A proton has an initial speed of 4.105 m/s. It is brought to rest by an electric
ID: 1536379 • Letter: A
Question
A proton has an initial speed of 4.105 m/s. It is brought to rest by an electric field. a.) What is the magnitude of the voltage difference needed to bring this charge to rest? b.) At the point when the proton comes to rest (momentarily): its potential energy is Select and it is at a Select voltage as compared to initially. c.) The electric field must be pointed ---Select--- the initial velocity of the proton, so that it is able to stop it. d.) Describe the motion of the proton after it comes to rest (momentarily). (Why this is problem like a "ball being thrown upwards"?)Explanation / Answer
the voltage difference is,
V = mv2/2e = 1.673x10-27 *[4x105]2 /2[1.6x10-19] = 836.5 V
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Its potential energy is larger and it is at a larger voltage.
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the electric field is pointed in the opposite direction.
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The proton after it comes to rest position, it will continues to move
toward the area with the lower electricpotential.