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julsineya [31]
3 years ago
11

Two adjacent students stand on stationary skateboards, face each other, and

Physics
1 answer:
notsponge [240]3 years ago
7 0
So the heavy skateboarder will move off with speed 2.45 m/s in the opposite direction
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A uniform electric field is created by two parallel plates separated by a distance of 0.04 m. What is the magnitude of the elect
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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

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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

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