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

296 ANSWERS AND SOLUTIONS
Upon solving the equations, we get
t= 2tlt2
t l-t9
400. It can, if we use the phenomenon
of electrostatic induction. Bring a
conductor on an insulated support up
to the charged body and connect the
conductor to the earth for a short time.
The conductor will retain a charge opposite
in sign to the given one, while
Fig. 411
the like charge will pass into the
earth.
The charge can be removed from the conductor by. introducing the latter
into a metallic space. The operation may be repeated many times with a
charge of any magnitude.
Electrostatic machines operate on a similar principle.
401. The energy is produced by the mechanical work that has to be performed
in moving the conductor from the oppositely charged body to the
body .that accumulates the charge.
402. They can, if the charge of one ball is much greater than that of the
other. The forces of attraction caused by the induced charges may exceed the
forces of repulsion.
403. Since Q ~ q, the interaction between the separate elements of the ring
can be neglected. Let us take a small element of the ring with a length RA.a,
(Fig. 410). From the side of the charge Q it is acted upon by the force
IJ.F= ~~q • where t1q= q:na.. The tension forces of the ring T balance IJ.F.
From the condition of equilibrium, and remembering that AeJ is small. we
have
t1F=2T sin (IJ.;) • rs«
The sought force is the tension T= ::.~. ·
404. Let us consider the case of opposite charges Q 1 > 0 and Q, < O. The
intensities created by the charges Q 1 and Qz are equal, respectively. to
£1 = ~l
,]
and £.= Q.- . A glance at Fig. 411 shows that
'1
EI=Er+ £:-2E 1E, cos cp
From triangle ABC
Hence,
cosq>
,I+,:-d l
2'1"