jee-main-electrostatics-important-questions-merged-pages-4

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22 
 
25. A particle of positive charge Q = 8q0, is having a fixed 
position P. Another charged particle of mass m and charge q 
= 10 C moves at a constant speed in a circle of radius r1 = 
2 cm with centre at P. Find the work that must be done to 
increase the radius of circle to r2 = 4 cm. 
Explanation: 
 Let q orbit round Q in a circle of radius r. 
 K.E. of orbiting particle = 1
2
 mv2. ... (i) 
 Where, v is orbital velocity. 
 Potential energy of q = 
o
1 qQ
4 r


 .... (ii) 
 P.E. is negative since q is negative. 
 Electrostatic attraction on q = 
2
o
1 Qq
4 r
 .... (iii) 
 This is used as centripetal force required for circular motion. 
 
2mv
r
 = 
2
o
1 Qq
4 r


 
 2
o
1 Qq
mv
4 r


 .... (iv) 
 From (1) and (4) 
 K.E. = 21
mv
2
= 
o
1 Qq
4 2r


 
 Total energy of the orbiting charge 
 
23 
 
 = K.E. + P.E.= 1
2 o o
1 Qq 1 Qq
4 r 4 r
 
    
 = – 1
2
. 
o
1 Qq
4 r

 
 The total energy of q when in orbit of radius r1 
 E1= – 1
2
.
o 1
1 Qq
4 r

 
 When it is in orbit of radius r2 
 E2= – 1
2
.
o 2
1 Qq
4 r

 
 The work done on q = change in energy 
 = E2  E1 
 = – 1
2
.
o 2 o 1
1 Qq 1 1 Qq
4 r 2 4 r
 
        = 
o 1 2
Qq 1 1
8 r r
 
   
 
 = 
2
0
0 1 2
8q 1 1
8 r r
 
   
= 90 J 
 
 
 
26. A ball of mass m = 100 gm with a 
charge q can rotate in a vertical plane at 
the end of a string of length l = 1 m in 
a uniform electrostatic field whose 
lines of force are directed upwards. 
What horizontal velocity must be 
 
E 
B 
v1 
T1 
mg 
T2 
qE 
A v2 
mg 
 
24 
 
imparted to the ball in the upper 
position so that the tension in the string 
in the lower position of the ball is 15 
times the weight of the ball? (given qE 
= 3 mg) 
 
 
Explanation: 
 As per principle of conservation of energy, 
 K.E. at B + P.E. at B = K.E. at A + P.E. at A. 
 Gain in K.E. = K.E. at A  K.E. at B 
 =  2 2
2 1
1
m v v
2
 … (i) 
 Loss in P.E. = P.E. at B  P.E. at A. 
 = loss in gravitational P.E. at B  gain in electrical energy at 
A 
 = mg(2 l )  (qE)  2l = (mg qE)2 l … (ii) 
 P.E. at B  P.E. at A = K.E. at A  K.E. at B 
 i.e., (mg qE) 2 l =  2 2
2 1
1
m v v
2
 … (iii) 
 Centripetal force at A = 
2
2mv

 = (T2 + qE mg) … (iv) 
 
25 
 
 From equation (3) 2
2mv = 2(mg qE)2 l + 2
1mv 
 From equation (4) 2
2mv = l (T2+ qE mg) 
 i.e., 2(mg qE)2 l + 2
1mv = l (T2 + qE mg) 
 i.e., 4 mg  4qE + 
2
1mv
l
= T2 + qE mg … (v) 
 Given in problem, T2= 15 mg 
  4 mg  4qE + 2
1
m
v
 
 
 l
 = 15 mg + qE mg 
 or 2
1
m
v
 
 
 l
 = 10 mg + 5qE 
 or 2
1v
m

l (10 mg + 5qE) 
 Horizontal velocity to be imparted to the ball, 
    
1/ 2
1
l
v (10 mg 5 qE)
m
= 50 m/s. 
 
 
27. Two capacitors are first connected in parallel and then in 
series. If the equivalent capacitances in the two cases are 16 
F and 3 F, respectively, then capacitance of each capacitor is 
 (A) 16 F, 3 F (B) 12 F, 4 F 
 (C) 6 F, 8 F (D) none of the above 
Explanation: B 
 
26 
 
 Let C1 and C2 be the individual capacitance 
then C1 + C2 = 16 . . . .(i) 
1 2
1 2
C C
C C
 = 3 . . . . (ii) 
From (i) and (ii) C1 = 12 F, C2 = 4F 
 
28. Two dielectrics of equal size are 
inserted inside a parallel plate capacitor 
as shown. By what factor does the 
effective capacitance increase? 
 (A) 1 2
1 2
k k
k k
 (B) 1 2k k
2
 
 
A 
k1 k2 d 
 
 (C) 1 2
1 2
2k k
k k
 (D) none of the above 
 
 
 
Explanation: B 
 Here two capacitors are formed, which are in parallel 
 Area of each = A/2 and thickness = d 
 Total capacitance 
c = 0 1 0 2 0 1 2k (A / 2) k (A / 2) A k k
d d d 2
        
 
 . 
 
27 
 
  Total capacitance increases by the factor 1 2k k
2
 
 
29. Two charged balls are attached by silk threads of length l to 
the same point. Their velocity is K
x
, where K is a constant 
and x is the distance between the balls, x is very small in 
comparison to l. Find the rate of leakage of charge in 10–5 
C/s. (take l
10
mg
 , k = 4 2 ) 
Explanation: 
Let T be the tension in each of the silk 
threads. 
 T sin  = F, T cos = mg 
 tan  = 
2
2
0
F q 1
mg 4 mg
 
 x
 
 Since  is small, tan  = sin  = 
2
x
l
 
 O 
 
F F 
mg mg 
x 
 
 x = 
2
2
0
2 F 2 q
mg mg 4
 

l l
x
 
 
2
3 2
0 0
2 q
q
mg 4 2 mg
   
 
l l
x 
 x = 
1/3
2/3
0
q
2 mg
 
  
l … (i) 
 
1/3 1
3
0
dx d dt 2
q
dq dq dt 2 mg 3
 
    
x l 
 
28 
 
 
1/3
1/3
0
dq d 2
q
dt 3 2 mgdt
 
   
x l … (ii) 
 It is given, 
1/6
1/3
0
d K K
dt
q
2 mg
 
 
  
x
x l
 … (iii) 
 From equations (2) and (3), we get, 
1/2
0
dq K
dt 2
3 2 mg

 
  
l
 = 
1/2
02 mg3K
2
    l
2  10–5 
  20 C/s 
 
 
 
30. Find the electric flux crossing the wire 
frame ABCD of length 
l = 1m width b and whose center is at 
a distance OP = d (=b/2) from an 
infinite line of charge with linear 
charge density 
 =  10–9 c/m. Consider that the 
plane of frame is perpendicular to line 
OP. 
 
 
D 
C 
l 
b 
P 
d 
B 
A 
O