(^272) Aptitude Test Problems in Physics
of the new equilibrium position be —x 1. In this
case, the charge on the capacitor is q 1 = =--
e 0 S16/(d — x 1 ), where S is the area of the capac-
itor plates. The field strength in the capacitor is
E = 10(d — xi), but it is produced by two plates.
Therefore, the field strength produced by one plate
is E 1 /2, and for the force acting on the plate B
we can write
Ea]. (^) eon 2
1
2 2 (d —x1)3
( )
where k is the rigidity of the spring.
Let us now consider the case when the key K
is closed for a short time. The capacitor acquires
a charge q 2 , = sonfil (the plates have no time
to shift), which remains unchanged. Let the dis-
placement of the plate B in the new equilibrium
position be x 2. Then the field strengthn the capa-
citor becomes E 2 = q 2 /[C 2 (d — x 2 )1 and C2 =
soS/(d — x 2 ). In this case, the equilibrium con-
dition for the plate B can be written in the formE2q2
—q: eirS12 = kx 5.
(2)
2 2e^0 S^ 2d^2
Dividing Eqs. (1) and (2) termwise, we obtain
xs = x 1 [(d — xi )/(11 2. Considering that x 1 = 0.1d
by hypothesis, we get
x 2 - ---0.08d.
3.40. Let us represent the central junction of wires
in the form of two junctions connected by the wireFig. 219