Question
Asked Dec 17, 2019
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A proton is released from rest in a uniform electric field of
magnitude 385 N/C. Find (a) the electric force on the pro-
ton, (b) the acceleration of the proton, and (c) the distance it
travels in 2.00 µs.
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A proton is released from rest in a uniform electric field of magnitude 385 N/C. Find (a) the electric force on the pro- ton, (b) the acceleration of the proton, and (c) the distance it travels in 2.00 µs.

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Expert Answer

Step 1

part A:

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Given info: The electric field strength (E) is 385 N/C. Explanation: Formula to calculate the electric force is, F = eE • e is the elementary charge. Substitute 1.6 × 10-19C for e and 375 N/C for E. F = (1.6 x 10-1ºC) (385 N/C) = 6.16 x 10-17N Conclusion: The electric force on the proton is 6.16 × 10-17N.

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Step 2

Part B:...

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Given info: The electric field strength (E) is 385 N/C. Explanation: Formula to calculate the acceleration is, m is the mass of the proton. Substitute 6.16 × 10-17N for Fand 1.67 × 10–27kg for m. 6.16x10-17N 1.67x10-27kg -2 = 3.69 x 101ºms Conclusion: The acceleration of proton is 3.69 × 10ºms-2. 10

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Electrostatic Potential and Capacitance

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