An electron is travelling in a straight line with a kinetic energy K = 1.60 x 10^-17J. What are (a) the magnitude and (b) the direction of the electric field that will stop the electron in a distance of 10.0 cm? For part (b), make a drawing showing the direction of motion of the electron, the direction of the electric field and the electric force on the electron

Respuesta :

Answer:

(a) 1000 N/C

Explanation:

Kinetic energy of electron, K = 1.6 x 10^-17 J

distance, d = 10 cm = 0.1 m

Let the potential difference is V and the electric field is E.

(a) The relation between the kinetic energy and the potential difference is

K = e V

V = K / e

Where, e be the electronic charge = 1.6 x 10^-19 C

V = [tex]\frac{1.6\times 10^{-17}}{1.6\times 10^{-19}}[/tex]

V = 100 V

The relation between the electric field and the potential difference is given by

V = E x d

100 = E x 0.1

E = 1000 N/C

(b) The force acting on the electron, F = q E

where q be the charge on electron

So, F = -e x E

It means the direction of electric field and the force are both opposite to each other.

The direction of electric field and the force on electron is shown in the diagram.

Ver imagen Vespertilio