Complete question:
A uniform electric field is created by two parallel plates separated by a
distance of 0.04 m. What is the magnitude of the electric field established
between the plates if the potential of the first plate is +40V and the second
one is -40V?
Answer:
The magnitude of the electric field established between the plates is 2,000 V/m
Explanation:
Given;
distance between two parallel plates, d = 0.04 m
potential between first and second plate, = +40V and -40V respectively
The magnitude of the electric field established between the plates is calculated as;
E = ΔV / d
where;
ΔV is change in potential between two parallel plates;
d is the distance between the plates
ΔV = V₁ -V₂
ΔV = 40 - (-40)
ΔV = 40 + 40
ΔV = 80 V
E = ΔV / d
E = 80 / 0.04
E = 2,000 V/m
Therefore, the magnitude of the electric field established between the plates is 2,000 V/m
When light passes through one transparent medium to another transperent medium it bends and that beding of light is know as refraction of light !
Answer: 27.21 V
Explanation:
The <u>electric potential</u> 
due to a point charge is expressed as:
Where:
is the <u>electric constant</u>
is the <u>electric charge of the hydrogen nucleus</u>, which is positive
is the <u>distance</u>
Rewritting the equation with the known values:
Finally:
It must be sliding friction, because the fish is already in motion.
