Question

A conductor has 14.4×10^-19 Coulomb positive charge.The conductor has (charge on electron
= 1.6×10^-19 C)
(1) 9 electrons in excess
(3) 27 electrons in excess
(2) 27 electrons in short
(4) 9 electrons in short
​

Answer 1

No. of electrons=

14.4*10^-19/ 1.6* 10^-19

= 9 electrons

Thank you.

Answer 2

Answer:

(1) 9 electrons in excess

Explanation:

It is given that,

Charge on the conductor, $q=14.4\times 10^{-19}\ C$

Charge on electron, $e=1.6\times 10^{-19}\ C$

We know that the charge is quantized. The quantisation of charge is given by :

q = n e

n is the number of electron

$n=\dfrac{q}{e}$

$n=\dfrac{14.4\times 10^{-19}}{1.6\times 10^{-19}}$    

n = 9

So, the conductor has 9 electrons in excess. Hence, this is the required solution.