An electron has an initial velocity v→0=(6.0×104m/s)j^. It enters a region where E^=(1.9i^+8.0j^)×104N/C. Determine the vector acceleration of the electron as a function of time. Express your...
An electron has an initial velocity v→0=(6.0×104m/s)j^. It enters a region where E^=(1.9i^+8.0j^)×104N/C.
The force experienced by electron is determined by the electric field and the charge of the electron:
`vecF = evecE`
According to the second Newton's Law, the electron's acceleration is proportional to the force:
`vecF = m_eveca`
Therefore, `evecE = m_eveca` and
`veca = e/m_e vecE`
Since the electric field is constant, the acceleration will also be constant, that is, it will not depend on time.
The x and y components of acceleration are
`a_x = e/m_e E_x ` and `a_y = e/m_e E_y` . Plugging in the charge of electron `e = -1.602*10^(-19) C` ,
mass of electron `m_e = 9.109*10^(-31) kg` , `E_x = 1.9*10^4 N/C` , and `E_y = 8*10^4 N/C` ,
the x- and y-components become
`a_x = -6.3*10^15 m/s^2`
and `a_y = -1.4*10^16 m/s^2` .
To find the angle at which the electron is moving at a given point in time, we need to find its velocity vector at that point of time. According to the equation of motion
`vecv(t) = vecv_0 + veca*t` , and the x- and y-coordinates of the velocity are
`v_x(t) = a_xt` (the initial velocity has only y-component) and
`v_y(t) = v_(0y) + a_yt` .
Plugging in acceleration found above, given initial velocity and the time `t = 1.3 ns = 1.3*10^(-9) s` , we get
`v_x = -8.2*10^6 m/s`
and `v_y =-1.2*10^7 m/s` .
The velocity vector is directed down and to the left. The angle between the velocity vector and the negative x-axis is
`tan^(-1) (v_y/v_x) = 56` degrees.
The original velocity vector is directed up, so the angle between the direction in which electron is moving at t = 1.3 ns and the original direction is 56 + 90 = 146 degrees, counting counterclockwise from the original direction.