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
Answer:
hi,
When one stands at a particular place, the point in the sky directly above the head is called Zenith.
Explanation:
hope it helps(≡^∇^≡)
good day
thank u ✅
<h2>
Answer:</h2>
The distance will be 1800 m
<h2>
Explanation</h2>
As in question
Time = 15 min
Time = 15 x 60 sec = 900 sec
Speed = 2 m/s
We know that
So, the answer is 1800 m
