jee-main-electrostatics-important-questions-merged-pages-2

Prévia do material em texto

8 
 
one ring to that of the other 
is 
 (a) zero 
 (b) q(Q1 Q2) ( 02 1) / 2 (4 R  ) 
 (c) 1 2 0q 2(Q Q ) / (4 R) 
 
 (d) 1 2 0q(Q Q )( 2 1) / 2(4 R)  
 
Explanation: VB = 2 1 1 2
A
KQ KQ KQ KQ
, V
R R2R 2R
   
  VA – VB = 1 2KQ KQ1 1
1 1
R R2 2
        
   
 
  VA–VB = 1 2
K 1
1 (Q Q )
R 2
   
 
, where 
0
1
K
4


 
  W = q (VA – VB) 
  (b) 
 
10. A and B are two concentric spheres If 
A is given a charge Q while B is 
earthed as shown in figure, then 
 (A) the charge densities of A and B are 
same 
 B 
A 
+ + 
+ 
+ 
+ 
+ + + 
+ 
+ + 
 
 
9 
 
 (B) the field inside and outside A is 
zero 
 (C) the field between A and B is not 
zero 
 (D) the field inside and outside B is 
zero 
Explanation: C 
 Since, B is grounded, therefore VB = 
0 
A BkQ kQ
R R
 = 0 (R = radius of shell B) 
 QB =  QA 
 Now take 
 QA = Q 
B = 
2
Q
4 R
 
But A = 
2
Q
4 r
 
 (A) is not correct 
 Apply Gauss's theorem. Only (C) is 
correct. 
 
 
10 
 
11. Electric charges q, q and 2q are placed at the corners of an 
equilateral triangle ABC of side L. The magnitude of 
electric dipole moment of the system is 
 (A) qL (B) 2qL 
 (C) (3)qL (D) 4qL 
Explanation: C 
 As shown the three charges 
are equivalent to two dipoles 
of magnitude q L. 
  Equivalent dipole 
moment 
= 2 2(qL) (qL) 2(qL)(qL)cos60   = 3 
qL 
 
 q 
q 
-q 
-q 
qL 
qL 
60 
 
12. If charges q/2 and 2q are placed at the centre of face and at 
the corner of a cube, then the total flux through the cube will 
be 
 (A) 
o
q
2
 (B) 
o
q

 
 (C) 
o
q
6
 (D) 
o
q
8
 
Explanation: A 
 
11 
 
 Flux through the cube when q/2 is placed at the centre of face 
 1
0 0
q 2 q
2 4
  
 
 
 Flux through the cube when 2q is placed at the corner of cube 
is 
 2
0 0
2q q
8 4
  
 
 
 Now, total flux through the cube 
 1 2
0
q
2
     

 
13. A solid insulating sphere of radius R is given a charge Q. If 
at a point inside the sphere the potential is 1.5 times the 
potential at the surface, this point will be 
 (A) at a distance of 2R/3 from the centre 
 (B) at the centre 
 (C) at a distance of 2R/3 from the surface 
 (D) data insufficient 
Explanation: B 
  2 2
in 3
0
Q 3R r1
V
4 2R



 
 s
0
1 Q
V
4 R


 
 
12 
 
 Given in s
3
V V
2
 
  
2 2
3
3 1 3R r
2 R 2R

 
 or r = 0 
 
14. A charge +q is fixed at each of the points x = x0, x = 3x0, x 
= 5x0, . . ., on the x-axis and a charge q is fixed at each of 
the points x = 2x0, x = 4x0, x = 6x0, . . ., . Here, x0 is a 
positive quantity. Take the electric potential at a point due to 
charge Q at a distance r from it to be
0
Q
4 r
. Then, the potential 
at origin due to the above system of charges is 
 (A) 0 (B)  
 (C) 
0 0
q
8 x ln 2
 (D) 
0 0
q ln 2
4 x
 
Explanation: D 
 V = 
0 0 0 0 0
q 1 1 1 1
4 x 2x 3x 4x
 
      
 = 
0 0 0 0
q 1 1 1 q
1 ln(2)
4 x 2 3 4 4 x
        
 
 
 
 
 
 
13 
 
16. Find the charge on an iron particle of mass 2.24 mg, if 0.02% 
of electrons are removed from it. 
 (A) 0.01996 C (B) 0.01996 C 
 (C) 0.02 C (D) 2.0 C 
Explanation: B 
 As mass
Atomic wt.
 = No. of atoms
Avogadro No.
 
 No. of atoms = 24  1018 atoms 
 = 24  1018 26 electrons. 
 n = No. of electrons removed = 24  1018 26  0.01
100
 = 1248 
 1014 electron 
 Q = ne = (+ve charge) = 0.01996 C 
 
17. Two small metallic spheres each of mass ‘m’ are 
suspended together with strings of length ‘’ and 
placed together. When a quantum of charge ‘q’ is 
transferred to each, the strings make an angle of 
90 with each other. The value of ‘q’ is 
 (A)  0mg (B)  02 mg 
 (C)  04 mg (D)  08 mg 
 
 
 
Explanation: D 
 
14 
 
 T cos 45 = FE 
 =
2
2 2
0
Q
4 (2 ) cos 45 
2 2
2 2
0 0
q q
16 (1/ 2) 8
 
  
 
 T sin 45 = mg 
 1 = 
2
0
2
mg8
q
  
  q =  08 mg . 
 
45
 
45
 
FE FE 
mg mg 
 cos 45 
T 
T 
 
18. Two connected charges of +q and q are at a 
fixed distance AB apart in a non-uniform 
electric field, whose lines of force are shown 
in the figure. The resultant effect on the two 
charges is 
 (A) a torque in the plane of the paper and no 
resultant force. 
 (B) a resultant force in the plane of the 
paper and no torque. 
 
 
A 
B 
-q 
+q 
 
 (C) a torque normal to the plane of the paper and no 
resultant force. 
 (D) a torque normal to the plane of the paper and a resultant 
force in the plane of the paper. 
Explanation: B