Question

An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after traveling a distance d. (a) What is the magnitude of the electric field? (Use any variable or symbol stated above along with the following as necessary: e for the charge of the electron.)

Answer 1

Answer:

$E=\frac{K}{ed}$

Explanation:

We are given that

Initial kinetic energy of an electron=K

Distance=d

Final velocity=v=0

Charge,q=-1e

We have to find the magnitude of electric field.

Work done=$Force\times displacement$

Using the formula

Work done=$qE\times d=-eEd$

Using work energy theorem

Work done=Final K.E-Initial K.E=0-K

Work done=-K

Substitute the values

-K=-eEd

K=eEd

$E=\frac{K}{ed}$

Hence, the magnitude of the electric field=$E=\frac{K}{ed}$