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

Answer:

The no. of electrons is $7.59\times 10^{21}$

Solution:

According to the question:

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

$\frac{dQ}{dt} = - 0.75$

Now,

$\int_{0}^{Q}dQ = - 0.75\int_{0}^{27 min} dt$

$Q = -0.75t|_{0}^{27 min}$

$Q= -0.75\times 27\times 60 = - 1215 C$

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

Q = ne

where

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

Thus

$n = \frac{Q}{e}$

$n= \frac{1215}{1.6\times 10^{- 19}}$

$n = 7.59\times 10^{21}$