492
11 Electrostatics
Fig. 11.26
11.21 Electric force F e =
q 2
4πε 0 r 2
(11.48)
Gravitational force F g =
GMm
r 2
(11.49)
F e
F g
=
q 2
4πε 0
1
GMm
=
(1.6 × 10 −19 ) 2 (9 × 10 9 )
(6.67 × 10 −11 )(1.66 × 10 −27 )(9.1 × 10 −31 )
(11.50)
= 2.29 × 10
39
(11.51)
The distance is immaterial. Note that the gravitational force at the atomic and
sub-atomic levels is small simply because the masses are small.
11.22 q 1 + q 2 = 15 μC
( 1 )
F =
q 1 q 2
4πε 0 r 2 =
9 × 10 9
(0.3) 2 q 1 q 2 = 5.4
or
q 1 q 2 = 54 (μC)
2
(2)
Solving (1) and (2), q 1 = 6 μC, q 2 = 9 μC.
11.23 At P, the electric field due to +q is
q
4πε 0 (a/
√
2) 2 or
2q
4πε 0 (a) 2 and points
towards +2q. The field due to +2q is
4q
4πε 0 a 2 and points towards +q. The
resultant field due to the pair (q, 2q) is
4q
4πε 0 a 2 −
2q
4πε 0 a 2 or E 1 =
2q
4πε 0 a 2
towards +q.
Similarly, the resultant filed due to the pair of charges (−q, −2q) will be
E 2 = −
2q
4πε 0 a 2 towards −2q or +
2q
4πε 0 a 2 towards −q.