Bukhovtsev-et-al-Problems-in-Elementary-Physics

CHAPTER 3
ELECTRICITY
AND MAGNETISM
3-1. Electrostatics
Q2
395. F=7=918 kgf.
The force is very great. It is impossible to impart a charge of one coulomb
to a small body since the electrostatic forces of repulsion are so high that
the charge cannot be retained on the body.
396. The balls will be arranged at the corners of an equilateral triangle
with a side ~31. The force acting from any two balls on the third is
4Q2
£=12 Y3'
The ball will be in equilibrium if tan a,=f- (where
mg
a=300). Hence,
I
Q=2 Y mg~ 100 CGS Q _
397. Since the threads do not deflect from the vertical, the coulombian
force of repulsion is balanced by the force of attraction between the balls
in conformity with the law of gravitation.
Therefore, in a vacuum
Q2 p2V.
-;:2=Y7
and in kerosene (taking into account the results of Problem 230)
Q2 (p-PO)2 Vi
E r2
r = 'V r 2
where V is the volume of the balls.
Hence,
P= Po ye r ~ 2.74 g/cm8
Ye,.-l
398. The conditions of equilibrium of the suspended ball give the following
equations for the two cases being considered:
- QQs Y2 0
T 1 sin (Xt - 2a 2 X -2-=
T +QQs Y2 QQ s 0
1 cos al 2a 2 X -2--7-mg=
· QQs V2 0
T 2 srn (X,2- 2a 2 X -2-=
T2cosa2+QQs-QQsx V2 -mg=O
a 2 2a 2 2