..il T-Mobile Wi-Fi 7:22 PM 86% webassign.net This question has several parts th

Question

..il T-Mobile Wi-Fi 7:22 PM 86% webassign.net This question has several parts that must be completed sequentlally. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Oppositely charged parallel plates are separated by 5.06 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? (b) What is the magnitude of the force on an electron between the plates? (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.75 mm from the positive plate? (a) A uniform electric field of magnitude E, directed perpendicular to the plates, exists in the region between the plates and exerts a constant force-of on a charged particle in this region. When the particle moves a distance d parallel to the field, going from one plate to the other, the work done by the field is w.oesge-of. Thus, the magnitude of the change in potential energy of the particle is-w-e, and the magnitude of the potential difference between the plates is We know the potential difference between the plates and the distance d = 5.06 mm between the plates. Therefore, the magnitude of the electric field between the plates is x 10° V/m 105 N/C Noed Help?

Explanation / Answer

E = V/d = (600 V)/(5.06×10-3 m) = 1.186×105 V/m = 1.186×105 N/C

So from the left in the vacant space you need to fill.

600

5.06

1.186

1.186

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