Question

Q. Proton is placed midway between two parallel metallic plates rich are 0.2 meter
apart. The plates are connected to an 80 volts battery. Its electric intensity will be?​

Answer 1

Answer:

Oh You mean A proton is placed midway between two parallel metallic plates which are 0.20.2 meter apart. The plates are connected to an 8080 volt battery. What is the magnitude of the electric force on the proton, in newtons?

Explanation:

That fine, Here answer:

6.4x$10^{-17}$

The plates were charged oppositely by the battery and this produces a uniform electric field between the plates. Its intensity is equal to the difference of potential between the plates to the distance between the plates :

$E=\frac{V}{D}=\frac{80V}{0.2m}=400N/C$

The field intensity is defined as the force on a unit positive charge placed in the field. So,

$E=\frac{F}{Q}\\\\F=qE=(1.6 x 10^{-19}C) x 400N/C= 6.4 x 10^{-17} N$

Your welcome

Answer 2

Answer:

dear Muhammad in which book this numerical is ,?