Answer:
it is 100 degree which is boiling point in water
Answer:
Area of the plates of a capacitor, A = 0.208 m²
Explanation:
It is given that,
Charge on the parallel plate capacitor, 
Electric field, E = 3.1 kV/mm = 3100000 V/m
The electric field of a parallel plates capacitor is given by :
A = 0.208 m²
So, the area of the plates of a capacitor is 0.208 m². Hence, this is the required solution.
Answer:
a. +10.9μC
b. 0.600N and downward
Explanation:
To determine the magnitude of the charge, we use the force rule that exist between two charges which us expressed as
F=(kq₁q₂)/r²
since q₁=-0.55μC and the force it applied on the charge above it is upward,we can conclude that the second charge is +ve, hence we calculate its magnitude as
q₂=Fr²/kq₁
q₂=(0.6N*0.3²)/(9*10⁹*0.55*10⁻⁶)
q₂=0.054/4950
q₂=1.09*10⁻⁵c
q₂=10.9μC.
Hence the second charge is +10.9μC
b. From the rule of charges which state that like charges repel and unlike charges attract, we can conclude that the two above charges will attract since they are unlike charges. Hence the direction of the force will be downward into the second charge and the magnitude of the force will remain the same as 0.600N
