Question

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.

Answer 1

`4.8*10^(9) m/s^(2) , West`

Step-by-step explanation:

Where given:

The mass (m) of the proton
`= 1.67*10^(-27) kg`

The charge (q) of proton
`= 1.609*10^(-19) C`

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:

`a = E*C/m = (50*1.602*10^(-19) )/(1.67*10^(-27) ) = 4.8*10^(9) m/s^(2)`

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
`4.8*10^(9) m/s^(2) , West`