Title 17. Electrostatics Hard Ans.pdf ✅

1. (c) r = (a2 + d2 ) 1/2 at angle 180 – 2  Enet = 2EA cos (180 – ) = 2EAcos(90 – ) = 2EA sin = = in –z direction. 2. (c) Initially the potential at centre of sphere is VC = After the sphere grounded, potential at centre becomes zero. Let the net charge on sphere finally be q.  + = 0 or q = The charge flowing out of sphere is 3. (C) V = V1 + V2 + V3 = = . 4. (c) P(r) = From Gauss law = = = E.4r1 2 = 0 E = 5. (d) W = q (V) = q (Vf – Vi) = – 100 × 1.6 × 10–19 (– 4 – 10) = 1.6 × 14 × 10–17 = 2.24 × 10–17 J 6. (a) Since net force on charge Q is zero (Placed at corner A). + = 0  =  = So option (1) is correct. 7. (b) Inside pressure must be greater than outside pressure in bubble. This excess pressure is provided by charge on bubble. = = ... Q = 8r 8. (c) F" = F F" = Z q A (0, 0, d) d (a, 0, 0) x d –q B(0, 0, –d)   EB EA r p Enet At P there are two fields (1) Due to A = EA = 2 r kq (2) Due to B = EB = 2 r kq 2 1 2 r 2kq (r) d 2 2 3/ 2 (a d ) 2kqd + x 3Q 4 1 x 2Q 4 1 x Q 4 1 0 0  0 =  +  0 4 1  r q x 3Q 4 1  0 x 3Qr x 3Qr        +     −  +  R 3Q 4 1 R 2Q 4 1 R Q . 4 1 0 0 0 0 4 1        R 2Q r R Q 4  r1 + + + + + + + + + + + + + + + + + + + + + + + + +  E.ds 0 qen  0 VdV        1 r O 4 0 2 / R Qr4 r dr 4 r 4 R Q 4 1 4   4 0 2 1 4 R Qr  – q B a A Q a D – q C Q 2a 2 2 ( 2a ) KQ       2 a kQq 2 2 2 2a kQ 2 a – 2kQq q Q – 2 2 r 4T r 4T 0 2 2  r 4T 0 2 4 2 16 r 2 Q            = 2 4 r Q 0 2rT (0,a) –q(a,a) q1 q F q F" F' F (a,0) 2 2 1 a 2kqq Pa Pa
F' = = FNet = F" – F' = 9. (d) Force due to (1), (2), (4) & (5) will cancel each other. Only force due to charge (3) will not get cancelled, hence FNet = . 10. (a) x = x = = 1 m 11. (c) ENet =0 12. (c) –q will experience a restoring force and will perform SHM. 13. (c) Net electric field will be zero at origin. At any co-ordinate (0, y) ENet = 2E cos 2/2 = For maximum electric field = 0 Solving y = ± a/ 14. (d) Due to induction correct answer is option (D). 15. (a) Field lines terminate into negative charge. 16. (a) = vA = uA + aAt vA = t vB = t  = 17. (b) FR = 2F cos /2 = 2 F cos 60/2 = F FR = 18. (b) q = ne  n = = = 6.25× 1018(In excess) 19. (c) Mass of positively charged sphere decreases where as negatively charged sphere increases. 20. (c) F = ........(i) 2 1 ( 2a) kq q 2 1 2a kq q       − 2 1 2 a kqq 2 1 q q q0 q q q F (1) (2) (3) (4) (5) F F F 2 0 a kqq 4Q –Q a E(–Q) E(4Q) x 4Q Q Q − 2 1 1 − Q Q Q Q E E E E E E Q Q  (0,y) (0,0) E    E (–a,0) (a,0) 2 2 3/ 2 (a y ) kQy + dy dE 2 F = 2qE 2q,m F = qE 2q, 2m E B A KE KE 2 B B 2 A A 1/ 2m v 1/ 2m v m 2qE 2m 2qE B A KE KE 2 2 ) 2m 2qEt 2m( 2 1 ) m 2qEt m( 2 1 F F q –2q q 60o 3 2 2 a 3 2kq e q –19 1.6 10 1  2 r k(10)(90)
charge on each when contact is made = = = –40μC  New force F =  = =  21. (c)  Force on charge q is zero F21 = F23 or = or = or = or 1 – x = 3x or x = 0.25 m from 1 × 10–9C  From 9 × 10–9C, distance = 1 – x = 1 – 0.25 = 0.75 22. (a) F12 = F14 = Resultant of F12 & F14 = = =  Net force on charge q is F = + F13 = + 23. (b) (–100C ice) (00C water) Given Fice = Fwater =  k1r1 2 = k2r2 2  k1 = k2 . = 80 × = 3.2 24. (b) ENet = E1 + E2 = + = 25. (d) Charge is quantized 26. (a) Let charges are q & (Q –q) force F = = (qQ – q 2 )  1 q  q2 2 10 + (–90) 2 r k(40)(40) F F  10 90 40 40   9 16 F 9 16 F  = 1 m q 1 2 3 q1 = 1× 10–9C F23 F21 q2 = 9 ×10–9C x  2 1 x kqq 2 2 (1 – x) kqq 2 –9 x 110 2 –9 (1 – x) 9 10 x 1 1 – x 3 2 2 a kq 2 14 2 F12 + F 2 12 2 F12 + F F12 2 q q q q a a (2) (4) (3) F12 F13 F14 F13 F12 2 (1) F12 2 2 a kq 2 2 2 2 (a 2) kq 2 1 1 1 q r 25cm k q = 2 2 2 1 q r 5cm k 80 q = = 0 4 1  2 1 1 1 2 k r q q 0 4 1  2 2 2 1 2 k r q q 2 1 2 r r         2 25 5       E2 Q E1 –Q d 2 d 1 2 2 2 d kQ       2 2 d kQ       2 4 0d 8Q  q =  ne r q (Q – q) 2 r kq(Q – q) 2 r k
For maximum force = 0  [Q – 2q] = 0   Charge of other part = Q – q = Q – = 27. (a) Since magnitude of each charges are same and situated at equal distance from centre O so all charge will produce same magnitude of electric field at centre. 28. (c) Equilibrium condition ... (1) where, F = ... (2) ABD and CAE are similar triangle = h : x ... (3) By using equations (1), (2) & (3) we can calculate x  Eelectro = = 2 mgh Total work done = 2 mgh + mgh = 3 mgh 29. (c) dF = dFx = dF cos  and Fx = 30. (b) Let give small displacement to q. The equilibrium will be stable if net force after giving displacement will be in direction of position of equilibrium Fe sin   mg sin 2 then net force will be directed towards vertex i.e. initial equilibrium position sin   mg sin 2 d = diameter of sphere and  is very small ×   mg × 2 Q  = 9.8 × 10–8 C 31. (c) dq dF 2 r k 2 Q q = 2 Q 2 Q 2E 2E O 2E A B(q) D(q) A(–q) q –q q –q C(q) E E E E E E E E q O ENet A h C  B q mg A D Q x 2 x kqQ F x mg  = 2 x kqQ :  2 x x kqQ  d d dFy dFy   y dF dF dFx dFx a .dy 2 a y 2 2 0 2 1  +     + − dFx  q q Q  q mg Fe sin  Fe mg sin 2 Fe cos  N 2 2 d kqQ 2 d kqQ Kq 2mgd2 y 3 (x, 3 x)  x