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Page 1 1. (c) ABCDE is an equipotential surface, on equipotential surface no work is done in shifting a charge from one place to another. 2. (c) Potential at centre O of the square Q Q Q Q O a 2 a 0 0 4 ( / 2) æö =ç÷ ç÷ pe èø Q V a Work done in shifting (– Q ) charge from centre to infinity 00 () W Q V V QV ¥ = = 2 00 422 . 4 QQ Q aa == pe pe 3. (b) Using 21 42 = Þµ Þ = == AA BB vQ QVq v vQ M v Qq 4. (a) Work done in moving a charge from P to L , P to M and P to N is zero while it is q (V P – V k ) > 0 for motion from P to k. 5. (a) 12 ( ) 2 (70 50) 40 KE q V V eV =  = ´  W= 6. (a) The electric potential 2 (, ,)4 Vxyzx = volt Now $ $ VVV E i jk x yz æö ¶ ¶¶ = ++ ç÷ ¶ ¶¶ èø ur $ Now 8,0 VV x xy ¶¶ == ¶¶ and 0 V z ¶ = ¶ Hence 8, = Ei ur $ so at point (1 ,0,2) mm 8 E xi = ur $ volt/meter or 8 along negative X  axis. 7. (b) Electric fields due to electrons on same line passing through centre cancel each other while electric potential due to each electron is negative at centre C. Therefore, at centre 0,0 EV ur =± 8. (a) By using ( .) W QEr =D urr $ $ $ 1 2 3 12 [( ).( )] ( ) W Q e i e j e k ai bj Qea eb Þ= + + + =+ $$ 9. ( b) Potential at A= Potential due to (+q) charge + Potential due to (– q) charge 2 2 22 00 1 1 () .0 44 qq a b ab  = += pe pe ++ 10. (c) Point P will lie near the charge which is smaller in magnitude i.e. 6 C m . Hence potential at P P 6 C  m 12 C m x 20 cm 66 00 1 ( 6 10 ) 1 (12 10 ) 0 0.2 4 4 (0.2) V xm xx  ´´ = + =Þ= pe pe + 11. (a) Work done 6 ( ); AB W q VV  = where 6 3 10 coulomb q  =´ where 66 106 22 ( 5 10 ) 2 10 1 10 10 15 1510 510 A V   éù ´´ = + =´ êú ´´ êú ëû volt and 66 10 22 (2 10 ) 5 10 10 15 10 5 10 B V   éù ´´ = êú ´´ êú ëû = – 6 13 10 15 ´ volt 6 66 1 13 3 10 10 10 15 15  éù æö \ =´ ´ ´ ç÷ êú èø ëû W = 2.8 J 12. (c) A B C l l l +q +q – 2q 60° p p net p 22 2 cos 60 3 3 ( ) =+ + == \= net P p p pp p ql p ql o Page 2 1. (c) ABCDE is an equipotential surface, on equipotential surface no work is done in shifting a charge from one place to another. 2. (c) Potential at centre O of the square Q Q Q Q O a 2 a 0 0 4 ( / 2) æö =ç÷ ç÷ pe èø Q V a Work done in shifting (– Q ) charge from centre to infinity 00 () W Q V V QV ¥ = = 2 00 422 . 4 QQ Q aa == pe pe 3. (b) Using 21 42 = Þµ Þ = == AA BB vQ QVq v vQ M v Qq 4. (a) Work done in moving a charge from P to L , P to M and P to N is zero while it is q (V P – V k ) > 0 for motion from P to k. 5. (a) 12 ( ) 2 (70 50) 40 KE q V V eV =  = ´  W= 6. (a) The electric potential 2 (, ,)4 Vxyzx = volt Now $ $ VVV E i jk x yz æö ¶ ¶¶ = ++ ç÷ ¶ ¶¶ èø ur $ Now 8,0 VV x xy ¶¶ == ¶¶ and 0 V z ¶ = ¶ Hence 8, = Ei ur $ so at point (1 ,0,2) mm 8 E xi = ur $ volt/meter or 8 along negative X  axis. 7. (b) Electric fields due to electrons on same line passing through centre cancel each other while electric potential due to each electron is negative at centre C. Therefore, at centre 0,0 EV ur =± 8. (a) By using ( .) W QEr =D urr $ $ $ 1 2 3 12 [( ).( )] ( ) W Q e i e j e k ai bj Qea eb Þ= + + + =+ $$ 9. ( b) Potential at A= Potential due to (+q) charge + Potential due to (– q) charge 2 2 22 00 1 1 () .0 44 qq a b ab  = += pe pe ++ 10. (c) Point P will lie near the charge which is smaller in magnitude i.e. 6 C m . Hence potential at P P 6 C  m 12 C m x 20 cm 66 00 1 ( 6 10 ) 1 (12 10 ) 0 0.2 4 4 (0.2) V xm xx  ´´ = + =Þ= pe pe + 11. (a) Work done 6 ( ); AB W q VV  = where 6 3 10 coulomb q  =´ where 66 106 22 ( 5 10 ) 2 10 1 10 10 15 1510 510 A V   éù ´´ = + =´ êú ´´ êú ëû volt and 66 10 22 (2 10 ) 5 10 10 15 10 5 10 B V   éù ´´ = êú ´´ êú ëû = – 6 13 10 15 ´ volt 6 66 1 13 3 10 10 10 15 15  éù æö \ =´ ´ ´ ç÷ êú èø ëû W = 2.8 J 12. (c) A B C l l l +q +q – 2q 60° p p net p 22 2 cos 60 3 3 ( ) =+ + == \= net P p p pp p ql p ql o 92 DPP/ P 33 13. (d) According to figure, potential at A and C are both equal to kQ. Hence work done in moving q  charge from A to C () AC qVV = = 0 A B C l l l –q +Q 14. (c) 19 9 10 ( 1.6 10 ) 9 10 27.2 0.53 10 Q VkV r   +´ =´=´´= ´ 15. (c) Potential will be zero at two points O M N x = 0 x = 4 x = 6 x = 12 1 2 q C =m 2 1 q C = m l ' l 6 At internal point (M) : 66 0 1 2 10 (110) 0 4 (6)ll  éù ´ ´ ´ += êú pe êú ëû 2l Þ= So distance of M from origin; 6 24 x== At exterior point (N): 66 0 1 2 10 ( 1 10 ) 0 4 (6 ')ll  éù ´ ´ ´ += êú ¢ pe êú ëû '6l Þ= So distance of N from origin, 6 6 12 x= += 16. (a) V = V AB + V BC + V CD 0 00 .5 .( 2 ) .(3 ) k Q k Q kQ R RR  = ++ 0 6kQ R = 0 0 3 2 Q R = pÎ 17. (a) 9 2 9 10. p V r =´ 19 10 9 102 (1.6 10 ) 1.28 10 9 10 0.13 (1210) V   ´ ´´ =´´= ´ 18. (b) 02 01 () = W qVV where 12 01 0 0 4 42 =+ pe pe QQ V R R and 21 02 0 0 4 42 =+ pe pe QQ V R R 21 02 01 0 () ( 2 1) () 4 2 qQQ W qVV R   Þ = = pe 19. (d) 00 00 1 1 111 1 ... ... 4 3 5 4 2 46 qq V xx é ù éù = + + +  + ++ ê ú êú pe pe ë û ëû 0 0 00 111 1 ... log2 4 2344 e qq xx éù =  +  += êú pe pe ëû 20. (b) Potential decreases in the direction of electric field. Dotted lines are equipotential surfaces AC VV \= and AB VV > E y C A B 21. (d) 3 .. equatorial kp E ieE p r =µ and 3  µ Er 22. (a) Suppose neutral point N lies at a distance x from dipole of moment p or at a distance x 2 from dipole of 64 p. 2 1 x 1 25 cm p 64 N p ® p ® At N E.F . due to dipole 1  = E.F. due to dipole 2  Þ 3 0 12 . 4 p x pe = ( ) ( ) 3 0 2 64 1 . 4 25 p x pe  Þ 3 1 x = ( ) 3 64 25 x  Þ x = 5 cm. 23. (a). BC = 2R sin 120 3R 2 æö = ç÷ èø Electric field at O = 22 0 0 1 2q/3q 4R R 6R æö = ç÷ èø pe pe along negative Xaxis. Page 3 1. (c) ABCDE is an equipotential surface, on equipotential surface no work is done in shifting a charge from one place to another. 2. (c) Potential at centre O of the square Q Q Q Q O a 2 a 0 0 4 ( / 2) æö =ç÷ ç÷ pe èø Q V a Work done in shifting (– Q ) charge from centre to infinity 00 () W Q V V QV ¥ = = 2 00 422 . 4 QQ Q aa == pe pe 3. (b) Using 21 42 = Þµ Þ = == AA BB vQ QVq v vQ M v Qq 4. (a) Work done in moving a charge from P to L , P to M and P to N is zero while it is q (V P – V k ) > 0 for motion from P to k. 5. (a) 12 ( ) 2 (70 50) 40 KE q V V eV =  = ´  W= 6. (a) The electric potential 2 (, ,)4 Vxyzx = volt Now $ $ VVV E i jk x yz æö ¶ ¶¶ = ++ ç÷ ¶ ¶¶ èø ur $ Now 8,0 VV x xy ¶¶ == ¶¶ and 0 V z ¶ = ¶ Hence 8, = Ei ur $ so at point (1 ,0,2) mm 8 E xi = ur $ volt/meter or 8 along negative X  axis. 7. (b) Electric fields due to electrons on same line passing through centre cancel each other while electric potential due to each electron is negative at centre C. Therefore, at centre 0,0 EV ur =± 8. (a) By using ( .) W QEr =D urr $ $ $ 1 2 3 12 [( ).( )] ( ) W Q e i e j e k ai bj Qea eb Þ= + + + =+ $$ 9. ( b) Potential at A= Potential due to (+q) charge + Potential due to (– q) charge 2 2 22 00 1 1 () .0 44 qq a b ab  = += pe pe ++ 10. (c) Point P will lie near the charge which is smaller in magnitude i.e. 6 C m . Hence potential at P P 6 C  m 12 C m x 20 cm 66 00 1 ( 6 10 ) 1 (12 10 ) 0 0.2 4 4 (0.2) V xm xx  ´´ = + =Þ= pe pe + 11. (a) Work done 6 ( ); AB W q VV  = where 6 3 10 coulomb q  =´ where 66 106 22 ( 5 10 ) 2 10 1 10 10 15 1510 510 A V   éù ´´ = + =´ êú ´´ êú ëû volt and 66 10 22 (2 10 ) 5 10 10 15 10 5 10 B V   éù ´´ = êú ´´ êú ëû = – 6 13 10 15 ´ volt 6 66 1 13 3 10 10 10 15 15  éù æö \ =´ ´ ´ ç÷ êú èø ëû W = 2.8 J 12. (c) A B C l l l +q +q – 2q 60° p p net p 22 2 cos 60 3 3 ( ) =+ + == \= net P p p pp p ql p ql o 92 DPP/ P 33 13. (d) According to figure, potential at A and C are both equal to kQ. Hence work done in moving q  charge from A to C () AC qVV = = 0 A B C l l l –q +Q 14. (c) 19 9 10 ( 1.6 10 ) 9 10 27.2 0.53 10 Q VkV r   +´ =´=´´= ´ 15. (c) Potential will be zero at two points O M N x = 0 x = 4 x = 6 x = 12 1 2 q C =m 2 1 q C = m l ' l 6 At internal point (M) : 66 0 1 2 10 (110) 0 4 (6)ll  éù ´ ´ ´ += êú pe êú ëû 2l Þ= So distance of M from origin; 6 24 x== At exterior point (N): 66 0 1 2 10 ( 1 10 ) 0 4 (6 ')ll  éù ´ ´ ´ += êú ¢ pe êú ëû '6l Þ= So distance of N from origin, 6 6 12 x= += 16. (a) V = V AB + V BC + V CD 0 00 .5 .( 2 ) .(3 ) k Q k Q kQ R RR  = ++ 0 6kQ R = 0 0 3 2 Q R = pÎ 17. (a) 9 2 9 10. p V r =´ 19 10 9 102 (1.6 10 ) 1.28 10 9 10 0.13 (1210) V   ´ ´´ =´´= ´ 18. (b) 02 01 () = W qVV where 12 01 0 0 4 42 =+ pe pe QQ V R R and 21 02 0 0 4 42 =+ pe pe QQ V R R 21 02 01 0 () ( 2 1) () 4 2 qQQ W qVV R   Þ = = pe 19. (d) 00 00 1 1 111 1 ... ... 4 3 5 4 2 46 qq V xx é ù éù = + + +  + ++ ê ú êú pe pe ë û ëû 0 0 00 111 1 ... log2 4 2344 e qq xx éù =  +  += êú pe pe ëû 20. (b) Potential decreases in the direction of electric field. Dotted lines are equipotential surfaces AC VV \= and AB VV > E y C A B 21. (d) 3 .. equatorial kp E ieE p r =µ and 3  µ Er 22. (a) Suppose neutral point N lies at a distance x from dipole of moment p or at a distance x 2 from dipole of 64 p. 2 1 x 1 25 cm p 64 N p ® p ® At N E.F . due to dipole 1  = E.F. due to dipole 2  Þ 3 0 12 . 4 p x pe = ( ) ( ) 3 0 2 64 1 . 4 25 p x pe  Þ 3 1 x = ( ) 3 64 25 x  Þ x = 5 cm. 23. (a). BC = 2R sin 120 3R 2 æö = ç÷ èø Electric field at O = 22 0 0 1 2q/3q 4R R 6R æö = ç÷ èø pe pe along negative Xaxis. DPP/ P 33 93 C A B 60° O 60° 60° 60° 30° 30° 120° q/3 q/3 –2q/3 The potential energy of the system is non zero Force between B & C = 2 22 0 0 1 (q / 3) ( 2q / 3) q 4 ( 3R) 54 R  = pe pe Potential at O = 0 1 q q 2q 0 4 3 33 æö += ç÷ èø pe 24. (d) The given graph is of charged conducting sphere of radius R 0 . The whole charge q distributes on the sur face of sphere 25 (b), 26 (b), 27 c + – F + F – (F + > F – as E + > E – ) Net force F : F – F + F y x Net torque immediately after it is released Þ clockwise A body cannot exert force on itself. 28. (d) When the bird perches on a single high power line, no current passes through its body because its body is at equipotential surface i.e., there is no potential difference. While when man touches the same line, standing bare foot on ground the electrical circuit is completed through the ground. The hands of man are at high potential and his feet’s are at low potential. Hence large amount of current flows through the body of the man and person therefore gets a fatal shock. 29. (a) Electron has negative charge, in electric field negative charge moves from lower potential to higher potential. 30. (b) Potential is constant on the surface of a sphere so it behaves as an equipotential surface. Free charges (electrons) are available in conductor. 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1. What is electrostatics? 
2. What are the types of electric charges? 
3. What is an electric field? 
4. How does the distance between charges affect the electric force? 
5. What is Coulomb's Law? 

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