Question

A cell phone battery uses chemistry to create a charge separation between the terminals (anode and cathode). Such a battery is listed as having a capacity of Q = 4.5E-08 C.

How many free electrons does the battery contain, N?

If there are 1.0 million electrons moving through the phone every second how long will the battery last in seconds?

Current, I, is given in amps which are coulombs per second. What is the current passing through the phone?

Answer 1

Solution:

Suppose There are N no of free electrons.

Battery is listed as having a capacity of Q = 4.5*10^-8 C

Number of free electrons contains in battery = N = (4.5*10^-8) / (1.6*10^-19) = 2.8125*10^11 [Answer]

B) There are 1 million = 10^6 electrons moving through the phone in 1 Second.

Therefore, 2.8125*10^11 electrons flow = (2.8125*10^11) / 10^6 = 2.8125*10^5 seconds. [Answer]

C) Now 1 million electrons flow through the phone every second.

Charge , Q = 4.5*10^-6 C

Therefore, Current passing through the phone, I = n*Q / t

Therefore, I = (10^6)*(4.5*10^-8) / 1 = 0.045 A = 45 mA [Answer]