Solutions (^251)
spheres for the case when they have different sur-
face charge densities:
n.R 2
F= kaia 2 — a las.
3.7. The density of charges induced on the sphere
is proportional to the electric field strength:
a a E. The force acting on the hemispheres is
proportional to the field strength:
F a uSE a R 2 E 2 ,
where S is the area of the hemisphere, and R is its
radius. As the radius of the sphere changes by a
factor of n, and the field strength by a factor of k,
the force will change by a factor of k 2 n 2. Since the
thickness of the sphere walls remains unchanged,
the force tearing the sphere per unit length must
remain unchanged, i.e. k 2 n 2 /n = 1 and k =
1111 n = — 1/ VT. Consequently, the minimum elec-
tric field strength capable of tearing the conducting
shell of twice as large radius is
E
Ei E0
V2 •
3.8. Let 1 be the distance from the large conducting
sphere to each of the small balls, d the separation
between the balls, and r the radius of each ball.
If we connect the large sphere to the first ball,
their potentials become equal:
1 I Q, qi
4n6 0 / Th -r
Here Q is the charge of the large sphere, and q is its
potential. If the large sphere is connected to the
second ball, we obtain a similar equation corre-
sponding to the equality of the potentials of the
large sphere and the second ball:
I (1_ 1 21^1 _ qa \ =9.^
4ne 0 klidir/
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