (2009 M) 1) V volts 2) kV volts 3) (k+1) V volts 4) 1 V k + volts Ans : 4 Sol: If the capacity of first capacitor is ‘c’ then the capacity of second capacitor is ‘KC’. ∴ common potential = 1 1 2 2 1 2 CV C V CV C C C KC + = + + =1 VK + volt 26. A charge q is placed at the mid-point of the line joining two equal charges each of Q. If the whole system is in equilibrium, then the value of q is (2008M) 1) 2 Q − 2) 2Q + 3) 4Q − 4) 4Q + Ans: 3 Sol: Potential energy of the system is equal to zero when the system is in equilibrium. ( ) 2 0 0 0 1 1 1 0 4 4 4 2 q Q Qq Q x x x π π π + + = ∈ ∈ ∈ ( )( ) 2 2 2 Q q Q x x − = 4Q q = − 27. A charge ‘Q’ is placed at each corner of a cube of side ‘a’. The potential at the centre of the cube is 1) 0 4 3 Qa ∈ 2) 0 4 3 Q a ∈ 3) 0 4 3 Q a π∈ Ans: 3 Sol: Length of the diagonal PQ of side a is ( ) ( ) 2 2 2 2 3 a a a a a + = + = Distance of midpoint from each corner 32a = Electrostatics 11 As 0 1 4 Q V r π = ∈ 0 1 8 2 4 3 Q V a π ⇒ = × × ∈ 0 4 3 Q a π = ∈ 28. The equivalent capacity between the points X and Y in the circuit with 1 C Fµ = (2007M) 1) 2 F µ 2) 3 F µ 3) 1 F µ 4) 0.5 F µ Ans: 1 Sol: The capacitor Q is short circuited and P and R in parallel. So the resultant capacitance is equal to 2 2 1 2 C Fµ = × = 29. Three charges 1 ,1 C C µ µ , and 2 C µ are kept at the vertices A, B and C of an equilateral triangle ABC of 10cm side, respectively. The resultant force on the charge at C is (2007M) 1) 0.9 N 2) 1.8 N 3) 2.72 N 4) 3.12 N Ans : 4 Sol: ( ) 6 6 1 2 2 0 1 10 2 10 4 10 10 F F π − − − × × = = ∈ × = 12 9 2 2 10 9 10 1.8 10 N − − × × × = The resultant force 2 2 1 1 2 2 2 cos RF F FF F θ = + + 2 2 0 1 2 1 2 2 cos 60 RF F F FF = + + = ( ) ( ) ( )( ) 2 2 1 1.8 1.8 2 1.8 1.8 2 + + × 1.8 3 1.8 1.732 3.12N = = × = Electrostatics 12 30. The electrical potential on the surface of a sphere of radius ‘r’ due to a charge 6 3 10 C − × is 500V. The intensity of electric field on the surface of the sphere is ( ) 9 2 2 1 0 1 9 10 4 Nm C in NC πε − −   = ×     (2006M) 1) <10 2) >20 3) Between 10 and 20 4) <5 Ans: 1 Sol: Potential on the surface of sphere 0 1 500 4 q V R π = = ∈ 2 0 0 1 4 1 4 qR E q ππ     ∈   = ∈ = 9 6 500 500 9 10 3 10− × × × × = 43 25 10 250 10 27 10 27 × = < × 31. Two unit negative charges are placed on a straight line. A positive charge ‘q’ is placed exactly at the midpooin between these unit charges. If the system of three charges is in equilibrium the value of ‘q’ (in C) is (2006M) 1) 1.0 2) 0.75 3) 0.5 4) 0.25 Ans :4 Sol: If the system of these three charges is in equilibrium if repulsive force between –Q and –Q is balanced by attraction forces between q and -Q =1 2 1 2 0 F F F F + = ⇒ =− = ( )( ) ( ) ( ) 2 2 0 0 1 1 4 4 2Q Q q Q r r π π − − −   =   ∈ ∈   On solving 0.25 4Q q = = 32 Three identical charges of magnitude 2 C µ are placed at the corners of a right angled triangle ABC whose base BC and height BA are respectively 4cm and 3cm. Forces on the charge at the right angled corner ’B’ due to the charges at ‘A’ and ‘C’ are respectively F1 and F2. The angle between their resultant force and F2 is (2005 M) 1) 1 9 16 Tan−       2) 1 16 9 Tan−       3) 1 16 9 Sin−       4) 1 16 9 Cos−       Ans : 2 Electrostatics 13 Sol : Angle made by the resultant tan resul t f with f2 is ( ) sin . tan cos b i e a b θ α θ = + 1 12 tan FF θ − = 22 0 1 2 2 2 0 1 4 3 16 1 9 4 4q F q F ππ∈ = = ∈ 1 16 tan 9 θ − = 33. Energy ‘E’ is stored in a parallel plate capacitor ‘C1’. An identical uncharged capacitor ‘C2’ is connected to it, kept in contact with it for a while and then disconnected, the energy stored in C2 is (2005 M) 1) 2E 2) 3E 3) 4E 4) Zero Ans :3 Sol : Energy stored in the 1st capacitor 2 12 E CV = If second similar capacitor is in contact with the 1st one the potential on the second capacitor is V/2. ∴Energy stored in second capacitor 2 12 2 4 V E C  = =     34. Capacitance of a capacitor becomes 76 times its original value if a dielectric slab of thickness, t= 23 d is introduced in between the plates. ‘d’ is the separation between the plates. The dielectric constant of the dielecctri slab is (2004 M) 1) 14 11 2) 11 14 3) 711 4) 11 7 Ans : 4 Sol. 0 0 A C d ∈ = ……….(1) 0 76 C C = …………(2) Electrostatics 14 0 0 /1 1 1 1 1 A A d C t d t K d K ∈ ∈ = =     − − − −         0 0 3 2 1 2 1 1 3C KCK K = +   − −     ………(3) Dividing (1) and (3) 0 3 7 2 6 C K C K ⇒ = = + 14 11 K ⇒ = 35. A parallel plate capacitor filled with a material of dielectric constant K is charged to a certain voltage. The dielectric material is removed. Then (2004M) a) The capacitance decreases by a factor K b) The electric field reduces by a factor K c) The voltage across the capacitor increases by a factor K d) The charge stored in the capacitor increases by a factor K 1) a and b are true 2) a and c are true 3) b and c are true 4) b and d are true Ans :2 Sol: i) Electric field increases by a factor K ii) Charge decreases by a factor K 36. Between the plates of a parallel plate capacitor of capacity C, two parallel plates of the same material and area same as the plate of original capacitor, are placed. If the thickness of these plates is equal to 15 th of the distance between the plates of the original capacitor, then the capacity of the new capacitor is (2003 M) 1) 53C 2) 35C 3) 3 10 C 4) 10 3 C       Ans :1 Sol: 0 A C d ∈ = After insertion of metal plates, the effective distance of separation becomes d-t ∴ 3 2 2 5 d d d − × = ' 0 0 5 5 3 3 3 5A A C C d d ∈ ∈ = = = Electrostatics 15 37. A charged sphere of diameter 4cm has a charge density of 10-4C/cm2. The work done in joules when a charge of 40nano-coulombs is moved from infinite to a point, which is at a distance of 2cm from the surface of the sphere is (2003 M) 1) 14.4 π 2) 28.8 π 3) 144 π 4) 288 π Ans :1 Sol: 2 2 4 4 q q R R σ π σ π = ⇒ = ………(1) Work done = potential energy at the given distance r = 2+2 = 4cm from the centre of the sphere. ' 2 0 0 4 ' 4 4 qq R q W r r π σ π π = = ∈ ∈ ……….(2) Sub. (1) in (2) ( ) ( ) ( ) 22 9 8 2 4 2 10 1 40 10 9 10 4 10 π − − − × × × × × × × = × = 14.4 J π 38. The capacities of three capacities are in the ratio 1 : 2 : 3. Their equivalent capacity when connected in parallel is 60 11 F µ more than that when connected in series. The individual capacities are ..... in F µ . 1. 4, 6, 7 2. 1, 2, 3 3. 2, 3, 4 4. 1, 3, 6 [2002 M] Ans :2 Sol: 1 0 2 0 3 0 , 2 , 3 C C C C C C = = = In parallel, 1 2 3 0 0 0 0 2 3 6 C C C C C C C C = + + = + + = In series, ' 1 2 3 1 2 2 3 3 1 CC C C CC C C C C = + + ( )( )( ) ( )( ) ( )( ) ( )( ) 0 0 0 0 0 0 0 0 0 0 2 3 6 2 2 3 3 11 C C C C C C C C C C = = + + Given that '0 0 6 60 6 11 11 C C C C − = − = 0 1 C Fµ ⇒ = 1 2 3 1 , 2 , 3 C FC FC F µ µ µ = = = 39. A capacitor of capacity 10µF is charged to 40 V and a second capacitor of capacity 15µF is charged to 30 V if the capactors are connected in parallel, the amount at change that flows from the smaller capacitor to higher capacitor in C µ is........ [ 2002 M] 1. 30 2. 60 3. 200 4. 250 Ans :2 Sol: Common potential 1 1 2 2 1 2 CV C V C C + = + 10 40 15 30 10 15 × + × = + = 34V Electrostatics 16 Amount of charge flowing = 10 40 10 34 × − × =60µC 40. A parallel plate capacitor of capacity 5 F µ and plate separation 6cm is connected to a 1V battery and is charged. A dielectric of dielectric constant 4 and thickness 4 cm is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is [2001 M] 1. 2 C µ 2. 3 C µ 3. 5 C µ 4. 10 C µ Ans : 3 Sol: 0 air A C d ∈ = , 0 medium A C t d t k ∈ = − + 6 2 4 6 4 4 medium air C d t C d t k ⇒ = = = − + − + ∴ ( ) 2 10 m air C C Fµ = = Charge q = CV = 10 x 1 = -1µC Additional charge = Final charge – Initial charge = 10µC – 5µC = 5µC 41. Two capacitors of capacity 4 F µ and 6 F µ are connected in series and a battery is connected to the combination. The energy stored is 1 E . If they are connected in parallel and if the same battery is connected to this combination the energy is . The ratio 1 2 : E E is [2001 M] 1. 4:9 2. 9:14 3. 6:25 4. 7:12 Ans: 3 Sol: As the capacitors are connected in series 2 1 2 1 1 2 12 series CC E E V C C   = =   +   As the capacitors are connected in parallel ( ) 2 2 1 2 12 parallel E E C CV = = + ∴ ( ) ( ) 1 1 2 2 2 1 2 6 4 6 4 E CC E C C × = = + + 1 2 : 6:25 E E ⇒ = Electrostatics 17 42. In a parallel plate capacitor, the capacitance [2001 M] 1. increases with increase in the distance between the plates 2. decreases if a dielectric material is put between the plates 3. increases with decrease in the distance between the plates 4. increases with decrease in the area of the plates Ans : 3 Sol: From the relation 0 1 A C C d d ∈ = ⇒ ∝ ∴ Capacity increases with decrease in distance 43. Two charges of 4 C µ each are placed at the corners of A and B of an equilateral triangle ABC of side length 0.2m in air. The electric potential at C is 9 0 1 9 10 4 π   = ×   ∈   [2000 M] 1. 4 9 10 V × 2. 4 18 10 V × 3. 4 36 10 V × 4. 4 72 10 V × Ans : 3 Sol: Electric potential of C 1 2 V V V ⇒ = + 1 2 0 0 1 1 . . 0 4 4 q q V r r πε πε = + = 4 0 1 2 . 36 10 4 q V r π   = = ×   ∈   􀂲 􀂲 􀂲