in this example, we’re given that the figure
shows 2 vertical smooth rails, ay b and c
d, seperated by a distance l. ends ay and
c are connected with a capacitor of capacitance
c. here a rod p q of mass m is horizontally
kept in touch with both the rails as shown.
and it is saying that if it is released at
t equal to zero and it remains in contact
with rails, during the fall, we’re required
to find the charge on capacitor as a function
of time. now in this situation we can see
on releasing due to its weight m g, the rod
will start falling, when it attains a speed
v, here we can see, across terminals p and
q a motional e m f is induced and we can directly
write, motional e m f across, p q is, this
we can directly write as, b v l if v is the
instantaneous speed of the rod, due to which
the capacitor will attain, a charge plus and
minus q on its plate, which is due to the
potential difference b v l, across the capacitor.
so we can directly write instantaneous charge,
on capacitor is, we can write it as, c v,
that is c b v l. and here we can see as motional
e m f is increasing the charge will also increase
due to which we can say a continuous current
will flow in the circuit. and here circuit
current can be directly given as, i is equal
to d q by d t. on differentiating this expression
we get, c b and l are constant, so d q by
dt is c b l, multiplied by d v by d t which
we write as, c b l ay, where ay can be written
as the acceleration of this rod, which is
d v by d t. and in this situation due to this
current it experiences, an upward magnetic
force b i l, the rod experience the force
in upward direction, so here we can write,
if rod acceleration, is ay, we can use acceleration
as m g minus, b i l divided by, its mass.
and in this situation if we substitute the
value of current as c b l ay, it’ll be,
m g minus b square, l square, c ay by, m.
and if this term is taken on the left hand
side, we can see, the value of acceleration
we’re getting is, i’ll write it here,
we can see it can be given as, m g divided
by, m plus b square l square c, that is acceleration
of this rod. and we can directly write after
time t, its velocity v we can write as ay
t, but can be written as, m g t divided by,
m plus b square l square c. and now if we
wish to write the charge on capacitor which
we’re required to find as a function of
time, this we already written as c b l v.
on substituting the value of v here we get,
c b l, m g t divided by, m plus b square l
square c. so this’ll be the answer to this
probem the charge on capacitor as a function
of time.
