Page 283 - Schaum's Outline of Theory and Problems of Applied Physics

268 ELECTRICITY [CHAP. 23

(a) Since a normal atom is electrically neutral, the number of negatively charged electrons it contains equals the
number of positively charged protons in its nucleus. The iron atom therefore contains 26 electrons.
(b) The symbol Fe 3+ represents an iron atom that has a net charge of +3e, which means that it has lost three of its
usual complement of electrons. The Fe 3+ ion therefore contains 23 electrons.

SOLVED PROBLEM 23.4
A plastic ball has a charge of +10 −12 C. (a) Does it contain an excess or a deficiency of electrons
compared with its normal state of electrical neutrality? (b) How many such electrons are involved?
(a) Since the plastic ball is positively charged, it has fewer electrons than are needed to balance the positive charge
of its nuclear protons.
(b) The charge on an electron is e = 1.6 × 10 −19 C. Hence

q 10 −12 C 6
Number of electrons = = = 6.25 × 10 electrons
e 1.6 × 10 −19 C/electron

SOLVED PROBLEM 23.5
What is the magnitude and direction of the force on a charge of +4 × 10 −9 C that is 5 cm from a charge
of +5 × 10 −8 C?
Since 5 cm = 5 × 10 −2 m, we have from Coulomb’s law

9
2
2
q 1 q 2 (9 × 10 N·m /C )(4 × 10 −9 C)(5 × 10 −8 C) −4
F = k = = 7.2 × 10 N
r 2 (5 × 10 −2 m) 2
The force is directed away from the +5 × 10 −8 C charge since both charges are positive.
SOLVED PROBLEM 23.6
Two charges, one of +5 × 10 −7 C and the other of −2 × 10 −7 C, attract each other with a force of 100 N.
How far apart are they?
From Coulomb’s law we have

9 2 2 −7 −7
kq 1 q 2 (9 × 10 N·m /C )(5 × 10 C)(2 × 10 C)
r = =
F 10 N
2

= 90 × 10 −7 m = 9 × 10 −6 m = 3 × 10 −3 m = 3mm
SOLVED PROBLEM 23.7
Two charges repel each other with a force of 10 −5 N when they are 20 cm apart. (a) What is the force on
each when they are 5 cm apart? (b) When they are 100 cm apart?

2
2
(a) Since F is proportional to 1/r , the force increases when the charges are brought closer together to (20/5) = 16
times what it was before, namely, to 1.6 × 10 −4 N.
2
(b) The force decreases when the charges are moved apart to (20/100) = 0.04 times what it was before, namely,
to 4 × 10 −7 N.
SOLVED PROBLEM 23.8
Under what circumstances, if any, is the gravitational attraction between two protons equal to their electric
repulsion?