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,