Question

In a vacuum, a proton (charge = +e, mass = 1.67 x 10-27 kg) is moving parallel to a uniform electric field that is directed along the +x axis (see figure below). The proton starts with a velocity of +4.40 x 104 m/s and accelerates in the same direction as the electric field, which has a value of +7.90 x 103 N/C. Find the velocity of the proton when its displacement is +2.0 mm from the starting point. +2.0 mm Number. Units

Answer 1

Relation between force and electric field is given by:

F = q*E

Since q = charge on proton (which is +ve) = +e = +1.6*10^-19 C

F = e*E

Since Electric field is towards right and charge in field is +ve, So force on proton will be in the same direction of electric field, which is towards right.

Now Using Force balance, From newton's 2nd law:

Fnet = m*a

So,

m*a = q*E

a = q*E/m

Now Using given values, acceleration will be

m = mass of proton = 1.67*10^-27 kg

a = 1.6*10^-19*7.90*10^3/(1.67*10^-27)

a = 7.57*10^11 m/sec^2

Vi = 4.40*10^4 m/sec

S = distance traveled = 2.0 mm = 2.0*10^-3 m

Using 3rd kinematic equation:

Vf^2 = Vi^2 + 2*a*S

Vf = sqrt ((4.40*10^4)^2 + 2*(7.57*10^11)*(2.0*10^-3))

Vf = 70455.66 m/sec

Vf = 7.05*10^4 m/sec

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