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 = 8.5E-08 C.

Part (a) How many free electrons does the battery contain, N?
N = ____.
Part (b) If there are 1.0 million electrons moving through the phone every second how long will the battery last in seconds?
t = ____.
Part (c) Current, I, is given in amps which are coulombs per second. What is the current passing through the phone?
I = ____.

Answer 1

The battery contains approximately 5.3125E+11 free electrons. The battery will last approximately 5.3125E+05 seconds. The current passing through the phone is approximately 8.5E-08 amps.

Step-by-step explanation:

Part (a):

To determine the number of free electrons in the battery, we can use the equation N = Q/e, where N is the number of electrons, Q is the charge, and e is the elementary charge. Plugging in the given charge of Q = 8.5E-08 C and the value of e = 1.6E-19 C, we can calculate:

N = 8.5E-08 C / 1.6E-19 C = 5.3125E+11 electrons

Therefore, the battery contains approximately 5.3125E+11 free electrons.

Part (b):

To find the time the battery will last in seconds, we can use the equation t = N/I, where t is the time, N is the number of electrons, and I is the current. Plugging in the given values of N = 5.3125E+11 electrons and I = 1.0 million electrons/second, we can calculate:

t = 5.3125E+11 electrons / 1.0 million electrons/second = 5.3125E+05 seconds

Therefore, the battery will last approximately 5.3125E+05 seconds.

Part (c):

The current passing through the phone can be calculated using the equation I = Q/t, where I is the current, Q is the charge, and t is the time. Plugging in the given values of Q = 8.5E-08 C and t = 1 second, we can calculate:

I = 8.5E-08 C / 1 second = 8.5E-08 amps

Therefore, the current passing through the phone is approximately 8.5E-08 amps.