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

(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 =

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.

real, smaller and upside-down.   so the answer should be b.

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..