Chapter 29 : Electric Field and Potential - HC Verma Solution, Physics Class 11 Notes | EduRev

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JEE : Chapter 29 : Electric Field and Potential - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


29.1
CHAPTER – 29
ELECTRIC FIELD AND POTENTIAL
EXERCISES
1. ?
0
= 
2
2
m Newton
Coulomb
= l
1
M
–1
L
–3
T
4
? F = 
2
2 1
r
q kq
2. q
1
= q
2
= q = 1.0 C distance between = 2 km = 1 × 10
3
m
so, force = 
2
2 1
r
q kq
F = 
2 3
9
) 10 2 (
1 1 ) 10 9 (
?
? ? ?
= 
6 2
9
10 2
10 9
?
?
= 2,25 × 10
3
N
The weight of body = mg = 40 × 10 N = 400 N
So, 
es arg ch between force
body of wt
= 
1
2
3
10 4
10 25 . 2
?
?
?
?
?
?
?
?
?
?
?
= (5.6)
–1
= 
6 . 5
1
So, force between charges = 5.6 weight of body.
3. q = 1 C, Let the distance be ?
F = 50 × 9.8 = 490
F = 
2
2
Kq
?
? 490 = 
2
2 9
1 10 9
?
? ?
or ?
2
= 
490
10 9
9
?
= 18.36 × 10
6 
? ? = 4.29 ×10
3
m ?
4. charges ‘q’ each, AB = 1 m
wt, of 50 kg person = 50 × g = 50 × 9.8 = 490 N
F
C
= 
2
2 1
r
q kq
?
2
2
r
kq
= 490 N
? q
2
= 
9
2
10 9
r 490
?
?
= 
9
10 9
1 1 490
?
? ?
? q = 
9
10 4 . 54
?
? = 23.323 × 10
–5
coulomb ˜ 2.3 × 10
–4
coulomb
5. Charge on each proton = a= 1.6 × 10
–19
coulomb
Distance between charges = 10 × 10
–15
metre = r
Force = 
2
2
r
kq
= 
30
38 9
10
10 6 . 1 6 . 1 10 9
?
?
? ? ? ?
= 9 × 2.56 × 10 = 230.4 Newton
6. q
1
= 2.0 × 10
–6
q
2
= 1.0 × 10
–6
r = 10 cm = 0.1 m
Let the charge be at a distance x from q
1
F
1
= 
2
1
q Kq
?
F
2
= 
2
2
) 1 . 0 (
kqq
? ?
= 
2
9 6
q 10 10 2 9 . 9
?
? ? ? ?
?
Now since the net force is zero on the charge q. ? f
1
= f
2
?
2
1
q kq
?
= 
2
2
) 1 . 0 (
kqq
? ?
? 2(0.1 – ?)
2
= ?
2
  ? 2 (0.1 – ?) = ?
? ? = 
2 1
2 1 . 0
?
= 0.0586 m = 5.86 cm ˜ 5.9 cm From larger charge
q 2
(0.1–x) m x m
q
q 1
10 cm
Page 2


29.1
CHAPTER – 29
ELECTRIC FIELD AND POTENTIAL
EXERCISES
1. ?
0
= 
2
2
m Newton
Coulomb
= l
1
M
–1
L
–3
T
4
? F = 
2
2 1
r
q kq
2. q
1
= q
2
= q = 1.0 C distance between = 2 km = 1 × 10
3
m
so, force = 
2
2 1
r
q kq
F = 
2 3
9
) 10 2 (
1 1 ) 10 9 (
?
? ? ?
= 
6 2
9
10 2
10 9
?
?
= 2,25 × 10
3
N
The weight of body = mg = 40 × 10 N = 400 N
So, 
es arg ch between force
body of wt
= 
1
2
3
10 4
10 25 . 2
?
?
?
?
?
?
?
?
?
?
?
= (5.6)
–1
= 
6 . 5
1
So, force between charges = 5.6 weight of body.
3. q = 1 C, Let the distance be ?
F = 50 × 9.8 = 490
F = 
2
2
Kq
?
? 490 = 
2
2 9
1 10 9
?
? ?
or ?
2
= 
490
10 9
9
?
= 18.36 × 10
6 
? ? = 4.29 ×10
3
m ?
4. charges ‘q’ each, AB = 1 m
wt, of 50 kg person = 50 × g = 50 × 9.8 = 490 N
F
C
= 
2
2 1
r
q kq
?
2
2
r
kq
= 490 N
? q
2
= 
9
2
10 9
r 490
?
?
= 
9
10 9
1 1 490
?
? ?
? q = 
9
10 4 . 54
?
? = 23.323 × 10
–5
coulomb ˜ 2.3 × 10
–4
coulomb
5. Charge on each proton = a= 1.6 × 10
–19
coulomb
Distance between charges = 10 × 10
–15
metre = r
Force = 
2
2
r
kq
= 
30
38 9
10
10 6 . 1 6 . 1 10 9
?
?
? ? ? ?
= 9 × 2.56 × 10 = 230.4 Newton
6. q
1
= 2.0 × 10
–6
q
2
= 1.0 × 10
–6
r = 10 cm = 0.1 m
Let the charge be at a distance x from q
1
F
1
= 
2
1
q Kq
?
F
2
= 
2
2
) 1 . 0 (
kqq
? ?
= 
2
9 6
q 10 10 2 9 . 9
?
? ? ? ?
?
Now since the net force is zero on the charge q. ? f
1
= f
2
?
2
1
q kq
?
= 
2
2
) 1 . 0 (
kqq
? ?
? 2(0.1 – ?)
2
= ?
2
  ? 2 (0.1 – ?) = ?
? ? = 
2 1
2 1 . 0
?
= 0.0586 m = 5.86 cm ˜ 5.9 cm From larger charge
q 2
(0.1–x) m x m
q
q 1
10 cm
Electric Field and Potential
29.2
7. q
1
= 2 ×10
–6
c q
2
= – 1 × 10
–6
c r = 10 cm = 10 × 10
–2
m
Let the third charge be a so, F-
AC
= – F-
BC
?
2
1
1
r
kQq
= 
2
2
2
r
KQq ?
?
2
6
) 10 (
10 2
? ?
?
?
= 
2
6
10 1
?
?
?
? 2 ?
2
= (10 + ?)
2
? 2 ? = 10 + ? ? ?( 2 - 1) = 10 ? ? =
1 414 . 1
10
?
?
= 24.14 cm ?
So, distance = 24.14 + 10 = 34.14 cm from larger charge ?
8. Minimum charge of a body is the charge of an electron
Wo, q = 1.6 × 10
–19
c ? = 1 cm = 1 × 10
–2
cm 
So, F = 
2
2 1
r
q kq
= 
2 2
19 19 9
10 10
10 10 6 . 1 6 . 1 10 9
? ?
? ?
?
? ? ? ? ?
= 23.04 × 10
–38+9+2+2
= 23.04 × 10
–25
= 2.3 × 10
–24
?
9. No. of electrons of 100 g water = 
18
100 10 ?
= 55.5 Nos Total charge = 55.5
No. of electrons in 18 g of H
2
O = 6.023 × 10
23
× 10 = 6.023 × 10
24
No. of electrons in 100 g  of H
2
O = 
18
100 10 023 . 6
24
? ?
= 0.334 × 10
26
= 3.334 × 10
25
Total charge = 3.34 × 10
25
× 1.6 × 10
–19
= 5.34 × 10
6
c
10. Molecular weight of H
2
O = 2 × 1 × 16 = 16
No. of electrons present in one molecule of H
2
O = 10
18 gm of H
2
O has 6.023 × 10
23
molecule
18 gm of H
2
O has 6.023 × 10
23
× 10 electrons
100 gm of H
2
O has 100
18
10 023 . 6
24
?
?
electrons
So number of protons = 
18
10 023 . 6
26
?
protons (since atom is electrically neutral)
Charge of protons = 
18
10 023 . 6 10 6 . 1
26 19
? ? ?
?
coulomb = 
18
10 023 . 6 6 . 1
7
? ?
coulomb
Charge of electrons = = 
18
10 023 . 6 6 . 1
7
? ?
coulomb
Hence Electrical force = 
2 2
7 7
9
) 10 10 (
18
10 023 . 6 6 . 1
18
10 023 . 6 6 . 1
10 9
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
? ?
?
= 
25
10 023 . 6 6 . 1
18
023 . 6 8
? ? ?
?
= 2.56 × 10
25
Newton
11. Let two protons be at a distance be 13.8 femi
F = 
30 2
38 9
10 ) 8 . 14 (
10 6 . 1 10 9
?
?
?
? ? ?
= 1.2 N
12. F = 0.1 N
r = 1 cm = 10
–2
(As they rubbed with each other. So the charge on each sphere are equal)
So, F = 
2
2 1
r
q kq
? 0.1 = 
2 2
2
) 10 (
kq
?
? q
2
= 
9
4
10 9
10 1 . 0
?
?
?
? q
2
= 
14
10
9
1
?
? ? q = 
7
10
3
1
?
?
1.6 × 10
–19
c Carries by 1 electron 1 c carried by 
19
10 6 . 1
1
?
?
0.33 × 10
–7
c carries by 
7
19
10 33 . 0
10 6 . 1
1
?
?
? ?
?
= 0.208 × 10
12
= 2.08 × 10
11
+
+
+
–
–
–
–
+
+
A
10 × 10
–10
m
C
B
a
2 × 10
–6
c
–1 × 10
–6
c
Page 3


29.1
CHAPTER – 29
ELECTRIC FIELD AND POTENTIAL
EXERCISES
1. ?
0
= 
2
2
m Newton
Coulomb
= l
1
M
–1
L
–3
T
4
? F = 
2
2 1
r
q kq
2. q
1
= q
2
= q = 1.0 C distance between = 2 km = 1 × 10
3
m
so, force = 
2
2 1
r
q kq
F = 
2 3
9
) 10 2 (
1 1 ) 10 9 (
?
? ? ?
= 
6 2
9
10 2
10 9
?
?
= 2,25 × 10
3
N
The weight of body = mg = 40 × 10 N = 400 N
So, 
es arg ch between force
body of wt
= 
1
2
3
10 4
10 25 . 2
?
?
?
?
?
?
?
?
?
?
?
= (5.6)
–1
= 
6 . 5
1
So, force between charges = 5.6 weight of body.
3. q = 1 C, Let the distance be ?
F = 50 × 9.8 = 490
F = 
2
2
Kq
?
? 490 = 
2
2 9
1 10 9
?
? ?
or ?
2
= 
490
10 9
9
?
= 18.36 × 10
6 
? ? = 4.29 ×10
3
m ?
4. charges ‘q’ each, AB = 1 m
wt, of 50 kg person = 50 × g = 50 × 9.8 = 490 N
F
C
= 
2
2 1
r
q kq
?
2
2
r
kq
= 490 N
? q
2
= 
9
2
10 9
r 490
?
?
= 
9
10 9
1 1 490
?
? ?
? q = 
9
10 4 . 54
?
? = 23.323 × 10
–5
coulomb ˜ 2.3 × 10
–4
coulomb
5. Charge on each proton = a= 1.6 × 10
–19
coulomb
Distance between charges = 10 × 10
–15
metre = r
Force = 
2
2
r
kq
= 
30
38 9
10
10 6 . 1 6 . 1 10 9
?
?
? ? ? ?
= 9 × 2.56 × 10 = 230.4 Newton
6. q
1
= 2.0 × 10
–6
q
2
= 1.0 × 10
–6
r = 10 cm = 0.1 m
Let the charge be at a distance x from q
1
F
1
= 
2
1
q Kq
?
F
2
= 
2
2
) 1 . 0 (
kqq
? ?
= 
2
9 6
q 10 10 2 9 . 9
?
? ? ? ?
?
Now since the net force is zero on the charge q. ? f
1
= f
2
?
2
1
q kq
?
= 
2
2
) 1 . 0 (
kqq
? ?
? 2(0.1 – ?)
2
= ?
2
  ? 2 (0.1 – ?) = ?
? ? = 
2 1
2 1 . 0
?
= 0.0586 m = 5.86 cm ˜ 5.9 cm From larger charge
q 2
(0.1–x) m x m
q
q 1
10 cm
Electric Field and Potential
29.2
7. q
1
= 2 ×10
–6
c q
2
= – 1 × 10
–6
c r = 10 cm = 10 × 10
–2
m
Let the third charge be a so, F-
AC
= – F-
BC
?
2
1
1
r
kQq
= 
2
2
2
r
KQq ?
?
2
6
) 10 (
10 2
? ?
?
?
= 
2
6
10 1
?
?
?
? 2 ?
2
= (10 + ?)
2
? 2 ? = 10 + ? ? ?( 2 - 1) = 10 ? ? =
1 414 . 1
10
?
?
= 24.14 cm ?
So, distance = 24.14 + 10 = 34.14 cm from larger charge ?
8. Minimum charge of a body is the charge of an electron
Wo, q = 1.6 × 10
–19
c ? = 1 cm = 1 × 10
–2
cm 
So, F = 
2
2 1
r
q kq
= 
2 2
19 19 9
10 10
10 10 6 . 1 6 . 1 10 9
? ?
? ?
?
? ? ? ? ?
= 23.04 × 10
–38+9+2+2
= 23.04 × 10
–25
= 2.3 × 10
–24
?
9. No. of electrons of 100 g water = 
18
100 10 ?
= 55.5 Nos Total charge = 55.5
No. of electrons in 18 g of H
2
O = 6.023 × 10
23
× 10 = 6.023 × 10
24
No. of electrons in 100 g  of H
2
O = 
18
100 10 023 . 6
24
? ?
= 0.334 × 10
26
= 3.334 × 10
25
Total charge = 3.34 × 10
25
× 1.6 × 10
–19
= 5.34 × 10
6
c
10. Molecular weight of H
2
O = 2 × 1 × 16 = 16
No. of electrons present in one molecule of H
2
O = 10
18 gm of H
2
O has 6.023 × 10
23
molecule
18 gm of H
2
O has 6.023 × 10
23
× 10 electrons
100 gm of H
2
O has 100
18
10 023 . 6
24
?
?
electrons
So number of protons = 
18
10 023 . 6
26
?
protons (since atom is electrically neutral)
Charge of protons = 
18
10 023 . 6 10 6 . 1
26 19
? ? ?
?
coulomb = 
18
10 023 . 6 6 . 1
7
? ?
coulomb
Charge of electrons = = 
18
10 023 . 6 6 . 1
7
? ?
coulomb
Hence Electrical force = 
2 2
7 7
9
) 10 10 (
18
10 023 . 6 6 . 1
18
10 023 . 6 6 . 1
10 9
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
? ?
?
= 
25
10 023 . 6 6 . 1
18
023 . 6 8
? ? ?
?
= 2.56 × 10
25
Newton
11. Let two protons be at a distance be 13.8 femi
F = 
30 2
38 9
10 ) 8 . 14 (
10 6 . 1 10 9
?
?
?
? ? ?
= 1.2 N
12. F = 0.1 N
r = 1 cm = 10
–2
(As they rubbed with each other. So the charge on each sphere are equal)
So, F = 
2
2 1
r
q kq
? 0.1 = 
2 2
2
) 10 (
kq
?
? q
2
= 
9
4
10 9
10 1 . 0
?
?
?
? q
2
= 
14
10
9
1
?
? ? q = 
7
10
3
1
?
?
1.6 × 10
–19
c Carries by 1 electron 1 c carried by 
19
10 6 . 1
1
?
?
0.33 × 10
–7
c carries by 
7
19
10 33 . 0
10 6 . 1
1
?
?
? ?
?
= 0.208 × 10
12
= 2.08 × 10
11
+
+
+
–
–
–
–
+
+
A
10 × 10
–10
m
C
B
a
2 × 10
–6
c
–1 × 10
–6
c
Electric Field and Potential
29.3
13. F = 
2
2 1
r
q kq
= 
2 10
19 19 9
) 10 75 . 2 (
10 10 6 . 1 6 . 1 10 9
?
? ?
?
? ? ? ? ?
= 
20
29
10 56 . 7
10 04 . 23
?
?
?
?
= 3.04 × 10
–9
14. Given: mass of proton = 1.67 × 10
–27
kg = M
p
k = 9 × 10
9
Charge of proton = 1.6 × 10
–19
c = C
p
G = 6.67 × 10
–11
Let the separation be ‘r’ 
Fe = 
2
2
p
r
) C ( k
, fg= 
2
2
p
r
) M ( G
Now, Fe : Fg = 
2
p
2
2
2
p
) M ( G
r
r
) C ( K
? = 
2 27 11
2 19 9
) 10 67 . 1 ( 10 67 . 6
) 0 ` 1 6 . 1 ( 10 9
? ?
?
? ? ?
? ? ?
= 9 × 2.56 × 10
38
˜ 1,24 ×10
38
15. Expression of electrical force F = 
2
r
kr
e C
?
?
Since e
–kr
is a pure number. So, dimensional formulae of F = 
2
r of formulae ensional dim
C of formulae ensional dim
Or, [MLT
–2
][L
2
] = dimensional formulae of C = [ML
3
T
–2
]
Unit of C = unit of force × unit of r
2
= Newton × m
2
= Newton–m
2
  
Since –kr is a number hence dimensional formulae of 
k = 
r of formulae entional dim
1
= [L
–1
] Unit of k = m
–1
16. Three charges are held at three corners of a equilateral trangle.
Let the charges be A, B and C. It is of length 5 cm or 0.05 m
Force exerted by B on A = F
1
force exerted by C on A = F
2
So, force exerted on A = resultant F
1
= F
2
? F = 
2
2
r
kq
= 
4
12 9
10 5 5
10 2 2 2 10 9
?
?
? ?
? ? ? ? ?
= 10
25
36
? = 14.4
Now, force on A = 2 × F cos 30° since it is equilateral ?.
? Force on A = 2 × 1.44 ×
2
3
= 24.94 N. ?
17. q
1
= q
2
= q
3
= q
4
= 2 × 10
–6
C
v = 5 cm = 5 × 10
–2
m
so force on c = 
CD CB CA
F F F ? ?
so Force along × Component = 0 45 cos F F
CA CD
? ? ?
= 
2 2
1
) 10 5 (
) 10 2 ( k
) 10 5 (
) 10 2 ( k
2 2
2 6
2 2
2 6
?
?
?
?
?
?
?
?
?
= 
?
?
?
?
?
?
?
?
?
?
?
? ? 4 4
2
10 2 50
1
10 25
1
kq
= 
?
?
?
?
?
?
?
?
?
?
? ? ?
?
?
2 2
1
1
10 24
10 4 10 9
4
12 9
= 1.44 (1.35) = 19.49 Force along % component = 19.49
So, Resultant R = 
2 2
Fy Fx ? = 19.49 2 = 27.56 
18. R = 0.53 A° = 0.53 × 10
–10
m
F = 
2
2 1
r
q Kq
= 
10 10
38 9
10 10 53 . 0 53 . 0
10 6 . 1 6 . 1 10 9
? ?
?
? ? ?
? ? ? ?
= 82.02 × 10
–9
N
19. Fe from previous problem No. 18 = 8.2 × 10
–8
N Ve = ?
Now, M
e
= 9.12 × 10
–31
kg r = 0.53 × 10
–10
m
Now, Fe = 
r
v M
2
e
? v
2
= 
e
m
r Fe ?
= 
31
10 8
10 1 . 9
10 53 . 0 10 2 . 8
?
? ?
?
? ? ?
= 0.4775 × 10
13
= 4.775 × 10
12
m
2
/s
2
? v = 2.18 × 10
6
m/s
0.05 m
60°
? ?
F 2 F 1
B
A
C
0.05 m
0.05 m
D
C
B
CD
F
A
CA
F
CB
F
Page 4


29.1
CHAPTER – 29
ELECTRIC FIELD AND POTENTIAL
EXERCISES
1. ?
0
= 
2
2
m Newton
Coulomb
= l
1
M
–1
L
–3
T
4
? F = 
2
2 1
r
q kq
2. q
1
= q
2
= q = 1.0 C distance between = 2 km = 1 × 10
3
m
so, force = 
2
2 1
r
q kq
F = 
2 3
9
) 10 2 (
1 1 ) 10 9 (
?
? ? ?
= 
6 2
9
10 2
10 9
?
?
= 2,25 × 10
3
N
The weight of body = mg = 40 × 10 N = 400 N
So, 
es arg ch between force
body of wt
= 
1
2
3
10 4
10 25 . 2
?
?
?
?
?
?
?
?
?
?
?
= (5.6)
–1
= 
6 . 5
1
So, force between charges = 5.6 weight of body.
3. q = 1 C, Let the distance be ?
F = 50 × 9.8 = 490
F = 
2
2
Kq
?
? 490 = 
2
2 9
1 10 9
?
? ?
or ?
2
= 
490
10 9
9
?
= 18.36 × 10
6 
? ? = 4.29 ×10
3
m ?
4. charges ‘q’ each, AB = 1 m
wt, of 50 kg person = 50 × g = 50 × 9.8 = 490 N
F
C
= 
2
2 1
r
q kq
?
2
2
r
kq
= 490 N
? q
2
= 
9
2
10 9
r 490
?
?
= 
9
10 9
1 1 490
?
? ?
? q = 
9
10 4 . 54
?
? = 23.323 × 10
–5
coulomb ˜ 2.3 × 10
–4
coulomb
5. Charge on each proton = a= 1.6 × 10
–19
coulomb
Distance between charges = 10 × 10
–15
metre = r
Force = 
2
2
r
kq
= 
30
38 9
10
10 6 . 1 6 . 1 10 9
?
?
? ? ? ?
= 9 × 2.56 × 10 = 230.4 Newton
6. q
1
= 2.0 × 10
–6
q
2
= 1.0 × 10
–6
r = 10 cm = 0.1 m
Let the charge be at a distance x from q
1
F
1
= 
2
1
q Kq
?
F
2
= 
2
2
) 1 . 0 (
kqq
? ?
= 
2
9 6
q 10 10 2 9 . 9
?
? ? ? ?
?
Now since the net force is zero on the charge q. ? f
1
= f
2
?
2
1
q kq
?
= 
2
2
) 1 . 0 (
kqq
? ?
? 2(0.1 – ?)
2
= ?
2
  ? 2 (0.1 – ?) = ?
? ? = 
2 1
2 1 . 0
?
= 0.0586 m = 5.86 cm ˜ 5.9 cm From larger charge
q 2
(0.1–x) m x m
q
q 1
10 cm
Electric Field and Potential
29.2
7. q
1
= 2 ×10
–6
c q
2
= – 1 × 10
–6
c r = 10 cm = 10 × 10
–2
m
Let the third charge be a so, F-
AC
= – F-
BC
?
2
1
1
r
kQq
= 
2
2
2
r
KQq ?
?
2
6
) 10 (
10 2
? ?
?
?
= 
2
6
10 1
?
?
?
? 2 ?
2
= (10 + ?)
2
? 2 ? = 10 + ? ? ?( 2 - 1) = 10 ? ? =
1 414 . 1
10
?
?
= 24.14 cm ?
So, distance = 24.14 + 10 = 34.14 cm from larger charge ?
8. Minimum charge of a body is the charge of an electron
Wo, q = 1.6 × 10
–19
c ? = 1 cm = 1 × 10
–2
cm 
So, F = 
2
2 1
r
q kq
= 
2 2
19 19 9
10 10
10 10 6 . 1 6 . 1 10 9
? ?
? ?
?
? ? ? ? ?
= 23.04 × 10
–38+9+2+2
= 23.04 × 10
–25
= 2.3 × 10
–24
?
9. No. of electrons of 100 g water = 
18
100 10 ?
= 55.5 Nos Total charge = 55.5
No. of electrons in 18 g of H
2
O = 6.023 × 10
23
× 10 = 6.023 × 10
24
No. of electrons in 100 g  of H
2
O = 
18
100 10 023 . 6
24
? ?
= 0.334 × 10
26
= 3.334 × 10
25
Total charge = 3.34 × 10
25
× 1.6 × 10
–19
= 5.34 × 10
6
c
10. Molecular weight of H
2
O = 2 × 1 × 16 = 16
No. of electrons present in one molecule of H
2
O = 10
18 gm of H
2
O has 6.023 × 10
23
molecule
18 gm of H
2
O has 6.023 × 10
23
× 10 electrons
100 gm of H
2
O has 100
18
10 023 . 6
24
?
?
electrons
So number of protons = 
18
10 023 . 6
26
?
protons (since atom is electrically neutral)
Charge of protons = 
18
10 023 . 6 10 6 . 1
26 19
? ? ?
?
coulomb = 
18
10 023 . 6 6 . 1
7
? ?
coulomb
Charge of electrons = = 
18
10 023 . 6 6 . 1
7
? ?
coulomb
Hence Electrical force = 
2 2
7 7
9
) 10 10 (
18
10 023 . 6 6 . 1
18
10 023 . 6 6 . 1
10 9
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
? ?
?
= 
25
10 023 . 6 6 . 1
18
023 . 6 8
? ? ?
?
= 2.56 × 10
25
Newton
11. Let two protons be at a distance be 13.8 femi
F = 
30 2
38 9
10 ) 8 . 14 (
10 6 . 1 10 9
?
?
?
? ? ?
= 1.2 N
12. F = 0.1 N
r = 1 cm = 10
–2
(As they rubbed with each other. So the charge on each sphere are equal)
So, F = 
2
2 1
r
q kq
? 0.1 = 
2 2
2
) 10 (
kq
?
? q
2
= 
9
4
10 9
10 1 . 0
?
?
?
? q
2
= 
14
10
9
1
?
? ? q = 
7
10
3
1
?
?
1.6 × 10
–19
c Carries by 1 electron 1 c carried by 
19
10 6 . 1
1
?
?
0.33 × 10
–7
c carries by 
7
19
10 33 . 0
10 6 . 1
1
?
?
? ?
?
= 0.208 × 10
12
= 2.08 × 10
11
+
+
+
–
–
–
–
+
+
A
10 × 10
–10
m
C
B
a
2 × 10
–6
c
–1 × 10
–6
c
Electric Field and Potential
29.3
13. F = 
2
2 1
r
q kq
= 
2 10
19 19 9
) 10 75 . 2 (
10 10 6 . 1 6 . 1 10 9
?
? ?
?
? ? ? ? ?
= 
20
29
10 56 . 7
10 04 . 23
?
?
?
?
= 3.04 × 10
–9
14. Given: mass of proton = 1.67 × 10
–27
kg = M
p
k = 9 × 10
9
Charge of proton = 1.6 × 10
–19
c = C
p
G = 6.67 × 10
–11
Let the separation be ‘r’ 
Fe = 
2
2
p
r
) C ( k
, fg= 
2
2
p
r
) M ( G
Now, Fe : Fg = 
2
p
2
2
2
p
) M ( G
r
r
) C ( K
? = 
2 27 11
2 19 9
) 10 67 . 1 ( 10 67 . 6
) 0 ` 1 6 . 1 ( 10 9
? ?
?
? ? ?
? ? ?
= 9 × 2.56 × 10
38
˜ 1,24 ×10
38
15. Expression of electrical force F = 
2
r
kr
e C
?
?
Since e
–kr
is a pure number. So, dimensional formulae of F = 
2
r of formulae ensional dim
C of formulae ensional dim
Or, [MLT
–2
][L
2
] = dimensional formulae of C = [ML
3
T
–2
]
Unit of C = unit of force × unit of r
2
= Newton × m
2
= Newton–m
2
  
Since –kr is a number hence dimensional formulae of 
k = 
r of formulae entional dim
1
= [L
–1
] Unit of k = m
–1
16. Three charges are held at three corners of a equilateral trangle.
Let the charges be A, B and C. It is of length 5 cm or 0.05 m
Force exerted by B on A = F
1
force exerted by C on A = F
2
So, force exerted on A = resultant F
1
= F
2
? F = 
2
2
r
kq
= 
4
12 9
10 5 5
10 2 2 2 10 9
?
?
? ?
? ? ? ? ?
= 10
25
36
? = 14.4
Now, force on A = 2 × F cos 30° since it is equilateral ?.
? Force on A = 2 × 1.44 ×
2
3
= 24.94 N. ?
17. q
1
= q
2
= q
3
= q
4
= 2 × 10
–6
C
v = 5 cm = 5 × 10
–2
m
so force on c = 
CD CB CA
F F F ? ?
so Force along × Component = 0 45 cos F F
CA CD
? ? ?
= 
2 2
1
) 10 5 (
) 10 2 ( k
) 10 5 (
) 10 2 ( k
2 2
2 6
2 2
2 6
?
?
?
?
?
?
?
?
?
= 
?
?
?
?
?
?
?
?
?
?
?
? ? 4 4
2
10 2 50
1
10 25
1
kq
= 
?
?
?
?
?
?
?
?
?
?
? ? ?
?
?
2 2
1
1
10 24
10 4 10 9
4
12 9
= 1.44 (1.35) = 19.49 Force along % component = 19.49
So, Resultant R = 
2 2
Fy Fx ? = 19.49 2 = 27.56 
18. R = 0.53 A° = 0.53 × 10
–10
m
F = 
2
2 1
r
q Kq
= 
10 10
38 9
10 10 53 . 0 53 . 0
10 6 . 1 6 . 1 10 9
? ?
?
? ? ?
? ? ? ?
= 82.02 × 10
–9
N
19. Fe from previous problem No. 18 = 8.2 × 10
–8
N Ve = ?
Now, M
e
= 9.12 × 10
–31
kg r = 0.53 × 10
–10
m
Now, Fe = 
r
v M
2
e
? v
2
= 
e
m
r Fe ?
= 
31
10 8
10 1 . 9
10 53 . 0 10 2 . 8
?
? ?
?
? ? ?
= 0.4775 × 10
13
= 4.775 × 10
12
m
2
/s
2
? v = 2.18 × 10
6
m/s
0.05 m
60°
? ?
F 2 F 1
B
A
C
0.05 m
0.05 m
D
C
B
CD
F
A
CA
F
CB
F
Electric Field and Potential
29.4
20. Electric force feeled by 1 c due to 1 × 10
–8
c.
F
1
= 
2 2
8
) 10 10 (
1 10 1 k
?
?
?
? ? ?
= k × 10
-6
N. electric force feeled by 1 c due to 8 × 10
–8
c.
F
2
= 
2 2
8
) 10 23 (
1 10 8 k
?
?
?
? ? ?
= 
9
10 10 8 k
2 8
? ? ? ?
?
= 
4
10 k 28
6 ?
?
= 2k × 10
–6
N.
Similarly F
3
= 
2 2
8
) 10 30 (
1 10 27 k
?
?
?
? ? ?
= 3k × 10
–6
N
So, F = F
1
+ F
2
+ F
3
+ ……+ F
10
= k × 10
–6
(1 + 2 + 3 +……+10) N
= k × 10
–6
× 
2
11 10 ?
= 55k × 10
–6
= 55 × 9 × 10
9
× 10
–6
N = 4.95 × 10
3
N 
21. Force exerted = 
2
2
1
r
kq
= 
2
16 9
1
10 2 2 10 9
?
? ? ? ?
= 3.6 × 10
–6
is the force exerted on the string
22. q
1
= q
2
= 2 × 10
–7
c m = 100 g
l = 50 cm = 5 × 10
–2
m d = 5 × 10
–2
m
(a) Now Electric force 
F = 
2
2
r
q
K = 
4
14 9
10 25
10 4 10 9
?
?
?
? ? ?
N = 14.4 × 10
–2
N = 0.144 N
(b) The components of Resultant force along it is zero, 
because mg balances T cos ? and so also.
F = mg = T sin ??
(c) Tension on the string
T sin ? = F T cos ? = mg
Tan ? = 
mg
F
= 
8 . 9 10 100
144 . 0
3
? ?
?
= 0.14693
But T cos ? = 10
2
× 10
–3
× 10 = 1 N
? T = 
? cos
1
= sec ? ? ?
? T = 
? sin
F
, ?
Sin ? = 0.145369 ; Cos ? = 0.989378; ?
23. q = 2.0 × 10
–8 
c n= ? T = ? Sin ? = 
20
1
Force between the charges 
F = 
2
2 1
r
q Kq
= 
2 2
8 8 9
) 10 3 (
10 2 10 2 10 9
?
? ?
?
? ? ? ? ?
= 4 × 10
–3
N
mg sin ? = F ? m = 
? sin g
F
= 
) 20 / 1 ( 10
10 4
3
?
?
?
= 8 × 10
–3
= 8 gm
Cos ??= ? ?
2
Sin 1 = 
400
1
1 ? = 
400
1 400 ?
= 0.99 ˜ 1
So, T = mg cos ?
Or T = 8 × 10
–3
10 × 0.99 = 8 × 10
–2
M ?
r = 1 m
q 1
q 1
90°
? ?
T
F
T Cos ? ?
??
? ?
90°
F
T Cos ? ?
T Sin ? ?
T Sin ? ?
20 cm
0
1 cm
20 T
mg
5 cm
1 cm 1 cm
T
20
Page 5


29.1
CHAPTER – 29
ELECTRIC FIELD AND POTENTIAL
EXERCISES
1. ?
0
= 
2
2
m Newton
Coulomb
= l
1
M
–1
L
–3
T
4
? F = 
2
2 1
r
q kq
2. q
1
= q
2
= q = 1.0 C distance between = 2 km = 1 × 10
3
m
so, force = 
2
2 1
r
q kq
F = 
2 3
9
) 10 2 (
1 1 ) 10 9 (
?
? ? ?
= 
6 2
9
10 2
10 9
?
?
= 2,25 × 10
3
N
The weight of body = mg = 40 × 10 N = 400 N
So, 
es arg ch between force
body of wt
= 
1
2
3
10 4
10 25 . 2
?
?
?
?
?
?
?
?
?
?
?
= (5.6)
–1
= 
6 . 5
1
So, force between charges = 5.6 weight of body.
3. q = 1 C, Let the distance be ?
F = 50 × 9.8 = 490
F = 
2
2
Kq
?
? 490 = 
2
2 9
1 10 9
?
? ?
or ?
2
= 
490
10 9
9
?
= 18.36 × 10
6 
? ? = 4.29 ×10
3
m ?
4. charges ‘q’ each, AB = 1 m
wt, of 50 kg person = 50 × g = 50 × 9.8 = 490 N
F
C
= 
2
2 1
r
q kq
?
2
2
r
kq
= 490 N
? q
2
= 
9
2
10 9
r 490
?
?
= 
9
10 9
1 1 490
?
? ?
? q = 
9
10 4 . 54
?
? = 23.323 × 10
–5
coulomb ˜ 2.3 × 10
–4
coulomb
5. Charge on each proton = a= 1.6 × 10
–19
coulomb
Distance between charges = 10 × 10
–15
metre = r
Force = 
2
2
r
kq
= 
30
38 9
10
10 6 . 1 6 . 1 10 9
?
?
? ? ? ?
= 9 × 2.56 × 10 = 230.4 Newton
6. q
1
= 2.0 × 10
–6
q
2
= 1.0 × 10
–6
r = 10 cm = 0.1 m
Let the charge be at a distance x from q
1
F
1
= 
2
1
q Kq
?
F
2
= 
2
2
) 1 . 0 (
kqq
? ?
= 
2
9 6
q 10 10 2 9 . 9
?
? ? ? ?
?
Now since the net force is zero on the charge q. ? f
1
= f
2
?
2
1
q kq
?
= 
2
2
) 1 . 0 (
kqq
? ?
? 2(0.1 – ?)
2
= ?
2
  ? 2 (0.1 – ?) = ?
? ? = 
2 1
2 1 . 0
?
= 0.0586 m = 5.86 cm ˜ 5.9 cm From larger charge
q 2
(0.1–x) m x m
q
q 1
10 cm
Electric Field and Potential
29.2
7. q
1
= 2 ×10
–6
c q
2
= – 1 × 10
–6
c r = 10 cm = 10 × 10
–2
m
Let the third charge be a so, F-
AC
= – F-
BC
?
2
1
1
r
kQq
= 
2
2
2
r
KQq ?
?
2
6
) 10 (
10 2
? ?
?
?
= 
2
6
10 1
?
?
?
? 2 ?
2
= (10 + ?)
2
? 2 ? = 10 + ? ? ?( 2 - 1) = 10 ? ? =
1 414 . 1
10
?
?
= 24.14 cm ?
So, distance = 24.14 + 10 = 34.14 cm from larger charge ?
8. Minimum charge of a body is the charge of an electron
Wo, q = 1.6 × 10
–19
c ? = 1 cm = 1 × 10
–2
cm 
So, F = 
2
2 1
r
q kq
= 
2 2
19 19 9
10 10
10 10 6 . 1 6 . 1 10 9
? ?
? ?
?
? ? ? ? ?
= 23.04 × 10
–38+9+2+2
= 23.04 × 10
–25
= 2.3 × 10
–24
?
9. No. of electrons of 100 g water = 
18
100 10 ?
= 55.5 Nos Total charge = 55.5
No. of electrons in 18 g of H
2
O = 6.023 × 10
23
× 10 = 6.023 × 10
24
No. of electrons in 100 g  of H
2
O = 
18
100 10 023 . 6
24
? ?
= 0.334 × 10
26
= 3.334 × 10
25
Total charge = 3.34 × 10
25
× 1.6 × 10
–19
= 5.34 × 10
6
c
10. Molecular weight of H
2
O = 2 × 1 × 16 = 16
No. of electrons present in one molecule of H
2
O = 10
18 gm of H
2
O has 6.023 × 10
23
molecule
18 gm of H
2
O has 6.023 × 10
23
× 10 electrons
100 gm of H
2
O has 100
18
10 023 . 6
24
?
?
electrons
So number of protons = 
18
10 023 . 6
26
?
protons (since atom is electrically neutral)
Charge of protons = 
18
10 023 . 6 10 6 . 1
26 19
? ? ?
?
coulomb = 
18
10 023 . 6 6 . 1
7
? ?
coulomb
Charge of electrons = = 
18
10 023 . 6 6 . 1
7
? ?
coulomb
Hence Electrical force = 
2 2
7 7
9
) 10 10 (
18
10 023 . 6 6 . 1
18
10 023 . 6 6 . 1
10 9
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
? ?
?
= 
25
10 023 . 6 6 . 1
18
023 . 6 8
? ? ?
?
= 2.56 × 10
25
Newton
11. Let two protons be at a distance be 13.8 femi
F = 
30 2
38 9
10 ) 8 . 14 (
10 6 . 1 10 9
?
?
?
? ? ?
= 1.2 N
12. F = 0.1 N
r = 1 cm = 10
–2
(As they rubbed with each other. So the charge on each sphere are equal)
So, F = 
2
2 1
r
q kq
? 0.1 = 
2 2
2
) 10 (
kq
?
? q
2
= 
9
4
10 9
10 1 . 0
?
?
?
? q
2
= 
14
10
9
1
?
? ? q = 
7
10
3
1
?
?
1.6 × 10
–19
c Carries by 1 electron 1 c carried by 
19
10 6 . 1
1
?
?
0.33 × 10
–7
c carries by 
7
19
10 33 . 0
10 6 . 1
1
?
?
? ?
?
= 0.208 × 10
12
= 2.08 × 10
11
+
+
+
–
–
–
–
+
+
A
10 × 10
–10
m
C
B
a
2 × 10
–6
c
–1 × 10
–6
c
Electric Field and Potential
29.3
13. F = 
2
2 1
r
q kq
= 
2 10
19 19 9
) 10 75 . 2 (
10 10 6 . 1 6 . 1 10 9
?
? ?
?
? ? ? ? ?
= 
20
29
10 56 . 7
10 04 . 23
?
?
?
?
= 3.04 × 10
–9
14. Given: mass of proton = 1.67 × 10
–27
kg = M
p
k = 9 × 10
9
Charge of proton = 1.6 × 10
–19
c = C
p
G = 6.67 × 10
–11
Let the separation be ‘r’ 
Fe = 
2
2
p
r
) C ( k
, fg= 
2
2
p
r
) M ( G
Now, Fe : Fg = 
2
p
2
2
2
p
) M ( G
r
r
) C ( K
? = 
2 27 11
2 19 9
) 10 67 . 1 ( 10 67 . 6
) 0 ` 1 6 . 1 ( 10 9
? ?
?
? ? ?
? ? ?
= 9 × 2.56 × 10
38
˜ 1,24 ×10
38
15. Expression of electrical force F = 
2
r
kr
e C
?
?
Since e
–kr
is a pure number. So, dimensional formulae of F = 
2
r of formulae ensional dim
C of formulae ensional dim
Or, [MLT
–2
][L
2
] = dimensional formulae of C = [ML
3
T
–2
]
Unit of C = unit of force × unit of r
2
= Newton × m
2
= Newton–m
2
  
Since –kr is a number hence dimensional formulae of 
k = 
r of formulae entional dim
1
= [L
–1
] Unit of k = m
–1
16. Three charges are held at three corners of a equilateral trangle.
Let the charges be A, B and C. It is of length 5 cm or 0.05 m
Force exerted by B on A = F
1
force exerted by C on A = F
2
So, force exerted on A = resultant F
1
= F
2
? F = 
2
2
r
kq
= 
4
12 9
10 5 5
10 2 2 2 10 9
?
?
? ?
? ? ? ? ?
= 10
25
36
? = 14.4
Now, force on A = 2 × F cos 30° since it is equilateral ?.
? Force on A = 2 × 1.44 ×
2
3
= 24.94 N. ?
17. q
1
= q
2
= q
3
= q
4
= 2 × 10
–6
C
v = 5 cm = 5 × 10
–2
m
so force on c = 
CD CB CA
F F F ? ?
so Force along × Component = 0 45 cos F F
CA CD
? ? ?
= 
2 2
1
) 10 5 (
) 10 2 ( k
) 10 5 (
) 10 2 ( k
2 2
2 6
2 2
2 6
?
?
?
?
?
?
?
?
?
= 
?
?
?
?
?
?
?
?
?
?
?
? ? 4 4
2
10 2 50
1
10 25
1
kq
= 
?
?
?
?
?
?
?
?
?
?
? ? ?
?
?
2 2
1
1
10 24
10 4 10 9
4
12 9
= 1.44 (1.35) = 19.49 Force along % component = 19.49
So, Resultant R = 
2 2
Fy Fx ? = 19.49 2 = 27.56 
18. R = 0.53 A° = 0.53 × 10
–10
m
F = 
2
2 1
r
q Kq
= 
10 10
38 9
10 10 53 . 0 53 . 0
10 6 . 1 6 . 1 10 9
? ?
?
? ? ?
? ? ? ?
= 82.02 × 10
–9
N
19. Fe from previous problem No. 18 = 8.2 × 10
–8
N Ve = ?
Now, M
e
= 9.12 × 10
–31
kg r = 0.53 × 10
–10
m
Now, Fe = 
r
v M
2
e
? v
2
= 
e
m
r Fe ?
= 
31
10 8
10 1 . 9
10 53 . 0 10 2 . 8
?
? ?
?
? ? ?
= 0.4775 × 10
13
= 4.775 × 10
12
m
2
/s
2
? v = 2.18 × 10
6
m/s
0.05 m
60°
? ?
F 2 F 1
B
A
C
0.05 m
0.05 m
D
C
B
CD
F
A
CA
F
CB
F
Electric Field and Potential
29.4
20. Electric force feeled by 1 c due to 1 × 10
–8
c.
F
1
= 
2 2
8
) 10 10 (
1 10 1 k
?
?
?
? ? ?
= k × 10
-6
N. electric force feeled by 1 c due to 8 × 10
–8
c.
F
2
= 
2 2
8
) 10 23 (
1 10 8 k
?
?
?
? ? ?
= 
9
10 10 8 k
2 8
? ? ? ?
?
= 
4
10 k 28
6 ?
?
= 2k × 10
–6
N.
Similarly F
3
= 
2 2
8
) 10 30 (
1 10 27 k
?
?
?
? ? ?
= 3k × 10
–6
N
So, F = F
1
+ F
2
+ F
3
+ ……+ F
10
= k × 10
–6
(1 + 2 + 3 +……+10) N
= k × 10
–6
× 
2
11 10 ?
= 55k × 10
–6
= 55 × 9 × 10
9
× 10
–6
N = 4.95 × 10
3
N 
21. Force exerted = 
2
2
1
r
kq
= 
2
16 9
1
10 2 2 10 9
?
? ? ? ?
= 3.6 × 10
–6
is the force exerted on the string
22. q
1
= q
2
= 2 × 10
–7
c m = 100 g
l = 50 cm = 5 × 10
–2
m d = 5 × 10
–2
m
(a) Now Electric force 
F = 
2
2
r
q
K = 
4
14 9
10 25
10 4 10 9
?
?
?
? ? ?
N = 14.4 × 10
–2
N = 0.144 N
(b) The components of Resultant force along it is zero, 
because mg balances T cos ? and so also.
F = mg = T sin ??
(c) Tension on the string
T sin ? = F T cos ? = mg
Tan ? = 
mg
F
= 
8 . 9 10 100
144 . 0
3
? ?
?
= 0.14693
But T cos ? = 10
2
× 10
–3
× 10 = 1 N
? T = 
? cos
1
= sec ? ? ?
? T = 
? sin
F
, ?
Sin ? = 0.145369 ; Cos ? = 0.989378; ?
23. q = 2.0 × 10
–8 
c n= ? T = ? Sin ? = 
20
1
Force between the charges 
F = 
2
2 1
r
q Kq
= 
2 2
8 8 9
) 10 3 (
10 2 10 2 10 9
?
? ?
?
? ? ? ? ?
= 4 × 10
–3
N
mg sin ? = F ? m = 
? sin g
F
= 
) 20 / 1 ( 10
10 4
3
?
?
?
= 8 × 10
–3
= 8 gm
Cos ??= ? ?
2
Sin 1 = 
400
1
1 ? = 
400
1 400 ?
= 0.99 ˜ 1
So, T = mg cos ?
Or T = 8 × 10
–3
10 × 0.99 = 8 × 10
–2
M ?
r = 1 m
q 1
q 1
90°
? ?
T
F
T Cos ? ?
??
? ?
90°
F
T Cos ? ?
T Sin ? ?
T Sin ? ?
20 cm
0
1 cm
20 T
mg
5 cm
1 cm 1 cm
T
20
Electric Field and Potential
29.5
24. T Cos ? = mg …(1)
T Sin ? = Fe …(2)
Solving, (2)/(1) we get, tan ? = 
mg
Fe
= 
mg
1
r
kq
2
?
?
1596
2
= 
8 . 9 02 . 0 ) 04 . 0 (
q 10 9
2
2 9
? ?
? ?
? q
2
= 
1596 10 9
2 8 . 9 02 . 0 ) 04 . 0 (
9
2
? ?
? ? ?
= 
95 . 39 10 9
10 27 . 6
9
4
? ?
?
?
= 17 × 10
–16
c
2
? q = 
16
10 17
?
? = 4.123 × 10
–8
c ?
25. Electric force = 
2
2
) Q sin Q sin (
kq
? ? ?
= 
2 2
2
sin 4
kq
?
So, T Cos ? = ms (For equilibrium) T sin ? = Ef
Or tan ? = 
mg
Ef
? mg = Ef cot ? = ?
?
cot
sin 4
kq
2 2
2
?
= 
0
2 2
2
E 16 sin
cot q
? ?
?
?
or m = 
g Sin E 16
cot q
2 2
0
2
? ?
?
?
unit.
26. Mass of the bob = 100 g = 0.1 kg
So Tension in the string = 0.1 × 9.8 = 0.98 N.
For the Tension to be 0, the charge below should repel the first bob.
? F = 
2
2 1
r
q kq
T – mg + F = 0 ? T = mg – f T = mg
? 0.98 = 
2
2
4 9
) 01 . 0 (
q 10 2 10 9 ? ? ? ?
?
? q
2
= 
5
2
10 2 9
10 1 98 . 0
? ?
? ?
?
= 0.054 × 10
–9
N
27. Let the charge on C = q
So, net force on c is equal to zero
So 
BA AC
F F ? = 0, But F
AC
= F
BC
?
2
x
kqQ
= 
2
) x d (
qQ 2 k
?
? 2x
2
= (d – x)
2
? 2 x = d – x 
? x = 
1 2
d
?
= 
) 1 2 (
) 1 2 (
) 1 2 (
d
?
?
?
?
= ) 1 2 ( d ?
For the charge on rest, F
AC
+ F
AB
= 0
2 2
2
d
) q 2 ( kq
d
kqQ
) 414 . 2 ( ? = 0 ? ] q 2 Q ) 414 . 2 [(
d
kq
2
2
? = 0 
? 2q = –(2.414)
2
Q 
? Q = q
) 1 2 (
2
2
? ?
= q
2 2 3
2
?
?
?
?
?
?
?
?
?
? = –(0.343) q = –(6 – 4 2 )
28. K = 100 N/m l = 10 cm = 10
–1
m q = 2.0 × 10
–8
c Find l = ?
Force between them F = 
2
2 1
r
q kq
= 
2
8 8 9
10
10 2 10 2 10 9
?
? ?
? ? ? ?
= 36 × 10
–5
N
So, F = – kx or x = 
K
F
?
= 
100
10 36
5 ?
?
= 36 × 10
–7
cm = 3.6 × 10
–6
m
4 cm
1596
20 g
B
A
? ?
20 g
40 cm
? ? v
2
q
E F
l
? ?
mg l sin ? ?
l
q
FBD for a mass 
(m)
T
E F
T cos ? ?
mg
a
T Sin ? ?
2 × 10
–4
C
10 cm mg
q
C
B
A
d
x d–x
2q
q 2
q 1
K
Read More
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