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(Solved): Correct answer given The following diagram depicts a side view of a mechanical separator. You can a ...



The following diagram depicts a side view of a mechanical separator. You can assume that the width of the machinery is uniforCorrect answer given

The following diagram depicts a side view of a mechanical separator. You can assume that the width of the machinery is uniformly (in other words, all of the components that you see in this side view extend into the page uniformly for a distance of ). For this machine, the surface of the drum had a radius , and the positively charged plate can be modelled as a section of a cylinder with radius 1.7 m. Assume the drum is uniformly charged, and that the region between the drum and the plate is defined by , and machine width < . If the charge on the drum is a uniform , what is the flux density at the surface defined by , and machine width ? 7.41 margin of error


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To determine the flux density at the specified surface, we can use Gauss's law. Since the charge on the drum is uniformly distributed, we can assume that the electric field is also uniformly distributed.

Let's consider a cylindrical Gaussian surface with radius 1.62 m and height dx along the width of the machine. The surface area of this cylinder is:
dA = 2?(1.62 m)(dx)
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