Answer:
The maximum no. of electrons- 
Solution:
As per the question:
Maximum rate of transfer of charge, I = 1.0 C/s
Time, t = 1.0 h = 3600 s
Rate of transfer of charge is current, I
Also,

Q = ne
where
n = no. of electrons
Q = charge in coulomb
I = current
Thus
Q = It
Thus the charge flow in 1. 0 h:

Maximum number of electrons, n is given by:

where
e = charge on an electron = 
Thus

Answer:
F-F(gr) = ma
a= {F-F(gr)}/m =
=(15-10)/15=0.33 m/s² (upward)
Oppositely charged objects form in induction but not in conduction.