Question

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

Answer 1

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 = $\frac{1.6\times 10^{-17}}{1.6\times 10^{-19}}$

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.