jee-main-electrostatics-important-questions-merged-pages-3

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15 
 
 UP =  0
0
1 1 1
qq
4 2R R
  
  
 = 0
0
qq1
4 2R
 
 Change in potential energy = 
gain in kinetic energy = 
0
0
qq1
4 2R
 
 
 
O 
R 
q 
3 R 
q0 
P 
 
 
19. A ring of radius R carries a charge +q. A test charge q0 is 
released on its axis at a distance 3R from its centre. How 
much kinetic energy will be acquired by the test charge when 
it reaches the centre of the ring? 
 (A) 0
0
qq1
4 R
 (B) 0
0
qq1
4 2R
 
   
 
 (C) 0
0
qq1
4 3R
 
   
 (D) 0
0
qq1
4 3R
 
   
 
Explanation: C 
 
3
q
E k r
r

  
 Electric field due to 10–9 C at (3, 1, 1) 
 
9
1
10
E k
11 11

 
  ˆ ˆ ˆ3i i k  
 Electric field due to Q at (3, 1, 1) 
 2
Q ˆ ˆ ˆE k (i i k)
3 3
   
 
 Ex = 0 
 
16 
 
 
93 10 Q
11 11 3 3

 = 0 
  Q = – 4.3  10–10C 
 
20. The figure shows a spherical capacitor with 
inner sphere earthed. The capacitance of the 
system is 
 (A) 4 ab
b a


 
(B) 
24 b
b a


 
 (C) 4 0 (b + a) 
(D) none of the above 
 
 a 
b 
 
 
Explanation: B 
 The potential on the outer 
sphere is V (assume). Thus we 
can consider two capacitors 
between the outer sphere and 
inner sphere C1 and outer sphere 
and earth C2. These two 
capacitors are in parallel. 
a b
1
C1C2
C22
C1
 
 
17 
 
 Thus, C1 = 40
ab
b a
 
 C2 = 40 b 
 C = 40
ab
b a
 + 40b 
  C = 
24 b
b a


 
 
INTEGERS TYPE QUESTIONS 
21. An electron is projected with a 
velocity of 1.186  107 m/s, at an 
angle  with the x-axis, towards a 
large metallic plate kept 0.44 mm 
away from the electron. The plate has 
a surface charge density 2  106 
C/m2. Find the minimum value of , 
for which it fails to strike the plate. 
 
 
0.44 mm 
v 
x 
y 
O 
 
 
Explanation: 
 60 
 E = 0  , F = 0e  
 By law of conservation of energy, 
 2 2 2
0
1 1 e x
mv mv sin
2 2

  

 
 
18 
 
  2 2
0
1 e x
mv cos
2

 

 
 Putting the values of , m, v and 0 , we get 
 cos2 = 0.248 
 cos = 0.5 
 andmin = 60 
 
22. A small ball of mass 2  103 kg having a charge of 1 C is 
suspended by a string of length 0.8 m. Another identical ball 
having the same charge is kept at the point of suspension. 
Determine the minimum horizontal velocity which should be 
imparted to the lower ball so that it can make complete 
revolution in a vertical circle. [Take g = 10 m/s2.] 
Explanation: 5.96 m/s 
 For looping the loop, at the highest 
point, 
 T + mg 
2 2
0
q mv
4

  
 
  v2 = 3.5 (m/s)2 . . . 
(i) 
 
mg 
Fe 
v 
 
q 
(m, q)u 
 
19 
 
 But the electrostatic work done when 
particle goes from bottom to top = 0 
 Hence, 1
2
mu2 = 1
2
mv2 + 2mg 
  u = 35.5 m/s = 5.96 m/s 
 
 
23. In the adjacent circuit, Each 
capacitor has a capacitance of 5 
F. Find the charge that will 
flow through MN when the 
switch S is closed. 
 
 C S 
C C 
M N 50 V 
Explanation: 
 When S is open: 
 Equivalent capacitance = C1 = 2C
3
 
 Q1 = C1E = 2C
3
E 
When S is closed: 
Equivalent capacitance = C2 = 2C 
 Q2 = C2E = 2CE 
 Charge flowing through MN = | Q1 Q2 | = | 2C
E 2CE
3
 | 
 
20 
 
 = 4
3
CE = 4
3
 5  50 C = 333.3 C 
24. Four charges + q, + q, – q and – q are 
placed respectively at the corners A, B, 
C and D of a square of side a =  5 1 cm 
arranged in the given order. If E and P 
are the midpoints of sides BC and CD 
respectively, what will be the work 
done in carrying a charge q0 from O to 
E and from O to P? 
 (take q = 10 C, q0 = 5 C) 
 
A 
D C 
B 
+q +q 
–q –q 
O E 
P 
Explanation: 
 ABCD is the given square of side a. The charges are placed 
at the corners as shown. O is the midpoint of square. 
 OA = OB = OC = OD = r (say) = a
2
 
 Potential at O due to the charges at the corners = 
o
1 q q q q
4 r r r r
 
     
 = 0 
 Therefore, O is at zero potential. The electric field at O due 
to charge at A 
 = 
2
o
1 q
4 r


 along OC. 
 To find the work done in carrying a charge e from O to E 
 
21 
 
 Potential at O = 0 
 Potential at E =
o
q 1 1 1 1
4 AE BE DE CE
 
     
 
 Since, AE = DE and BE = CE the summation in bracket 
vanishes. So potential at E = 0. 
 Hence no work is done in moving the charge from O to E. 
 To find the work done in carrying the charge from O to P 
 Potential at P = 
o
q 1 1 1 1
4 AP BP DP CP
 
     
 = 
o
2q 1 1
4 AP DP
    
 
 Now, AP= 2 2AD DP = 
2
2 a
a
2
   
 
 = 5
a
2
 
 DP = a
2
 
 Potential at P 
 V =
o
2q 2 2
4 a5a
 
 
   
 = 
o
2 2 52q
4 5a



 = 
o
4(1 5)q
4 5a


 
 Potential difference between O and P =  
o
4 1 5q
0
4 5a



 = 
 
o
4 5 1q
4 5a


 
 Work done in carrying a charge e from O to P =  
0
4qe 5 11
4 5 a


J = 36 J