250 Aptitude Test Problems in Physicschoice of the system of units. If the hemispheres
were charged with the same surface density a,
the corresponding force of interaction between the
hemispheres would be
F = kers,where the coefficient k is the same as in the previous
formula.
Let us determine the force F. For this purpose,
we consider the "upper" hemisphere. Its small
surface element of area AS carries a charge Aq =
a AS and experiences the action of the electric
field whose strength E' is equal to half the electric
field strength produced by the sphere having a
radius R and uniformly charged with the surface
density a. (We must exclude the part produced by
the charge Aq itself from the electric field strength.)
The force acting on the charge Aq is
102=-- AS — a (^) AS
4aso Rs 2^ 2a^0
end is directed along the normal to the surface
element. In order to find the force acting on the
upper hemisphere, it should be noted that according
to the expression for the force AF the hemisphere
as if experiences the action of an effective pressure
p = 04280 ). Hence the resultant force acting on
the upper hemisphere is
i.pnR 2 =
02
—
ztRa(although not only the "lower" hemisphere, but
all the elements of the "upper" hemisphere make
a contribution to the expression AF for the force
acting on the surface element AS, the forces of
interaction between the elements of the upper hemi-
sphere will be cancelled out in the general expres-
sion for the force of interaction between the hemi-
spheres obtained above).
Since F.= kat, - we obtain the following expres-
sion for the force of interaction between the hemi-