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:
1-d
2- a
3-e
4-b
5-c
This is the correct sequence
The product label which Mateo should place in the marked cell is that it: B. provides electrical energy.
<h3>What is a product label?</h3>
A product label can be defined as a paper-document with specific information about the chemical elements, nutrients, and other chemical compounds that are present in a product and its general use or function.
In this scenario, the product label which Mateo should place in the marked cell shown in the image attached below is that, this product was designed and developed to provide electrical energy, based on the electrical symbol for battery.
Read more on product label here: brainly.com/question/14446515
#SPJ4
Answer:
B) Force = 7.5, Time = 2 is equal to an impulse of 15 units
I think it’s b... not sure tho sorry
