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
Is there options for this??
Answer:
When an electric current flows, the shape of the magnetic field is very similar to the field of a bar magnet
Explanation:
Hi there!
Great question!
Basketballs have air inside them. A special pump is used to insert the air. That's why you can lift the basketballs off the ground easily. If it was a solid, though, you'd hardly be able to lift the ball up! Basketballs can float, too, because anything with air inside can float. If it were solid, it would sink in the water easily.
Hope this helps! :D
Answer:
The electric flux is zero because charge is zero.
Explanation:
Given that,
Positive charge 
Negative charge 
We need to calculate the total charged
Using formula of charge
Put the value into the formula
We need to calculate the electric flux
Using formula of electric flux
Put the value into the formula
Hence, The electric flux is zero because charge is zero.
