(a)
In a velocity selector, the speed of the beam is related to the magnitude of the electric field and of the magnetic field by the formula:
where
E is the magnitude of the electric field
B is the magnitude of the magnetic field
In this problem, we have
(magnetic field)
(speed of the particles)
Solving the equation for E, we find the electric field:
(b) 3.2 kV
The relationship between electric field and potential difference between the two plates is:
where, in this problem:
is the magnitude of the electric field
is the separation between the plates
Substituting into the equation, we find the potential difference:
Shein. but i heard that kids work in a workshop and are forced to make the clothing. and i bought from them but i ddi not know and ya but thank you timmy for the t shirt
Answer:
<h2>The work done is 0.882 Joules.</h2>
Explanation:
To calculate the work, we need to find all forces that are involved in the movement.
As you can analyse, the body is the force that make the box to move, and the friction force is opposite to it, we cannot forget about the friction. So, we have to calculate the resultant force to this context.
So, to find we use: ; where
The friction force would be:
Then, the resultant force is:
Now, we calculate the work:
Therefore, the work done is 0.882 Joules.
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