Question
Asked Dec 17, 2019
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Oppositely charged parallel plates are separated by 5.33 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 2.90 mm from the positive plate?

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Expert Answer

Step 1

Part A:

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Given info: Potential difference is 600 V. The distance between the plates is 5.33 mm. Explanation: Formula to calculate the electric field between the plates is, E =|-| AV Az Ax is the distance. AV is the potential difference. Substitute 5.33 mm for Ax and 600 V for AV E =|- 600 V 5,33 mm 600 V 5.33x10-³m = 1.13 x 10°N/C %3D Conclusion: The magnitude of electric field is 1.13 x 105N/C.

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Step 2

Part B:

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Given info: Potential difference is 600 V. The distance between the plates is 5.33 mm. Explanation: Formula to calculate the electric force is, F = eE • e is the elementary charge. Substitute 1.6 x 10-19C for q and 1.13 x 10°N/C for E. F = (1.6 x 10-19C) (1.13 × 10°N/C) = 1.80 x 10¬14N Conclusion: The electric force on the electron is 1.80 x 10-14N.

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Step 3

Part C:...

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Given info: Potential difference is 600 V. The distance between the plates is 5.33 mm. The displacement of the electron is 2.90 mm. Explanation: Formula to calculate the work done is, W = FAx cos 0 O is the angle between the force and displacement. Aæ is the total displacement. The total displacement is, Ax = (5.33 mm) – (2.90 mm) = 2.43 mm Substitute 1.80 × 10-14N for F, 2.43 mm for Ax and 0° for 0. W = (1.80 x 10-14N) (2.43 mm) cos 0° = (1.80 x 10-14N) (2.43 × 10–³m) cos 0° = 4.37 × 10-17J

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Electrostatic Potential and Capacitance

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