Question

A body has a charge of +3.2×10^-7 Coulumb .Calculate the less number of electrons as compared to neutral state..

answer fast not in short way and in a page

Answer 1

GIVEN : q=3.2×10^-7c
q=ne
n=q/e
n=3.2×10^-7 / 1.6x10^-19
n=2x10^12

Answer 2

Answer:

The less number of electrons is $2\times 10^{12}$ while comparing to a neutral state/body.

Explanation:

• If a body has more protons than electrons, the body is positively charged.
• A neutral body has an equal number of electrons and protons.

The formula for the number of electrons is:

$q=ne$

where $q$ is the total charge on the body

           $n$ is the number of electrons.

           $e$ is the charge on an electron.

It is given as $q=+3.2\times 10^{-7}$ $C$

                     $e= -1.6\times 10^{-19}$ $C$

The less number of electrons $=\frac{+3.2\times 10^{-7}C}{ (-1.6\times 10^{-19} C)}$

                                                  $=2\times 10^{12}$

So, as compared to a neutral state, the less number of electrons are $2\times 10^{12}$.