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Page 1 1. (d) 12 Q QQ += .....(i) and 12 2 QQ Fk r = ..... (ii) From (i) and (ii) 11 2 () - = kQ QQ F r For F to be maximum 12 1 0 2 dFQ QQ dQ = Þ == 2. (a) The position of the balls in the satellite will become as shown below + Q L L 180° + Q Thus angle 180 q= o and force 2 2 0 1 . 4 (2) = pÎ Q L 3. (b) A F = Force on C due to charge placed at A 120° A B 10 cm C + 1µC +2µC 1µC - A F B F 66 9 22 10 2 10 9 10 1.8 (10 10 ) N -- - ´´ =´´= ´ B F = force on C due to charge placed at B 66 9 2 10 2 10 9 10 1.8 (0.1) N -- ´´ =´´= Net force on C 22 ( ) ( ) 2 cos120 1.8 = ++= net A B AB FFFFFN o 4. (c) 22 23 . ee F k r kr rr =- =- urr $ r r r æö \= ç÷ ç÷ èø r $ 5. (c) After following the guidelines mentioned above A B D C + Q + Q + Q A F AC F D F C F 22 net AC D A CD F F F F FF = += ++ Since 2 2 AC kq FF a == and 2 2 ( 2) D kq F a = 2222 2222 0 2 1 1 22 2 22 24 æö + æö = + = += ç÷ ç÷ èø èø pÎ net kq kq kq q F aaaa 6. (a) 6 6 12 22 00 1 ( 7 10 ( 5 10 ) 1 35 10 44 FN rr --- +´-´´ = =- pe pe 6 6 12 22 00 1 ( 5 10 ( 7 10 ) 1 35 10 44 FN rr --- +´-´´ ¢ = =- p e pe 7. (c) Electric field outside of the sphere E out = 2 kQ r ...(i) Ele ctric field inside the dielec tric s hphere E in = 3 kQx R ...(ii) From (i) and (ii), E in = 2 out rx E R ´ Þ At 3 cm, E = () 2 3 3 20 100 10 ´ = 120 V/m Page 2 1. (d) 12 Q QQ += .....(i) and 12 2 QQ Fk r = ..... (ii) From (i) and (ii) 11 2 () - = kQ QQ F r For F to be maximum 12 1 0 2 dFQ QQ dQ = Þ == 2. (a) The position of the balls in the satellite will become as shown below + Q L L 180° + Q Thus angle 180 q= o and force 2 2 0 1 . 4 (2) = pÎ Q L 3. (b) A F = Force on C due to charge placed at A 120° A B 10 cm C + 1µC +2µC 1µC - A F B F 66 9 22 10 2 10 9 10 1.8 (10 10 ) N -- - ´´ =´´= ´ B F = force on C due to charge placed at B 66 9 2 10 2 10 9 10 1.8 (0.1) N -- ´´ =´´= Net force on C 22 ( ) ( ) 2 cos120 1.8 = ++= net A B AB FFFFFN o 4. (c) 22 23 . ee F k r kr rr =- =- urr $ r r r æö \= ç÷ ç÷ èø r $ 5. (c) After following the guidelines mentioned above A B D C + Q + Q + Q A F AC F D F C F 22 net AC D A CD F F F F FF = += ++ Since 2 2 AC kq FF a == and 2 2 ( 2) D kq F a = 2222 2222 0 2 1 1 22 2 22 24 æö + æö = + = += ç÷ ç÷ èø èø pÎ net kq kq kq q F aaaa 6. (a) 6 6 12 22 00 1 ( 7 10 ( 5 10 ) 1 35 10 44 FN rr --- +´-´´ = =- pe pe 6 6 12 22 00 1 ( 5 10 ( 7 10 ) 1 35 10 44 FN rr --- +´-´´ ¢ = =- p e pe 7. (c) Electric field outside of the sphere E out = 2 kQ r ...(i) Ele ctric field inside the dielec tric s hphere E in = 3 kQx R ...(ii) From (i) and (ii), E in = 2 out rx E R ´ Þ At 3 cm, E = () 2 3 3 20 100 10 ´ = 120 V/m DPP/ P 32 89 8. (c) Electric lines force due to negative charge are radially inward. – 9. (a) In non-uniform electric field, intensity is more, where the lines are more denser. 10. (b) According to the question, mg eE mgE e = Þ= 11. (b) Because E points along the tangent to the lines of force. If initial velocity is zero, then due to the force, it always moves in the direction of E. Hence will always move on some lines of force. 12. (b) The field produced by charge - 3Q at A , this is E as mentioned in the example. 2 3Q E x \= (along AB directed towards negative charge) A x B Q –3Q Now field at location of 3Q - i.e. field at B due to charge Q will be 2 ' 3 QE E x == (along AB directed away from positive charge) 13. (c) The electric field is due to all charges present whether inside or outside the given surface. 14. (b) q q q q q q E E E E E E 0 net E Þ= –q q q –q q q E E E E E E Þ 2E 2E 2E 120° 2 net E E = 2q 2q q 2q 2q q E 2E 2E E 2E 2E 0 net E Þ= q 2q 2q q 2q q E 2E 2E E 2E E Þ E 2E 2E 0 net E = E 15. (b) A B D C O A E B E C E D E Þ A B D C 2E 2E net E ,, 2, 34 A B CD E EE EE EEE = = == 16. (b) $ $ $ $ 000 0 22 2 22 E k k kk s s ss =- - - =- eee e ur 17. (b) 18. (c) According to Gauss law . E ds ò Ñ = 1 0 q e ds ò Ñ = 2 prl; (E is constant) \ E. 2prl = 1 0 q e Þ E = 0 2 q r pe i.e. E µ 1 r 19. (c) Let sphere has uniform chare density r 3 3 4 Q R æö ç÷ p èø and E is the electric field at distance x from the centre of the sphere. Page 3 1. (d) 12 Q QQ += .....(i) and 12 2 QQ Fk r = ..... (ii) From (i) and (ii) 11 2 () - = kQ QQ F r For F to be maximum 12 1 0 2 dFQ QQ dQ = Þ == 2. (a) The position of the balls in the satellite will become as shown below + Q L L 180° + Q Thus angle 180 q= o and force 2 2 0 1 . 4 (2) = pÎ Q L 3. (b) A F = Force on C due to charge placed at A 120° A B 10 cm C + 1µC +2µC 1µC - A F B F 66 9 22 10 2 10 9 10 1.8 (10 10 ) N -- - ´´ =´´= ´ B F = force on C due to charge placed at B 66 9 2 10 2 10 9 10 1.8 (0.1) N -- ´´ =´´= Net force on C 22 ( ) ( ) 2 cos120 1.8 = ++= net A B AB FFFFFN o 4. (c) 22 23 . ee F k r kr rr =- =- urr $ r r r æö \= ç÷ ç÷ èø r $ 5. (c) After following the guidelines mentioned above A B D C + Q + Q + Q A F AC F D F C F 22 net AC D A CD F F F F FF = += ++ Since 2 2 AC kq FF a == and 2 2 ( 2) D kq F a = 2222 2222 0 2 1 1 22 2 22 24 æö + æö = + = += ç÷ ç÷ èø èø pÎ net kq kq kq q F aaaa 6. (a) 6 6 12 22 00 1 ( 7 10 ( 5 10 ) 1 35 10 44 FN rr --- +´-´´ = =- pe pe 6 6 12 22 00 1 ( 5 10 ( 7 10 ) 1 35 10 44 FN rr --- +´-´´ ¢ = =- p e pe 7. (c) Electric field outside of the sphere E out = 2 kQ r ...(i) Ele ctric field inside the dielec tric s hphere E in = 3 kQx R ...(ii) From (i) and (ii), E in = 2 out rx E R ´ Þ At 3 cm, E = () 2 3 3 20 100 10 ´ = 120 V/m DPP/ P 32 89 8. (c) Electric lines force due to negative charge are radially inward. – 9. (a) In non-uniform electric field, intensity is more, where the lines are more denser. 10. (b) According to the question, mg eE mgE e = Þ= 11. (b) Because E points along the tangent to the lines of force. If initial velocity is zero, then due to the force, it always moves in the direction of E. Hence will always move on some lines of force. 12. (b) The field produced by charge - 3Q at A , this is E as mentioned in the example. 2 3Q E x \= (along AB directed towards negative charge) A x B Q –3Q Now field at location of 3Q - i.e. field at B due to charge Q will be 2 ' 3 QE E x == (along AB directed away from positive charge) 13. (c) The electric field is due to all charges present whether inside or outside the given surface. 14. (b) q q q q q q E E E E E E 0 net E Þ= –q q q –q q q E E E E E E Þ 2E 2E 2E 120° 2 net E E = 2q 2q q 2q 2q q E 2E 2E E 2E 2E 0 net E Þ= q 2q 2q q 2q q E 2E 2E E 2E E Þ E 2E 2E 0 net E = E 15. (b) A B D C O A E B E C E D E Þ A B D C 2E 2E net E ,, 2, 34 A B CD E EE EE EEE = = == 16. (b) $ $ $ $ 000 0 22 2 22 E k k kk s s ss =- - - =- eee e ur 17. (b) 18. (c) According to Gauss law . E ds ò Ñ = 1 0 q e ds ò Ñ = 2 prl; (E is constant) \ E. 2prl = 1 0 q e Þ E = 0 2 q r pe i.e. E µ 1 r 19. (c) Let sphere has uniform chare density r 3 3 4 Q R æö ç÷ p èø and E is the electric field at distance x from the centre of the sphere. 90 DPP/ P 32 Applying Gauss law, E.4p x 2 = 0 q e = 0 V r¢ e = 3 0 4 3 x r ´p e (V = V olume of dotted sphere) x R \ E = 0 3 x r e Þ E = µ x 20. (b) T sin q = qE + + + + + + + q T mg T cos q qE T sin q and T cos q = mg Þ tan q = qE mg = q mg 0 2 æö s ç÷ e èø Þs µ tan q. 21. (d) Next flux through the cube 0 net Q f= e ; so flux through one face face 0 6 f= e q 22. (d). For A : Power consumed P = I 2 R But 0 q =f e , so 2 00 dq q t I 2t dt = ae Þ = = ae 2 22 0 P 4 Rt Þ = ae For B : Assuming initial charge in reservoir be q 0 then electric flux through a closed Spherical surface around S 2 will be 2 00 S 2 0 qt - ae f= e For C : S 2 d 2t dt f = -a 23. (c). The time period will change only when the additional electrostatic force has a component along the direction of the displacement, which is always perpendicular to the string. 24. () 25. (b) Net charge inside the sphere sphere dV =r ò Due to spherical symmetry , we get 2 0 4 () R Q r r dr = pr ò 2 0 4 () R A r R r dr =p- ò 44 4 34 RR A æö =p- ç÷ ç÷ èø \ 4 3Q A R = p 26. (a) According to Gauss law 2 . 4 () s EdS r Er =p ò uuruur Ñ 2 0 0 4 () r r r dr pr = Î ò Þ 2 2 0 0 4 () 4 () r A r R r dr r Er p- p= Î ò 34 0 4 34 A rRr æö p =- ç÷ ç÷ Î èø Hence, 2 0 () 34 A rRr Er æö =- ç÷ ç÷ Î èø , for 0 rR << But 4 3Q A R = p \ We get, 2 2 0 3 11 () 34 Q rr Er RR R éù æ ö æö =- êú ç ÷ ç÷ è ø èø Îêú ëû 27. (b) The electric field outside the sphere is given by : 2 (), kQ Er r = for rR ³ 28. (c) If electric lines of forces cross each other, then the electric field at the point of intersection will have two direction simultaneously which is not possible physically. 29. (c) Electric field at the nearby point will be resultant of existing field and field due to the charge brought. It may increase or decrease if the charge is positive or negative depending on the position of the point with respect to the charge brought. 30. ()Read More
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