Solutions (^249)
from the positive plate is
q i r 2q 11
E (x)=
4neo L x2 (x+ 1) 2 4ne0x3
(x 1).
The force .acting on the second capacitor situated
at a distance d from the first is
1
F (12 IE (d)— E (d+01=.
qi(121 r^
4ne 0 [ co (d+ 1) 3 J
3 q142/ 2
2neod 4 •
Therefore, the capacitors will repel each other in
this case.
A similar analysis can be carried out for the
case when the capacitors are oriented so that the
plates with unlike charges face each other. Then
the capacitors will attract each other with the
same force
(^3 419212)
F =
2 as 0 d 4 •
3.6. We choose two small arbitrary elements be-
longing to the surfaces of the first and second hemi-
_spheres and having the areas AS 1 and AS 2. Let the
separation between the two elements be r 12. The
force of interaction between the two elements can
be determined from Coulomb's law:
1 1 AS 1 A
2
AF12
4ne (^) 0 12 1
AS 1 U 22 AS2=
4=0
S2 Cria2.
r 2 /
In order to determine the total force of interac-
tion between the hemispheres, we must, proceeding
from the superposition principle, sum up the forces
AF 12 for all the interacting pairs of elements so
that the resultant force of interaction between the
hemispheres is
F = kala 2 ,
where the coefficient k is determined only by the
geometry of the charge distribution and by the
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