Answer: the radial distance between the 500-v equipotential surface and the 1000 v surface will be 8.91*106 times the charge Q.
Explanation: To find the answer, we have to know more about the equipotential surfaces.
<h3>What are equipotential surfaces?</h3>
- An equipotential surface is the locus of all points which have the same potential due to the charge distribution.
- Any surface in an electric field, at every point of which, the direction of electric field is normal to the surface can be regarded as equipotential.
- We have the equation for electric potential as,
, where k is equal to 1/(4π∈₀) = .
- equation for radial distance will be,
<h3>How to solve the problem?</h3>
- For the first surface, we can write the equation of potential as,
- For the second surface, we can write the equation of potential as,
- Thus, the radial distance will be,
Thus, we can conclude that, the radial distance between the equipotential surface of 500V and 1000V will be,8.91*106 times the charge Q.
Learn more about the equipotential surface here:
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Power is defined as the rate at which the body is doing work:
Work is defined as displacement done by the force times that displacement:
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
of work.
Now we just divide that by 5s to get how much power is required:
Answer:
The mass of the block m is:
Explanation:
Let's analyze the block by parts
For the block M
(1)
Where:
- T is the tension
- W(x) is the component of the weight in the x-direction
- F(f) is the friction force
For the block m
(2)
Now, let's combines equation (1) and (2):
Finally, let's solve it for block m.
I hope it helps you!
Explanation:
A. Up along the page
Using the law of the right hand