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:

Explanation:
= Permittivity of free space = 
= Surface charge density = 
= 0.57-0.26
q = Charge = 
m = Mass of object = 
Electric field due to a sheet is given by

Electric field is given by

Voltage is given by

Kinetic energy is given by


The initial speed of the object is 
1. Roll a ball across a table into an object
2. Drop something
Answer:
this description is valid for mediadle displacement, bone is an acceptable description
Explanation:
The description of a person's position must be done with a position vector. These vectors must have magnitude, a given direction and a starting point.
In the description this has a starting point corner NO of pine and 675.
Each displacement occurs with respect to the previous one, indicating the magnitude of the displacement and its direction.
After analyzing this description is valid for mediadle displacement, bone is an acceptable description