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CHAPTER – 29

ELECTRIC FIELD AND POTENTIAL
EXERCISES
1. 0 =

Coulomb 2 Newton m
kq1q2
2

=l M L T

1

–1 –3 4

F= 2.

r2 3 q1 = q2 = q = 1.0 C distance between = 2 km = 1 × 10 m kq1q2 r 2

so, force =

F=

(9  10 9 )  1 1 (2  10 )
3 2

=

9  10 9 2 2  10 6

= 2,25 × 10 N

3

The weight of body = mg = 40 × 10 N = 400 N So,

 2.25  10 3 wt of body =   4  10 2 force between ch arg es 

   

1

= (5.6)

–1

=

1 5 .6

3.

So, force between charges = 5.6 weight of body. q = 1 C, Let the distance be  F = 50 × 9.8 = 490 F=

Kq 2 2

 490 =
3

9  10 9  12 2

or  =

2

9  10 9 6 = 18.36 × 10 490

4.

  = 4.29 ×10 m charges ‘q’ each, AB = 1 m wt, of 50 kg person = 50 × g = 50 × 9.8 = 490 N FC = 2

kq1q2 r
2

 =

kq2 r2

= 490 N

q = q= 5.

490  r 2 9  10
9

490  1 1 9  10 9
–5

54.4  10 9 = 23.323 × 10

coulomb ≈ 2.3 × 10
–19

–4

coulomb

Charge on each proton = a= 1.6 × 10 coulomb –15 Distance between charges = 10 × 10 metre = r = 9 × 2.56 × 10 = 230.4 Newton r 10  30 –6 –6 q2 = 1.0 × 10 r = 10 cm = 0.1 m q1 = 2.0 × 10 Let the charge be at a distance x from q1 kqq 2 Kq1q q1 F1 = F2 = 2  (0.1  )2 2

Force =

kq2

=

9  10 9  1.6  1.6  10 38

6.

q xm (0.1–x) m 10 cm q2

=

9.9  2  10 6  10 9  q 2
 f 1 = f2

Now since the net force is zero on the charge q. 

kq1q 2

=

kqq 2 (0.1  )2
2 2

 2(0.1 – ) =  =

 2 (0.1 – ) =  From larger charge 29.1

0 .1 2 1 2

= 0.0586 m = 5.86 cm ≈ 5.9 cm

Electric Field and Potential 7. q1 = 2 ×10 c q2 = – 1 × 10 c r = 10 cm = 10 × 10 Let the third charge be a so, F-AC = – F-BC  –6 –6 –2

m
10 × 10–10 m

kQq1 r12
2

=

KQq2 r2
2
2



2  10 6 (10  )
2

=

1  10 6 
2

A 2 × 10–6 c

C B –1 × 10–6 c

a

 2 = (10 + )  8.

2  = 10 +   ( 2 - 1) = 10   =

10 = 24.14 cm  1.414  1

So, distance = 24.14 + 10 = 34.14 cm from larger charge  Minimum charge of a body is the charge of an electron –19 –2  = 1 cm = 1 × 10 cm Wo, q = 1.6 × 10 c So, F =

kq1q2 r2

=

9  10 9  1.6  1.6  10 19  10 19 10  2  10  2

= 23.04 × 10

–38+9+2+2

= 23.04 × 10

–25

= 2.3 × 10

–24



9.

10  100 = 55.5 Nos Total charge = 55.5 18 23 24 No. of electrons in 18 g of H2O = 6.023 × 10 × 10 = 6.023 × 10 No. of electrons of 100 g water = No. of electrons in 100 g of H2O =

6.023  10 24  100 26 25 = 0.334 × 10 = 3.334 × 10 18 25 –19 6 Total charge = 3.34 × 10 × 1.6 × 10 = 5.34 × 10 c 10. Molecular weight of H2O = 2 × 1 × 16 = 16 No. of electrons present in one molecule of H2O = 10 23 18 gm of H2O has 6.023 × 10 molecule 23 18 gm of H2O has 6.023 × 10 × 10 electrons 100 gm of H2O has

6.023  10 24  100 electrons 18

So number of protons = Charge of protons =

6.023  10 26 protons (since atom is electrically neutral) 18

1.6  10 19  6.023  10 26 1.6  6.023  10 7 coulomb = coulomb 18 18

Charge of electrons = =

1.6  6.023  10 7 coulomb 18
   

 1.6  6.023  10 7   1.6  6.023  10 7  9  10 9     18 18    Hence Electrical force = 2 2 (10  10 )

8  6.023 25  1.6  6.023  10 25 = 2.56 × 10 Newton 18 11. Let two protons be at a distance be 13.8 femi = F=

9  10 9  1.6  10 38 (14.8)2  10  30

= 1.2 N

– + – + + – + + –

12. F = 0.1 N –2 r = 1 cm = 10 (As they rubbed with each other. So the charge on each sphere are equal) So, F = 1.6 × 10

kq1q2 r
–19

2

 0.1 =

kq 2 (10
2 2

)

q =

2

0.1  10 4 9  10 9

q =

2

1 1  10 14  q =  10  7 9 3

c

Carries by 1 electron

1 c carried by

1 1.6  10 19

0.33 × 10

–7

c carries by

1 1.6  10 19

12 11  0.33  10  7 = 0.208 × 10 = 2.08 × 10

29.2

Electric Field and Potential 13. F =

kq1q2 r
2

=

9  10 9  1.6  1.6  10 19  10 19 (2.75  10 10 2

)

=

23.04...
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