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3 years ago

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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.)

1 answer:

geniusboy [140]3 years ago

3 0

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

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