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A proton is moving through a uniform electric field of 50 N/C that points West. What are the magnitude and direction of the proton’s acceleration? A) 9.6 x 107 m/s2, South B) 1.6 x 10-19 m/s2, North C) 4.8 x 109 m/s2, West D) 1.9 x 105 m/s2, West E) zero Assume that the mass of the proton is 1.67 x 10-27 kg. Choose the closest answer below.

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aaregbe

Answer:

[tex]4.8*10^{9} m/s^{2} , West[/tex]

Explanation:

Where given:

The mass (m) of the proton [tex] = 1.67*10^{-27} kg[/tex]

The charge (q) of proton [tex]= 1.609*10^{-19} C[/tex]

The electric field (E) = 50 N/C

Let the acceleration of proton = a

Using the equation that relates electric field to the mass, charge, and acceleration of proton:

[tex]a = E*C/m = \frac{50*1.602*10^{-19} }{1.67*10^{-27} } = 4.8*10^{9} m/s^{2}[/tex]

The direction of the proton's acceleration is the same as the direction of the electric field. Thus:

The magnitude and direction of the proton’s acceleration is [tex]4.8*10^{9} m/s^{2} , West[/tex]

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