Question

An electron is launched with a constant horizontal speed of 2.0times 10+m/s into a region between two large parallel plated. The magnitude of the constant electric field between the plates is 50,000N/C, Calculate the distance that the electron will be deflected by the Electric field.

Answer 1

a) We need to calculate the time in the horizontal axis.

$x = v_o t \rightarrow t = \frac{x}{v_0}= \frac{0.2}{2.0\times 10^{7}}= 1\times 10^{-8} s$

in the vertical axis:

$y = \frac{at^2}{2}= \frac{qEt^2}{2m} = \frac{1.6\times 10^{-19}(50000)(1\times 10^{-8})^2}{2(9.11\times 10^{-31})}= 0.439 m$

b) The magnetic field required to undeflected the electron the net force must be zero:

so the magnetic field is:

$B = \frac{E}{v}= 2.5\times 10^{-3} T$

the direction of the magnetic force must be in the opposite direction of the electric field so using the right hand rule

the magnetic field must be out of the page.