(^246) Aptitude Test Problems in Physics
from the charge -1-q% within the angle 2a is equal
to the number of lines entering the charge —q 2
at an angle 2P. Consequently,
r qi^1 (1 — cos a) = I q^2 I (1 — cos p),
whence
13 cc 1 / 1 q 1 1
sin — sin —
(^2 2) 42'1 •
If VI q I / I 1 q2 I sin (a/2) > 1, an electric field
line will not enter the charge —q 2.
3.3. Before solving this problem, let us formulate
the theorem which will be useful for solving this
and more complicated problems. Below we shall
give the proof of this theorem applicable to the
specific case considered in Problem 3.3.
If a charge is distributed with a constant den-
sity a over a part of the spherical surface of radius
R, the projection of the electric field strength due
to this charge at the centre of the spherical surface
on an arbitrary direction a is
Ea
_
I a
4neo R 2 s la,
where Sia is the area of the projection of the part
of the surface on the plane perpendicular to the
direction a.
Let us consider a certain region of the spherical
surface ("lobule") and orient it as shown in Fig. 207,
i.e. make the symmetry plane of the lobule coincide
with the z- and x-axes. From the symmetry of
charge distribution it follows that the total field
strength at the origin of the coordinate system
(point 0) will be directed against the z-axis (if
a > 0), and the field strength components along
the x- and y-axes will be zero.
Let us consider a small region of the surface
AS of the lobule. The vertical component of the
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