A proton is shot into an electric field of magnitude 1x10^4 NC^-1.if the velocity of the proton as it enters the field is 1x10^6 ms^-1 and is perpendicular to the field,determine the magnitude and direction of the velocity in 2 seconds of motion in the field

To determine the magnitude and direction of the velocity of the proton after 2 seconds in the electric field, we need to consider the effect of the electric field on the proton's motion.

The electric field exerts a force on the charged proton. According to Newton's second law of motion, the force on a charged particle in an electric field is given by:

F = qE

where F is the force, q is the charge of the particle, and E is the electric field strength. In this case, the proton has a charge of +e (elementary charge), and the electric field strength is given as 1x10^4 NC^-1.

Given that the velocity of the proton as it enters the field is perpendicular to the field, we can assume that the force it experiences is perpendicular to its motion. This means that the force does not affect the magnitude of the velocity but changes its direction.

To find the force acting on the proton, we can plug in the values into the equation:

F = qE = (+e)(1x10^4) = 1.6x10^-19 C * 1x10^4 NC^-1 = 1.6x10^-15 N

Now, we can determine the acceleration of the proton using Newton's second law of motion, where acceleration is defined as the force divided by the mass:

a = F/m

The mass of a proton is approximately 1.67x10^-27 kg. So,

a = 1.6x10^-15 N / 1.67x10^-27 kg = 9.58x10^11 m/s^2

Now, we can use the equation of motion to find the displacement of the proton after 2 seconds in the electric field:

d = ut + 0.5at^2

where d is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

In this case, the initial velocity (u) is given as 1x10^6 m/s, acceleration (a) is 9.58x10^11 m/s^2, and time (t) is 2 seconds.

d = (1x10^6 m/s)(2 s) + 0.5(9.58x10^11 m/s^2)(2 s)^2
= 2x10^6 m + 0.5(9.58x10^11 m/s^2)(4 s^2)
= 2x10^6 m + 1.916x10^12 m^2/s^2
= 1.916x10^12 m^2/s^2 + 2x10^6 m
= 1.916x10^12 m^2/s^2 + 0.000002x10^6 m
= 1.916x10^12 m^2/s^2 + 2 m

Therefore, the displacement of the proton after 2 seconds in the electric field is approximately 1.916x10^12 m^2/s^2 + 2 m.

To find the magnitude and direction of the velocity, we need to divide the displacement by the time:

v = d / t

In this case, the displacement is 1.916x10^12 m^2/s^2 + 2 m, and the time is 2 seconds.

v = (1.916x10^12 m^2/s^2 + 2 m) / 2 s
= 9.58x10^11 m^2/s^2 + 1 m/s

Therefore, the magnitude of the velocity of the proton after 2 seconds in the electric field is approximately 9.58x10^11 m^2/s^2 + 1 m/s.

Since the electric field is perpendicular to the initial velocity of the proton, the resulting direction of the velocity will be perpendicular to both the initial velocity and electric field.