JEE  >  HC Verma Solutions: Chapter 29 - Electric Field & Potential

# HC Verma Solutions: Chapter 29 - Electric Field & Potential Notes | Study HC Verma and Irodov Solutions - JEE

## Document Description: HC Verma Solutions: Chapter 29 - Electric Field & Potential for JEE 2022 is part of HC Verma and Irodov Solutions preparation. The notes and questions for HC Verma Solutions: Chapter 29 - Electric Field & Potential have been prepared according to the JEE exam syllabus. Information about HC Verma Solutions: Chapter 29 - Electric Field & Potential covers topics like and HC Verma Solutions: Chapter 29 - Electric Field & Potential Example, for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solutions: Chapter 29 - Electric Field & Potential.

Introduction of HC Verma Solutions: Chapter 29 - Electric Field & Potential in English is available as part of our HC Verma and Irodov Solutions for JEE & HC Verma Solutions: Chapter 29 - Electric Field & Potential in Hindi for HC Verma and Irodov Solutions course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: HC Verma Solutions: Chapter 29 - Electric Field & Potential Notes | Study HC Verma and Irodov Solutions - JEE
``` 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
```

## HC Verma and Irodov Solutions

76 docs
 Use Code STAYHOME200 and get INR 200 additional OFF

## HC Verma and Irodov Solutions

76 docs

### Top Courses for JEE

Track your progress, build streaks, highlight & save important lessons and more!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;