Question

A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75. How many electrons are delivered to the phone during 27 min of charging?

Answer 1

The no. of electrons is
`7.59* 10^(21)`

Solution:

According to the question:

The rate at which the charge is delivered is given by:

`(dQ)/(dt) = - 0.75`

Now,

`\int_(0)^(Q)dQ = - 0.75\int_(0)^(27 min) dt`

`Q = -0.75t|_(0)^(27 min)`

`Q= -0.75* 27* 60 = - 1215 C`

No. of electrons, n can be calculated from the following relation:

Q = ne

where

e = electronic charge =
`1.6* 10^(- 19) C`

Thus

`n = (Q)/(e)`

`n= (1215)/(1.6* 10^(- 19))`

`n = 7.59* 10^(21)`